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Mathematics Part Ii Chapter 9 Differential Equations

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NCERT Solution For Class 12 Mathematics Mathematics Part Ii

Differential Equations Here is the CBSE Mathematics Chapter 9 for Class 12 students. Summary and detailed explanation of the lesson, including the definitions of difficult words. All of the exercises and questions and answers from the lesson's back end have been completed. NCERT Solutions for Class 12 Mathematics Differential Equations Chapter 9 NCERT Solutions for Class 12 Mathematics Differential Equations Chapter 9 The following is a summary in Hindi and English for the academic year 2021-2022. You can save these solutions to your computer or use the Class 12 Mathematics.

Question 1

Wired Faculty App

At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, – 3). Find the equation of the curve given that it passes through ( 2, 1).

Solution

Let y = f(x) be equation of curve.
Now

dy over dx

is the slope of the tangent to the curve at the point (x, y)
From the given condition,

dy over dx space equals space 2 open parentheses fraction numerator negative 3 minus straight y over denominator negative 4 minus straight x end fraction close parentheses space space or space space space dy over dx space equals space 2 open parentheses fraction numerator straight y plus 3 over denominator straight x plus 4 end fraction close parentheses

Separating the variables, we get,

fraction numerator 1 over denominator straight y plus 3 end fraction dy space equals space fraction numerator 2 over denominator straight x plus 4 end fraction dx

Integrating,

integral fraction numerator 1 over denominator straight y plus 3 end fraction space dy space equals space 2 space integral fraction numerator 1 over denominator straight x plus 4 end fraction dx

therefore space space space log space open vertical bar straight y plus 3 close vertical bar space equals space 2 space log space open vertical bar straight x plus 4 close vertical bar plus straight c

...(1)
Since curve passe through (-2, 1)

therefore space space space log space open vertical bar 1 plus 3 close vertical bar space equals space 2 space log space open vertical bar negative 2 plus 4 close vertical bar space plus space straight c therefore space space log space 4 space equals space 2 space log space 2 space plus straight c space space space rightwards double arrow space space space space 2 space log space 2 space space equals space 2 space log space 2 plus space straight c space space rightwards double arrow space space straight c space equals space 0 therefore space space from space left parenthesis 1 right parenthesis comma space space log space open vertical bar straight y plus 3 close vertical bar space equals space 2 space log space open vertical bar straight x plus 4 close vertical bar or space space log space open vertical bar straight y plus 3 close vertical bar space equals space log space open vertical bar straight x plus 4 close vertical bar squared therefore space space space space open vertical bar straight y plus 3 close vertical bar space equals space open vertical bar straight x plus 4 close vertical bar squared space space space or space space space straight y plus 3 space equals space left parenthesis straight x plus 4 right parenthesis squared

which is required equation of curve.

Question 2

Wired Faculty App

Find the equation of the curve passing through the point $open parentheses 0 comma space straight pi over 4 close parentheses$ whose differential equation is sin x cos y dx + cos x sin y dy = 0.

Solution

The given differential equation is
sin x cos y dx + cos x sin y dy = 0
or sin x cos y dx = - cos x sin y dy
$therefore space space space space space space space space space fraction numerator negative sin space straight y over denominator cos space straight y end fraction dy space equals space fraction numerator sin space straight x over denominator cos space straight x end fraction dx$
Integrating $integral fraction numerator negative sin space straight y over denominator cos space straight y end fraction dy space equals space fraction numerator sin space straight x over denominator cos space straight x end fraction dx$
$therefore space space log space open vertical bar cos space straight y close vertical bar space plus space log space open vertical bar cos space straight x close vertical bar space equals space log space straight A space space space space space space space rightwards double arrow space space space space space log space open vertical bar cosx space cosy close vertical bar space equals space log space straight A therefore space space space open vertical bar cosx space cosy close vertical bar space equals space straight A space space space space space space space space space space space space space space space rightwards double arrow space space space space cos space straight x space cos space straight y space equals space straight c space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis$
Since the curve passes through $open parentheses 0 comma space straight pi over 4 close parentheses$
$therefore space space space cos space 0 space cos space straight pi over 4 space equals straight c space space space space space space rightwards double arrow space space space 1 space cross times space fraction numerator 1 over denominator square root of 2 end fraction space equals space straight c space space space space space rightwards double arrow space space space straight c space equals space fraction numerator 1 over denominator square root of 2 end fraction$
$therefore space space from space left parenthesis 1 right parenthesis comma space cosx space cos space straight y space space equals space fraction numerator 1 over denominator square root of 2 end fraction comma space space which space is space required space equation.$

Question 3

Wired Faculty App

Find the equation of a curve passing through the point (0, 0) and whose differential equation is y' = e^x sin x.

Solution

The given differential equation is
y' = e^x sin x or

dy over dx space equals space straight e to the power of straight x space sin space straight x

Separating the variables, we get,

dy space equals space straight e to the power of straight x space sinx space dx

Integrating,

integral space dy space equals space integral straight e to the power of straight x space sinx space dx

therefore space space space space space space space space space space space space space space space space space space straight y space equals space fraction numerator 1 over denominator 1 plus 1 end fraction straight e to the power of straight x space left parenthesis sinx minus space cosx right parenthesis space plus space straight c

open square brackets because space space integral straight e to the power of ax space sin space straight b space straight x space dx space equals space fraction numerator 1 over denominator straight a squared plus straight b squared end fraction straight e to the power of ax space left parenthesis straight a space sin space bx space minus space straight b space cos space bx right parenthesis close square brackets

therefore space space space space space straight y space space equals 1 half straight e to the power of straight x left parenthesis sinx space minus space cosx right parenthesis space plus straight c

...(1)
Now curve passes through (0, 0)

therefore space space space 0 space equals space 1 half straight e to the power of 0 left parenthesis sin space 0 space minus cos space 0 right parenthesis space space plus straight c space space space rightwards double arrow space space space 0 space equals space 1 half cross times space left parenthesis 0 minus 1 right parenthesis space plus straight c

therefore space space space space space space space space straight e space equals space 1 half

therefore space space space from space left parenthesis 1 right parenthesis comma space space straight y space equals space 1 half straight e to the power of straight x left parenthesis sinx space minus space cosx right parenthesis space plus space 1 half comma space which space is space required space solution. space

Question 4

Wired Faculty App

The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

Solution

Let v be volume of spherical balloon of radius r.

therefore space space space space space space space straight v space equals space 4 over 3 πr cubed

...(1)
From given condition,

dv over dt equals space straight k space space or space space straight d over dt open parentheses 4 over 3 πr cubed close parentheses space equals space straight k

open square brackets because space space of space left parenthesis 1 right parenthesis close square brackets

therefore space space fraction numerator 4 straight pi over denominator 3 end fraction. space 3 space straight r squared space dr over dt space equals space straight k space space space or space space 4 πr squared dr over dt space equals space straight k

Separating the variables and integrating, we get.

4 straight pi integral straight r squared space dr space equals space straight k space integral space dt space space space or space space space 4 space straight pi space straight r cubed over 3 space equals space straight k space straight t space plus straight c

...(2)
Now t = 0 when r = 3

therefore space space space space 4 straight pi fraction numerator left parenthesis 3 right parenthesis cubed over denominator 3 end fraction space equals space straight k cross times 0 space plus space straight c space space space rightwards double arrow space straight c space equals space 36 space straight pi

...(3)
Again t = 3 when r = 6

therefore space space fraction numerator 4 straight pi over denominator 3 end fraction left parenthesis 6 right parenthesis cubed space equals space 3 straight k space plus space 36 straight pi

open square brackets because space space of space left parenthesis 3 right parenthesis close square brackets

therefore space space space space 288 space straight pi space equals space 3 space straight k space plus space 36 space straight pi space space or space space 3 space straight k space equals space 252 space straight pi

therefore space space space straight k space equals space 84 space straight pi

Putting

straight k space equals 84 space straight pi comma space space straight c space equals space 36 space straight pi space in space left parenthesis 2 right parenthesis comma space we space get

fraction numerator 4 straight pi over denominator 3 end fraction straight r cubed equals space space 84 space straight pi space straight t space plus space 36 space straight pi space space space or space space space straight r cubed over 3 space equals space 21 space straight t space plus space 9

therefore space space space straight r cubed space equals space 63 space straight t space plus space 27 space space space space rightwards double arrow space space space space straight r space equals space left square bracket 9 space left parenthesis 7 space straight t space plus space 3 right parenthesis right square bracket to the power of 1 third end exponent

Question 5

Wired Faculty App

In a bank, principal increases continuously at the rate of 5% per year. In how many years Rs. 1000 double itself ?

Solution

Let P be the principal at any time t. According to the given problem,

dP over dt space equals space open parentheses 5 over 100 close parentheses space cross times space straight P

therefore space space space space space space space dP over dt space equals space straight P over 20

Separating the variables, we get

dP over straight P space equals dt over 20

Integrating,

integral dP over straight P equals space integral dt over 20

therefore space space space space space space space space space space space space space space space log space straight P space equals space straight t over 20 plus straight c subscript 1 space space space or space straight P space space equals space straight e to the power of straight t over 20 plus straight c subscript 1 end exponent

straight P space equals space straight e to the power of straight t over 20 end exponent. space straight e to the power of straight c subscript 1 end exponent

straight P space equals space straight c space straight e to the power of straight t over 20 end exponent space left parenthesis where space straight e to the power of straight c subscript 1 end exponent space equals space straight c right parenthesis

...(1)
Now

straight P space equals space 1000 comma space space space when space straight t space equals space 0

therefore space space space space space space 1000 space equals space ce to the power of 0 space space space space rightwards double arrow space space space space 1000 space equals space space straight c

Putting c = 1000 in (1), we get

straight P space equals space 1000 space straight e to the power of straight t over 20 end exponent

Let t years be the time required to double the principal. Then

2000 space equals space 1000 space straight e to the power of straight t over 20 end exponent space space space space space rightwards double arrow space space space space 2 space equals space straight e to the power of straight t over 20 end exponent space space space rightwards double arrow space space straight t over 20 space equals space log subscript straight e squared space space space rightwards double arrow space space space straight t space equals space 20 space log subscript straight e 2.

Question 6

Wired Faculty App

In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs. 100 double itself in 10 years.

Solution

Let P denote the principal at any time t and r% per annum be the interest rate..
From the given condition,

dP over dt space equals space left parenthesis 0.0 space straight r right parenthesis thin space straight P

Separating the variables and integrating, we get,

integral 1 over straight P dP space equals space integral 0.0 space straight r space dt

therefore space space space space space space log space straight P space equals 0.0 space straight r space straight t space plus space log space straight c space space or space space space log space straight P over straight c space equals space 0.0 space straight r space straight t

therefore

straight P space equals space straight c space straight e to the power of 0.00 straight r space straight t end exponent

...(1)
Now

straight P space equals space 100

when t = 0

therefore

100 space equals space straight c space straight e to the power of 0

rightwards double arrow space space space straight c space equals space 100

therefore space space space from space left parenthesis 1 right parenthesis comma space space space straight P space equals space 100 straight e to the power of 0.0 straight r space straight t end exponent

Now, Rs. 100 become Rs.200 in 10 years.

therefore space space space space space space space space space space space 200 space equals space 100 space straight e to the power of left parenthesis 0.0 space straight r right parenthesis space space 10 end exponent space space space space space space space space rightwards double arrow space space space space 2 space equals space straight e to the power of straight r over 10 end exponent therefore space space space space space space log subscript straight e 2 space equals space straight r over 10 space space space space space rightwards double arrow space space space space space space 0.6931 space equals space straight r over 10 space space rightwards double arrow space space space straight r space equals space 6.931 therefore space space space space space required space rate space of space interest space equals space 6.9 percent sign

Question 7

Wired Faculty App

In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs. 100 double itself in 10 years. (e^0·5 = 1·648).

Solution

Let P denote the principal at any time t.
From given condition,

dP over dt space equals space left parenthesis 0.05 right parenthesis thin space straight P

Separating the variables and integrating,

integral 1 over straight P dP space equals space 0.05 space integral 1. space dt

therefore

log space straight P space equals space left parenthesis 0.05 right parenthesis space straight t space plus space log space straight c space space space space space space rightwards double arrow space space space space space space log space straight P over straight c space equals space left parenthesis 0.05 right parenthesis space straight t

therefore

straight P space equals space straight c space straight e to the power of 0.05 space straight t end exponent

...(1)
Now

straight P space equals space 1000 space space space space space space space space space space space when space space space straight t space equals space 0

therefore space space space space space space 1000 space equals space straight c space straight e to the power of 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space straight c space equals space 1000 therefore space space space space from space left parenthesis 1 right parenthesis comma space space space straight P space equals space 1000 space straight e to the power of 0.05 space straight t end exponent

When

straight t space equals space 10 space years comma space then

straight P space equals space 1000 space cross times space straight e to the power of 0.05. space 10 end exponent

equals space 1000 space cross times straight e to the power of 0.5 end exponent space equals space 1000 space cross times space 1.648 space equals space 1648

therefore space space Rs. space 1000

will become Rs. 1648 after 10 years.

Question 8

Wired Faculty App

In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

Solution

Let y denote the bacteria at any time t.
From the given condition,

dy over dt space equals space straight k space straight y comma space

where k is the constant of proportionality.
Separting the variables and integrating, we get,

integral 1 over straight y dy space equals space straight k integral dt

therefore space space space space space space space log space straight y space equals space straight k space straight t space space plus space straight c

...(1)
Let

straight y subscript 0

be bacteria at time t = 0

therefore space space space space space space space space space space space log space straight y subscript 0 space equals space 0 plus straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space straight c space equals space log space straight y subscript 0 therefore space space space space from space left parenthesis 1 right parenthesis comma space log space straight y space equals space straight k space straight t space plus space log space straight y subscript 0 space space space space space space space space space space space space space space rightwards double arrow space space log space open parentheses straight y over straight y subscript 0 close parentheses space equals space straight k space straight t space space space space space... left parenthesis 2 right parenthesis

Now,

straight y space equals space straight y subscript 0 plus 10 over 100 straight y subscript 0 space equals space 11 over 10 straight y subscript 0 space space space when space straight t space equals space 2

therefore space space space from space left parenthesis 2 right parenthesis comma space space log open parentheses fraction numerator 11 space straight y subscript 0 over denominator 10 space straight y subscript 0 end fraction close parentheses space equals space 2 space straight k space space rightwards double arrow space space space straight k space equals space 1 half log space open parentheses 11 over 10 close parentheses

therefore space space from space left parenthesis 2 right parenthesis comma space space log space open parentheses straight y over straight y subscript 0 close parentheses space equals space 1 half log space open parentheses 11 over 10 close parentheses. space straight t subscript 1

...(3)
Let the bacteria be 2,00,000 from 1,00,000 in

straight t subscript 1

hours
i.e.

straight y space equals space 2 space straight y subscript 0 space space space space when space straight t space space equals space straight t subscript 1

therefore space space space space from space left parenthesis 3 right parenthesis comma space space log space open parentheses fraction numerator 2 space straight y subscript 0 over denominator straight y subscript 0 end fraction close parentheses space equals space 1 half log space open parentheses 11 over 10 close parentheses. space straight t subscript 1

therefore space space space space space space log space 2 space equals space 1 half log open parentheses 11 over 10 close parentheses space straight t subscript 1 therefore space space space space space space space space space straight t subscript 1 space equals space fraction numerator 2 space log space 2 over denominator log space open parentheses begin display style 11 over 10 end style close parentheses end fraction space hours. space

Question 9

Wired Faculty App

The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20.000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009 ?

Solution

Let y denote the population at any t.
From the given condition,

dy over dt space equals space straight k space straight y. space space

where k is constant of proportionality.
Separating the variables and integrating,

integral 1 over straight y dy space equals space straight k integral space dt

therefore space space space space space space space space space space log space straight y space equals space straight k space straight t space space plus space straight c

...(1)
Let

straight y subscript 0 space equals space 20000 space

be the population at t = 0.

therefore space space space space space space space space space log space straight y subscript 0 space equals space 0 plus space straight c space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space space straight c space equals space log space straight y subscript 0 therefore space space from space left parenthesis 1 right parenthesis comma space space log space straight y space equals space straight k space straight t space plus space log space straight y subscript 0 space space space space space space space rightwards double arrow space space log space straight y space minus space log space straight y subscript 0 minus space straight k space straight t therefore space space space space space space log space open parentheses straight y over straight y subscript 0 close parentheses space equals space straight k space straight t space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis

Now y = 25000, when t = 5

therefore space space space log space open parentheses 25000 over 20000 close parentheses space equals space 5 space straight k space space space space space space rightwards double arrow space straight k space equals space 1 fifth log space open parentheses 5 over 4 close parentheses therefore space space from space left parenthesis 2 right parenthesis comma space space log space open parentheses straight y over straight y subscript 0 close parentheses space equals space 1 fifth straight t space log space open parentheses 5 over 4 close parentheses

Let

straight y subscript 1

be the population in 2004 i.e. after 10 years

therefore space space space space log space open parentheses straight y subscript 1 over straight y subscript 0 close parentheses space equals space 1 fifth cross times 10 cross times log space open parentheses 5 over 4 close parentheses therefore space log space open parentheses straight y subscript 1 over straight y subscript 0 close parentheses space equals space 2 space log space open parentheses 5 over 4 close parentheses space space space space rightwards double arrow space space space space log open parentheses straight y subscript 1 over straight y subscript 0 close parentheses space equals space log space 25 over 16 therefore space space space space space space space straight y subscript 1 over straight y subscript 0 equals space 25 over 16 space space space space rightwards double arrow space space space space space straight y subscript 1 space equals space 25 over 16 cross times 20000 space equals space 31250

therefore

required population in 2004 is 31250.

Question 10

Wired Faculty App

The general solution of the differential equation $dy over dx space equals space straight e to the power of straight x plus straight y end exponent$ is
e^x + e^{– y} = C
e^x+ e^y = C
e^{– x} + e^y = C
e^{– x}+ e^–y = C

Solution

e^x + e^{– y} = C The given differential equation is

dy over dx space equals space straight e to the power of straight x plus straight y end exponent space space space space space or space space dy over dx space equals space straight e to the power of straight x. space straight e to the power of straight y

Separating the variables, we get,

1 over straight e to the power of straight y dy space equals space straight e to the power of straight x space dx

Integrating,

integral straight e to the power of negative straight y end exponent dy space equals space integral straight e to the power of straight x space dx

therefore space space space space fraction numerator straight e to the power of negative straight y end exponent over denominator negative 1 end fraction space equals space straight e to the power of straight x plus straight c apostrophe therefore space space space space space space minus straight e to the power of negative straight y end exponent space equals space straight e to the power of straight x plus straight c apostrophe space space or space space straight e to the power of straight x plus straight e to the power of negative straight y end exponent space equals space minus straight c apostrophe therefore space space space space straight e to the power of straight x plus straight e to the power of negative straight y end exponent space equals space straight C comma space which space is space required space solution. space therefore space space left parenthesis straight A right parenthesis space is space correct space answer.

Question 11

Wired Faculty App

Solve $dy over dx space equals space left parenthesis 4 straight x plus straight y plus 1 right parenthesis squared.$

Solution

The given differential equation is

dy over dx space equals space space left parenthesis 4 straight x plus straight y plus 1 right parenthesis squared

Put

4 straight x plus straight y plus 1 space equals space straight t comma

therefore space space 4 plus dy over dx space equals space dt over dx comma space space or space space dy over dx space equals space dt over dx minus 4

therefore space space space given space equation space becomes comma space dt over dx minus 4 space equals straight t squared

dt over dx space equals space straight t squared plus 4

Separating the variables, we get,

fraction numerator dt over denominator straight t squared plus 4 end fraction space equals space dx

Integrating both sides,

integral fraction numerator dt over denominator straight t squared plus left parenthesis 2 right parenthesis squared end fraction space equals space integral 1. space dx

1 half tan to the power of negative 1 end exponent straight t over 2 space equals space straight x plus straight c space space space or space space tan to the power of negative 1 end exponent straight t over 2 space equals space 2 space left parenthesis straight x plus straight c right parenthesis comma space space or space space straight t over 2 space equals space tan space open square brackets 2 space left parenthesis straight x plus straight c right parenthesis close square brackets

therefore

straight t space equals space 2 space tan space left square bracket space 2 space left parenthesis straight x space plus straight c right parenthesis right square bracket

4 straight x plus straight y plus 1 space equals space 2 space tan space left square bracket 2 space left parenthesis straight x plus straight c right parenthesis right square bracket comma space which space is space the space required space solution.

Question 12

Wired Faculty App

Solve $left parenthesis straight x minus straight y right parenthesis squared space dy over dx space equals space straight a squared.$

Solution

The given differential equation is

left parenthesis straight x minus straight y right parenthesis squared dy over dx space equals space straight a squared

...(1)
Put x - y = t,

therefore space space space space 1 space minus space dy over dx space equals space dt over dx space space or space space dy over dx equals space 1 minus dt over dx

therefore space space space from space left parenthesis 1 right parenthesis comma space space space straight t squared space open parentheses 1 minus dt over dx close parentheses space equals space straight a squared space space space or space space space 1 minus dt over dx space equals space straight a squared over straight t squared

therefore space space space space space space space space space space space space space space space space space minus dt over dx space equals space straight a squared over straight t squared minus 1 space space space space space or space space space space dt over dx space equals space minus fraction numerator straight a squared minus straight t squared over denominator straight t squared end fraction

Separating the variables, we get,

fraction numerator straight t squared over denominator straight a squared minus straight t squared end fraction dt space equals space minus dx

negative straight t squared plus straight a squared space fraction numerator 1 over denominator long division enclose straight t squared end enclose end fraction space space space space space space space space space space space space space space space space space space straight t squared minus straight a squared space space space space space space space space space space space space space space space space space space space minus space space plus space space space space space space space space space space space space space space space space space _______ space space space space space space space space space space space space space space space space space space space space space space straight a squared

Integrating,

integral fraction numerator straight t squared over denominator straight a squared minus straight t squared end fraction dt space equals space minus integral space 1 space dx

therefore space space space integral open parentheses negative 1 plus fraction numerator straight a squared over denominator straight a squared minus straight t squared end fraction close parentheses dt space equals space minus integral 1 space dx therefore space space space minus straight t plus straight a to the power of 2 space end exponent fraction numerator 1 over denominator 2 straight a end fraction space log space open vertical bar fraction numerator straight a plus straight t over denominator straight a minus straight t end fraction close vertical bar space equals space minus straight x plus straight c therefore space space minus straight t plus straight a over 2 log space open vertical bar fraction numerator straight a plus straight x minus straight y over denominator straight a minus straight x plus straight y end fraction close vertical bar space equals space minus straight x plus straight c therefore space space straight y minus straight x minus straight a over 2 log space open vertical bar fraction numerator straight a plus straight x minus straight y over denominator straight a minus straight x plus straight y end fraction close vertical bar space equals space minus straight x plus straight c therefore space space straight y minus straight x minus straight a over 2 log space open vertical bar fraction numerator straight a plus straight x minus straight y over denominator straight a minus straight x plus straight y end fraction close vertical bar space equals space minus straight x plus straight c

straight y minus straight a over 2 log space open vertical bar fraction numerator straight a plus straight x minus straight y over denominator straight a minus straight x plus straight y end fraction close vertical bar space equals space straight c comma

which is required solution.

Question 13

Wired Faculty App

Solve the following differential equation:
$open parentheses straight x plus straight y close parentheses squared dy over dx space equals straight a squared$

Solution

The given differential equation is

left parenthesis straight x plus straight y right parenthesis squared space dy over dx space equals space straight a squared

Put

straight x plus straight y space equals straight t comma space space space space space therefore space space 1 plus dy over dx space space equals space dt over dx space space or space space dy over dx equals space dt over dx minus 1

therefore space space space space from space left parenthesis 1 right parenthesis comma space space straight t squared space open parentheses dt over dx minus 1 close parentheses space equals straight a squared space or space space straight t squared space dt over dx space equals space straight a squared plus straight t squared space space or space space space fraction numerator straight t squared over denominator straight a squared plus straight t squared end fraction dt space equals space dx

integral fraction numerator straight t squared over denominator straight a squared plus straight t squared end fraction dt space equals space integral dx

integral open parentheses 1 minus fraction numerator straight a squared over denominator straight a squared plus straight t squared end fraction close parentheses dt space equals space straight x space space or space straight t minus straight a squared space 1 over straight a tan to the power of negative 1 end exponent straight t over straight a space equals straight x plus straight c

straight x plus straight y space minus straight a space tan to the power of negative 1 end exponent open parentheses fraction numerator straight x plus straight y over denominator straight a end fraction close parentheses space equals space straight x plus straight c

straight y minus straight a space tan to the power of negative 1 end exponent open parentheses fraction numerator straight x plus straight y over denominator straight a end fraction close parentheses space equals space straight c space is space the space required space solution. space

Question 14

Wired Faculty App

Solve the following differential equation:
$open parentheses straight x plus straight y plus 2 close parentheses space dy over dx space equals 2$

Solution

The given differential equation is

open parentheses straight x plus straight y plus 2 close parentheses space dy over dx space equals 2

...(1)
Put x+y+2 = t,

therefore space space space space space 1 plus dy over dx equals dt over dx space space or space space dy over dx equals dt over dx minus 1 therefore space space from space left parenthesis 1 right parenthesis comma space space straight t open parentheses dt over dx minus 1 close parentheses space equals space 2 space space or space space straight t dt over dx space equals space 2 plus straight t space space or space space fraction numerator straight t over denominator 2 plus straight t end fraction dt space equals space dx

integral fraction numerator straight t over denominator 2 plus straight t end fraction dt space equals space integral dx

or space space integral open parentheses 1 minus fraction numerator 2 over denominator 2 plus straight t end fraction close parentheses space dt space equals space straight x space space or space space straight t minus 2 space log space open vertical bar 2 plus straight t close vertical bar space equals space straight x plus straight c or space space straight x plus straight y plus 2 minus 2 space log space space open vertical bar straight x plus straight y plus 4 close vertical bar space equals space straight x plus straight c space or space space straight y minus 2 space log space open vertical bar straight x plus straight y plus 4 close vertical bar space equals space straight c minus 2 space equals space straight C comma space say comma space is space the space required space solution.

Question 15

Wired Faculty App

Solve the following differential equation:
$dy over dx plus 1 space equals space straight e to the power of straight x plus straight y end exponent$

Solution

The given differential equation is

dy over dx plus 1 space equals straight e to the power of straight x plus straight y end exponent

...(1)
Put

straight x plus straight y space equals straight t comma space space space space space space space space space space space therefore space space space space 1 plus dy over dx equals space dt over dx space space or space space space dy over dx space equals dt over dx minus 1

therefore space space space from space left parenthesis 1 right parenthesis comma space space dt over dx minus 1 plus 1 space equals space straight e to the power of straight t space space space space space or space space space dt over dx space equals straight e to the power of straight t rightwards double arrow space space space space 1 over straight e to the power of straight t dt space equals space dx space space space rightwards double arrow space space space integral straight e to the power of negative straight t end exponent space dt space equals space integral 1. space dx therefore space space space space space space space space space space fraction numerator straight e to the power of negative straight t end exponent over denominator negative 1 end fraction space equals space straight x plus straight c space space space space rightwards double arrow space space space minus straight e to the power of negative left parenthesis straight x plus straight y right parenthesis end exponent space equals space straight x space plus straight c space space which space is space required space solution. space space space

Question 16

Wired Faculty App

Solve the differential equation $dy over dx space equals sin left parenthesis straight x plus straight y right parenthesis space space or space sin to the power of negative 1 end exponent open parentheses dy over dx close parentheses space equals space straight x plus straight y$

Solution

The given differential equation is

dy over dx space equals space sin space left parenthesis straight x plus straight y right parenthesis

...(1)
Put x + y = t,

therefore space space space 1 plus dy over dx space equals space dt over dx space space or space space space dy over dx space equals space dt over dx minus 1

therefore space space space space space space from space left parenthesis 1 right parenthesis comma space space dt over dx minus 1 space equals space space sin space straight t space space space or space space dt over dx space equals space 1 plus sin space straight t

Separating the variables, we get,

fraction numerator 1 over denominator 1 plus sin space straight t end fraction dt space equals space dx

Integrating both sides, we get,

integral fraction numerator 1 over denominator 1 plus sin space straight t end fraction dt space equals space integral 1. space dx space space space space or space space space space integral fraction numerator 1 over denominator 1 plus sin space straight t end fraction cross times fraction numerator 1 minus sin space straight t over denominator 1 minus sin space straight t end fraction dt space equals space integral 1. space dx

integral fraction numerator 1 minus sint over denominator 1 minus sin squared straight t end fraction dt space equals space integral 1. space dx space space space space space space space space space space space or space space space space space fraction numerator 1 minus sin space straight t over denominator cos squared straight t end fraction dt space equals space integral 1. space dx

integral open parentheses fraction numerator 1 over denominator cos squared straight t end fraction minus fraction numerator sin space straight t over denominator cos squared straight t end fraction close parentheses dt space equals space integral 1. space dx space space or space space integral open parentheses fraction numerator 1 over denominator cos squared straight t end fraction minus fraction numerator 1 over denominator cos space straight t end fraction. fraction numerator sin space straight t over denominator cos space straight t end fraction close parentheses dt space equals space integral 1. space dx

open parentheses sec squared straight t minus sect space tant close parentheses space dt space equals space 1. space dx

tant space minus sect space equals space straight x plus straight c

tan left parenthesis straight x plus straight y right parenthesis space minus space sec left parenthesis straight x plus straight y right parenthesis space equals space straight x plus straight c

which is required solution.

Question 17

Wired Faculty App

Solve: sin (x+y) $dy over dx space equals 1$ .

Solution

The given differential equation is

sin space left parenthesis straight x plus straight y right parenthesis space dy over dx space equals space 1 space space space space space or space space dy over dx space equals space fraction numerator 1 over denominator sin space left parenthesis straight x plus straight y right parenthesis end fraction

...(1)
Put x+ y = t,

therefore space space space 1 plus dy over dx space equals space dt over dx space space space or space space space dy over dx space equals space dt over dx minus 1

therefore space space space space space from space left parenthesis 1 right parenthesis comma space dt over dx minus 1 space equals space fraction numerator 1 over denominator sin space straight t end fraction space space rightwards double arrow space space space dt over dx space equals space 1 plus fraction numerator 1 over denominator sin space straight t end fraction therefore space space space space dt over dx space equals space fraction numerator 1 plus sint over denominator sint end fraction space space space space space space space space space space space space space space space space space space rightwards double arrow space space space space fraction numerator sint over denominator 1 plus sint end fraction dt space equals space dx therefore space space space space integral fraction numerator sin space straight t over denominator 1 plus sin space straight t end fraction dt space equals space integral 1. space dx space space space space rightwards double arrow space space space space integral fraction numerator left parenthesis 1 plus sin space straight t right parenthesis minus 1 over denominator 1 plus sint end fraction dt space equals space integral 1. space dx therefore space space space space integral space open parentheses 1 minus fraction numerator 1 over denominator 1 plus sin space straight t end fraction close parentheses dt space equals space integral 1. space dx rightwards double arrow space space integral open parentheses 1 minus fraction numerator 1 over denominator 1 plus sin space straight t end fraction cross times fraction numerator 1 minus sin space straight t over denominator 1 minus sin space straight t end fraction close parentheses space dt space equals space integral space 1. space dx rightwards double arrow space space integral open parentheses 1 minus fraction numerator 1 minus sin space straight t over denominator 1 minus sin squared straight t end fraction close parentheses space dt space equals space integral 1 space dx space space space space space space space space space rightwards double arrow space space space space space space space integral open parentheses 1 minus fraction numerator 1 minus sin space straight t over denominator cos squared straight t end fraction close parentheses dt space equals space integral 1. dx rightwards double arrow space space space integral open parentheses 1 minus fraction numerator 1 over denominator cos squared straight t end fraction plus fraction numerator 1 over denominator cos space straight t end fraction. fraction numerator sin space straight t over denominator cos space straight t end fraction close parentheses space dt space equals space integral space 1 space. space dx rightwards double arrow space integral left parenthesis 1 minus sec squared straight t space plus space sect space tant right parenthesis space dt space equals space integral 1. space dx rightwards double arrow space space space straight t minus tan space straight t space plus sec space straight t space equals space straight x space plus straight c therefore space space space straight x plus straight y minus tan space left parenthesis straight x plus straight y right parenthesis plus sec left parenthesis straight x plus straight y right parenthesis space equals space straight x plus straight c therefore space space straight y minus tan left parenthesis straight x plus straight y right parenthesis plus sec left parenthesis straight x plus straight y right parenthesis space equals space straight c space is space the space required space solution.

Sponsor Area

Question 18

Wired Faculty App

Solve:
$dy over dx space equals space cos space left parenthesis straight x plus straight y right parenthesis space space or space space cos to the power of negative 1 end exponent open parentheses dy over dx close parentheses space equals space straight x plus straight y$

Solution

The given differential equation is

dy over dx space equals cos left parenthesis straight x plus straight y right parenthesis

...(1)
Put x + y = t,

therefore space space space 1 plus dy over dx space equals space dt over dx space space or space space dy over dx space equals dt over dx minus 1

therefore space space from space left parenthesis 1 right parenthesis comma space space dt over dx minus 1 space equals space cos space straight t space space or space space dt over dx equals space 1 plus space cos space straight t therefore space space space space fraction numerator 1 over denominator 1 plus cos space straight t end fraction dt space space equals space dx space space space space rightwards double arrow space space space space integral fraction numerator 1 over denominator 1 plus cos space straight t end fraction dt space equals space integral 1. space dx rightwards double arrow space space space space integral fraction numerator 1 over denominator 1 plus cos space straight t end fraction cross times fraction numerator 1 minus cos space straight t over denominator 1 minus cos space straight t end fraction dt space equals space integral 1. space dx rightwards double arrow space space space space integral fraction numerator 1 minus cos space straight t over denominator 1 minus cos squared straight t end fraction dt space equals space integral 1. space dx space space space rightwards double arrow space space space space integral fraction numerator 1 minus cos space straight t over denominator sin squared straight t end fraction dt space equals space integral 1. space dx rightwards double arrow space integral open parentheses fraction numerator 1 over denominator sin squared straight t end fraction minus fraction numerator cos space straight t over denominator sin squared straight t end fraction close parentheses space dt space equals space integral 1. space dx space space space rightwards double arrow space space space space integral left parenthesis cosec squared straight t minus cosect space cot space straight t right parenthesis space dx space equals space integral 1. space dx rightwards double arrow space space space minus cot space straight t space plus space cosec space straight t space equals straight x plus straight c rightwards double arrow space space space space space fraction numerator cos space straight t over denominator sin space straight t end fraction plus fraction numerator 1 over denominator sin space straight t end fraction space equals space straight x plus straight c space space space space rightwards double arrow space space space space space fraction numerator 1 minus cos space straight t over denominator sin space straight t end fraction space equals space straight x plus straight c rightwards double arrow space space space fraction numerator 2 space sin squared begin display style straight t over 2 end style over denominator 2 space sin begin display style straight t over 2 end style space cos begin display style straight t over 2 end style end fraction space equals space straight x plus straight c space space space rightwards double arrow space space space tan straight t over 2 space equals space straight x plus straight c space space space rightwards double arrow space space space tan fraction numerator straight x plus straight y over denominator 2 end fraction space equals straight x plus straight c space space space which space is space required space solution. space

Question 19

Wired Faculty App

Solve:
$cos space left parenthesis straight x plus straight y right parenthesis space dy over dx space equals space 1.$

Solution

The given differential equation is

dy over dx space equals space fraction numerator 1 over denominator cos space left parenthesis straight x plus straight y right parenthesis end fraction

...(1)
Put x+y = t,

therefore space space space 1 plus dy over dx space equals space dt over dx space space or space space dy over dx space equals space dt over dx minus 1

therefore space space from space left parenthesis 1 right parenthesis comma space space space dt over dx minus 1 space equals space fraction numerator 1 over denominator cos space straight t end fraction space space space space space space space space space rightwards double arrow space space space space space dt over dx space equals space 1 plus fraction numerator 1 over denominator cos space straight t end fraction therefore space space space space dt over dx equals space fraction numerator 1 plus cos space straight t over denominator cos space straight t end fraction space space space rightwards double arrow space space space space space fraction numerator cos space straight t over denominator 1 plus space cos space straight t end fraction dt space equals space dx therefore space space space integral fraction numerator cos space straight t over denominator 1 plus space cos space straight t end fraction dt space equals space integral 1. space dx space space space rightwards double arrow space space space integral fraction numerator left parenthesis 1 plus cost right parenthesis minus 1 over denominator 1 plus cost end fraction dt space equals space integral 1. space dx therefore space space space space integral open parentheses 1 minus fraction numerator 1 over denominator 1 plus cost end fraction close parentheses dt space equals space integral 1. space dx space space space rightwards double arrow space space space integral open parentheses 1 minus fraction numerator 1 over denominator 2 space cos squared begin display style straight t over 2 end style end fraction close parentheses dt space equals space integral 1. space dx therefore space space space integral open parentheses 1 minus 1 half sec squared straight t over 2 close parentheses space dt space equals space integral 1. space dx space space space space rightwards double arrow space space space straight t minus tan straight t over 2 space equals space straight x plus straight c apostrophe therefore space space space space straight x plus straight y minus tan fraction numerator straight x plus straight y over denominator 2 end fraction space equals space straight x plus straight c apostrophe space space space space rightwards double arrow space space space tan fraction numerator straight x plus straight y over denominator 2 end fraction space equals space straight y plus straight c comma space which space is space required space solution. comma space

Question 20

Wired Faculty App

Solve:
$dy over dx space equals space tan space left parenthesis straight x plus straight y right parenthesis$

Solution

The given differential equation is

dy over dx space equals space tan left parenthesis straight x plus straight y right parenthesis

Put

straight x plus straight y space equals straight t comma space space space space space space space space space space space space space space space space space therefore space space space 1 plus dy over dx space equals dt over dx space space space or space space space space dy over dx equals space dt over dx minus 1

therefore space space space space from space left parenthesis 1 right parenthesis comma space space space dt over dx minus 1 space equals space tan space straight t space space space space space or space space space space dt over dx space equals space 1 plus tan space straight t

Separating the variables, we get,

fraction numerator 1 over denominator 1 plus tan space straight t end fraction dt space equals dx space space space or space space space fraction numerator 1 over denominator 1 plus begin display style fraction numerator sin space straight t over denominator cos space straight t end fraction end style end fraction dt space equals space dx

fraction numerator cos space straight t over denominator sin space straight t plus cos space straight t end fraction dt space equals space dx space space space space space rightwards double arrow space space space space integral fraction numerator cos space straight t over denominator sin space straight t space plus space cos space straight t end fraction dt space equals space integral 1 space dx

rightwards double arrow space space space space 1 half space integral fraction numerator left parenthesis sint space plus space cost right parenthesis space plus space left parenthesis cost space minus space sint right parenthesis over denominator sint plus cost end fraction dt space equals space integral 1 space dx or space space space 1 half integral open parentheses 1 plus fraction numerator cost space minus sint over denominator sint plus cost end fraction close parentheses dt space equals space integral 1 space dx or space space space 1 half integral open square brackets straight t space plus space log space open vertical bar sin space straight t space plus space cos space straight t close vertical bar close square brackets space equals space straight x plus straight c or space space space 1 half open square brackets straight x plus straight y plus log space open vertical bar sin space left parenthesis straight x plus straight y right parenthesis plus cos space left parenthesis straight x plus straight y right parenthesis close vertical bar close square brackets space equals space straight x plus straight c

which is required solution.

Question 21

Wired Faculty App

Solve the following initial value problem:
(x + y + 1)² dy = dx, y ( –1) = 0

Solution

The given differential equation is
(x + y + 1)²dy = dx
or

dy over dx space equals space fraction numerator 1 over denominator left parenthesis straight x plus straight y plus 1 right parenthesis squared end fraction

Put

straight x plus straight y plus 1 space equals space straight t comma space space space space therefore space space space space 1 plus dy over dx space equals dt over dx space space space or space space space dy over dx equals space dt over dx minus 1

therefore space space space space given space equation space becomes space dt over dx minus 1 space equals 1 over straight t squared

therefore space space space space space space space space space space space dt over dx equals 1 plus 1 over straight t squared

dt over dx space equals fraction numerator straight t squared plus 1 over denominator straight t squared end fraction

Separating the variables and integrating, we get,

integral fraction numerator straight t squared over denominator straight t squared plus 1 end fraction dt space equals space integral dx space space space space or space space space integral open parentheses 1 minus fraction numerator 1 over denominator 1 plus straight t squared end fraction close parentheses space dt space equals space integral 1. space dx

therefore space space space space space space straight t minus tan to the power of negative 1 end exponent straight t space equals space straight x plus straight c space space space space or space space space left parenthesis straight x plus straight y plus 1 right parenthesis space minus space tan to the power of negative 1 end exponent left parenthesis straight x plus straight y plus 1 right parenthesis space equals space straight x plus straight c therefore space space space space straight y plus 1 minus tan to the power of negative 1 end exponent left parenthesis straight x plus straight y plus 1 right parenthesis space equals space straight c

Now

straight y left parenthesis negative 1 right parenthesis space equals space 0 space space space space space space space space rightwards double arrow space space space space straight y space equals space 0 space space space when space straight x space equals space minus 1

therefore space space space space 0 plus 1 space minus space tan to the power of negative 1 end exponent left parenthesis negative 1 plus 0 plus 1 right parenthesis space equals space straight c space space space space space space rightwards double arrow space space space space 1 minus tan to the power of negative 1 end exponent 0 space equals space straight c therefore space space space space space space space space space space space space space space 1 minus 0 space equals space straight c space space space space space space space space rightwards double arrow space space space space straight c space equals space 1 therefore space space space space space from space left parenthesis 1 right parenthesis comma space space straight y plus 1 space minus space tan to the power of negative 1 end exponent left parenthesis straight x plus straight y plus 1 right parenthesis space equals space 1 therefore space space tan to the power of negative 1 end exponent left parenthesis straight x plus straight y plus 1 right parenthesis space equals space straight y space space space space space or space space space straight x plus straight y plus 1 space equals space tan space straight y

which is required solution.

Question 22

Wired Faculty App

Solve the following initial value problem
cos (x + y) dy = dx, y (0) = 0

Solution

The given differential equation is

cos space left parenthesis straight x plus straight y right parenthesis space dy space equals space dx space space space space or space space space space dy over dx space equals space fraction numerator 1 over denominator cos space left parenthesis straight x plus straight y right parenthesis end fraction

...(1)
Put x + y = t,

therefore space space space 1 plus dy over dx equals dt over dx space space or space space dy over dx space equals space dt over dx minus 1

therefore space space space space from space left parenthesis 1 right parenthesis comma space space dt over dx minus 1 space equals space fraction numerator 1 over denominator cos space straight t end fraction space space space rightwards double arrow space space space space dt over dx space equals space 1 plus fraction numerator 1 over denominator cos space straight t end fraction therefore space space space space dt over dx space equals space fraction numerator 1 plus cos space straight t over denominator cost end fraction space space space rightwards double arrow space space space fraction numerator cos space straight t over denominator 1 plus cos space straight t end fraction dt space equals space dx therefore space space space space integral fraction numerator cos space straight t over denominator 1 plus cost space end fraction dt space equals space integral 1. space dx space space space rightwards double arrow space space integral fraction numerator left parenthesis 1 plus cost right parenthesis minus 1 over denominator 1 plus cost end fraction dt space equals space integral 1. space dx therefore space space space integral open parentheses 1 minus fraction numerator 1 over denominator 1 plus cos space straight t end fraction close parentheses dt space equals space integral 1. space dx space space rightwards double arrow space space integral open parentheses 1 minus fraction numerator 1 over denominator 2 space cos squared begin display style straight t over 2 end style end fraction close parentheses dt space equals space integral 1. space dx therefore space space space integral space open parentheses 1 minus 1 half sec squared straight t over 2 close parentheses space dt space equals space integral 1. space dx space space space rightwards double arrow space space space 1 minus tan straight t over 2 space equals space straight x plus straight c apostrophe therefore space space space straight x plus straight y minus tan fraction numerator straight x plus straight y over denominator 2 end fraction space equals space straight x plus straight c apostrophe space space space space space space rightwards double arrow space space space space tan fraction numerator straight x plus straight y over denominator 2 end fraction space equals space straight y plus straight c Now space space space space space space space space space space space space space space space space space space space straight y left parenthesis 0 right parenthesis space equals space 0 space space space space space space space space space space space space rightwards double arrow space space space space space space space space straight y space equals space 0 space space space space space when space straight x space equals space 0 therefore space space space space space space space space space space space space space space space space space space tan space 0 space equals space 0 plus space straight c space space space space space rightwards double arrow space space space straight c space equals space 0 therefore space space space space space tan space open parentheses fraction numerator straight x plus straight y over denominator 2 end fraction close parentheses space equals space straight y space is space the space required space solution. space

Question 23

Wired Faculty App

For the following differential equation, given below, find particular solution satisfying the given condition:
$open parentheses straight x cubed plus straight x squared plus straight x plus 1 close parentheses dy over dx space equals space 2 straight x squared plus straight x semicolon space space straight y space equals space 1 space space when space straight x space equals space 0$

Solution

The given differential equation is

open parentheses straight x cubed plus straight x squared plus straight x plus 1 close parentheses space dy over dx space equals space 2 straight x squared plus straight x

dy over dx space equals space fraction numerator 2 straight x squared plus straight x over denominator straight x cubed plus straight x squared plus straight x plus 1 end fraction

dy over dx space equals space fraction numerator 2 straight x squared plus 1 over denominator straight x squared left parenthesis straight x plus 1 right parenthesis space plus space 1 left parenthesis straight x plus 1 right parenthesis end fraction

dy over dx space equals space fraction numerator 2 straight x squared plus straight x over denominator left parenthesis straight x plus 1 right parenthesis thin space left parenthesis straight x squared plus 1 right parenthesis end fraction

Separting the variables, we get,

dy space equals fraction numerator 2 straight x squared plus straight x over denominator left parenthesis straight x plus 1 right parenthesis thin space left parenthesis straight x squared plus 1 right parenthesis end fraction

dx
Integrating,

integral dy space equals space integral fraction numerator 2 straight x squared plus straight x over denominator left parenthesis straight x plus 1 right parenthesis thin space left parenthesis straight x squared plus 1 right parenthesis end fraction dx

...(1)
Put

fraction numerator 2 straight x squared plus straight x over denominator left parenthesis straight x minus 1 right parenthesis thin space left parenthesis straight x squared plus 1 right parenthesis end fraction space equals fraction numerator straight A over denominator straight x plus 1 end fraction plus fraction numerator Bx plus straight C over denominator straight x squared plus 1 end fraction

...(2)
Mutiplying both sides by

open parentheses straight x plus 1 close parentheses space open parentheses straight x squared plus 1 close parentheses comma space space we space get

2 straight x squared plus straight x space identical to space straight A space left parenthesis straight x squared plus 1 right parenthesis space plus space left parenthesis Bx plus straight C right parenthesis thin space left parenthesis straight x plus 1 right parenthesis

2 straight x squared plus straight x space identical to space straight A left parenthesis straight x squared plus 1 right parenthesis space plus Bx left parenthesis straight x plus 1 right parenthesis space plus space straight C left parenthesis straight x plus 1 right parenthesis

...(3)
Put x + 1 = 0 or x = -1 in (3)

therefore

2 - 1 = A(1+1)+0+0

rightwards double arrow space space space space space straight I space equals space 2 straight A space space rightwards double arrow space space straight A space equals space 1 half

straight x squared right parenthesis space 2 space equals space straight A plus straight B space space space rightwards double arrow space 2 space equals space 1 half plus straight B space space space rightwards double arrow space space space straight B space equals space 3 over 2

straight x right parenthesis space space 1 space equals space straight B plus straight C space space space space rightwards double arrow space space space space 1 space equals space 3 over 2 plus straight C space space space space rightwards double arrow space space space straight C space equals space minus 1 half

therefore space space space from space left parenthesis 2 right parenthesis comma space space fraction numerator 2 straight x squared plus straight x over denominator left parenthesis straight x plus 1 right parenthesis thin space left parenthesis straight x squared plus 1 right parenthesis end fraction space equals space fraction numerator begin display style 1 half end style over denominator straight x plus 1 end fraction plus fraction numerator begin display style 3 over 2 end style straight x minus begin display style 1 half end style over denominator straight x squared plus 1 end fraction

fraction numerator 2 straight x squared plus straight x over denominator left parenthesis straight x plus 1 right parenthesis thin space left parenthesis straight x squared plus 1 right parenthesis end fraction space identical to space fraction numerator 1 over denominator 2 left parenthesis straight x plus 1 right parenthesis end fraction plus 3 over 2 open parentheses fraction numerator straight x over denominator straight x squared plus 1 end fraction close parentheses space minus space 1 half open parentheses fraction numerator 1 over denominator straight x squared plus 1 end fraction close parentheses

therefore space space space from space left parenthesis 1 right parenthesis comma space space space integral dy space equals space 1 half integral fraction numerator 1 over denominator straight x plus 1 end fraction dx plus 3 over 2 integral fraction numerator straight x over denominator straight x squared plus 1 end fraction dx minus 1 half integral fraction numerator 1 over denominator straight x squared plus 1 end fraction dx therefore space space space space integral 1 space dy space equals space 1 half integral fraction numerator 1 over denominator straight x plus 1 end fraction dx plus 3 over 4 integral fraction numerator 2 straight x over denominator straight x squared plus 1 end fraction dx minus 1 half integral fraction numerator 1 over denominator straight x squared plus 1 end fraction dx therefore space space space space space straight y space equals space 1 half log space open vertical bar straight x plus 1 close vertical bar plus 3 over 4 log space left parenthesis straight x squared plus 1 right parenthesis space minus 1 half tan to the power of negative 1 end exponent straight x plus straight c space space space space space space space space space space... left parenthesis 4 right parenthesis

Now y = 1 when x = 0

therefore space space space space straight I space equals space 1 half log space left parenthesis 1 right parenthesis space plus space 3 over 4 log space 1 space minus space 1 half tan to the power of negative 1 end exponent 0 plus straight c therefore space space space space straight I space equals space 1 half left parenthesis 0 right parenthesis plus 3 over 4 left parenthesis 0 right parenthesis minus 1 half left parenthesis 0 right parenthesis space plus space straight c space space space space space rightwards double arrow space space space straight c space equals space 1 therefore space space space space from space left parenthesis 4 right parenthesis comma space space straight y space equals space 1 half log space open vertical bar straight x plus 1 close vertical bar plus 3 over 4 log space left parenthesis straight x squared plus 1 right parenthesis minus 1 half tan to the power of negative 1 end exponent straight x plus 1

Question 24

Wired Faculty App

For the following differential equation, given below, find particular solution satisfying the given condition:
$straight x open parentheses straight x squared minus 1 close parentheses space dy over dx space equals space 1 space semicolon space space straight y space equals space 0 space space when space straight x space equals space 2$

Solution

The given differential equation is

straight x open parentheses straight x squared minus 1 close parentheses dy over dx space equals space 1 space space space or space space dy over dx space equals space fraction numerator 1 over denominator straight x left parenthesis straight x squared minus 1 right parenthesis end fraction

Separating the variables, we get.

dy space equals space fraction numerator 1 over denominator straight x left parenthesis straight x minus 1 right parenthesis thin space left parenthesis straight x plus 1 right parenthesis end fraction dx

Integrating,

integral space dy space equals space integral fraction numerator 1 over denominator straight x space left parenthesis straight x minus 1 right parenthesis space left parenthesis straight x plus 1 right parenthesis end fraction space dx

integral dy space equals space integral open square brackets fraction numerator 1 over denominator straight x left parenthesis 0 minus 1 right parenthesis thin space left parenthesis 0 plus 1 right parenthesis end fraction plus fraction numerator 1 over denominator left parenthesis 1 right parenthesis space left parenthesis straight x minus 1 right parenthesis thin space left parenthesis 1 plus 1 right parenthesis end fraction plus fraction numerator 1 over denominator left parenthesis negative 1 right parenthesis thin space left parenthesis negative 1 minus 1 right parenthesis space left parenthesis straight x plus 1 right parenthesis end fraction close square brackets dx therefore space space space space integral space 1 space dy space equals space integral open square brackets negative 1 over straight x plus fraction numerator 1 over denominator 2 left parenthesis straight x minus 1 right parenthesis end fraction plus fraction numerator 1 over denominator 2 left parenthesis straight x plus 1 right parenthesis end fraction close square brackets dx therefore space space space straight y space equals space minus log space open vertical bar straight x close vertical bar plus 1 half space log space open vertical bar straight x minus 1 close vertical bar space plus space 1 half log space open vertical bar straight x plus 1 close vertical bar plus straight c space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis

Now y = 0 when x = 2

therefore space space space space 0 space equals space minus log space 2 space plus 1 half log space 1 space plus space 1 half log space 3 space plus straight c rightwards double arrow space space space space 0 space equals space minus space log space 2 plus 1 half cross times 0 plus 1 half log space 3 space plus straight c rightwards double arrow space 0 space equals space minus 1 half log space 4 plus 1 half space log 3 space plus straight c rightwards double arrow space space space straight c space equals space minus 1 half left parenthesis log space 3 space minus space log space 4 right parenthesis space space space space rightwards double arrow space space space straight c space equals space minus 1 half log open parentheses 3 over 4 close parentheses therefore space space from space left parenthesis 1 right parenthesis comma space space space space space space space straight y space equals space minus log space open vertical bar straight x close vertical bar plus 1 half log space open vertical bar straight x minus 1 close vertical bar plus 1 half log space open vertical bar straight x plus 1 close vertical bar minus 1 half log space open parentheses 3 over 4 close parentheses or space space space space straight y space equals space 1 half open square brackets log space open vertical bar straight x minus 1 close vertical bar plus log space open vertical bar straight x plus 1 close vertical bar minus 2 space log open vertical bar straight x close vertical bar space close square brackets minus 1 half log space open parentheses 3 over 4 close parentheses or space space space space straight y space equals space 1 half open square brackets log space open vertical bar space left parenthesis straight x minus 1 right parenthesis thin space left parenthesis straight x plus 1 right parenthesis close vertical bar minus log space open vertical bar straight x close vertical bar squared close square brackets space minus space 1 half log space 3 over 4 or space space space straight y space equals space 1 half log space open vertical bar fraction numerator straight x squared minus 1 over denominator straight x squared end fraction close vertical bar space minus space 1 half log space 3 over 4 space is space the space required space solution. space

Question 25

Wired Faculty App

For the following differential equation, given below, find particular solution satisfying the given condition:
$cos space open parentheses dy over dx close parentheses space space equals straight a space space space left parenthesis straight a space element of space straight R right parenthesis semicolon space space space straight y space equals space 1 space space when space straight x space equals space 0$

Solution

The given differential equation is

cos space open parentheses dy over dx close parentheses space equals space straight a space space space space or space space space dy over dx space equals space cos to the power of negative 1 end exponent straight a

therefore space space space space space space space space space dy space equals space cos to the power of negative 1 end exponent straight a. space dx space space space rightwards double arrow space space space space integral space dy space equals space cos to the power of negative 1 end exponent space straight a integral space dx

therefore space space space space space space space space straight y space equals space left parenthesis cos to the power of negative 1 end exponent straight a right parenthesis. space straight x plus space straight c

...(1)
Now y = 1 when x = 0

therefore space space space 1 space equals space left parenthesis cos to the power of negative 1 end exponent straight a right parenthesis space. space 0 space plus straight c space rightwards double arrow space space 1 space equals space straight c therefore space space space space from space left parenthesis 1 right parenthesis comma space space straight y space equals space straight x space cos to the power of negative 1 end exponent straight a plus 1

Question 26

Wired Faculty App

For the following differential equation, given below, find particular solution satisfying the given condition:
$dy over dx equals straight y space tanx space semicolon space space straight y space equals space 1 space space when space straight x space equals space 0$

Solution

The given differential equation is

dy over dx space equals space straight y space tanx

Separating the variables, we get,

1 over straight y dy space equals space tanx space dx

Integrating,

integral 1 over straight y dy space equals space integral tanx space dx

therefore space space space space log space open vertical bar straight y close vertical bar space equals space log space open vertical bar cosx close vertical bar space plus space straight c apostrophe space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis therefore space space space space log space open vertical bar straight y close vertical bar plus log space open vertical bar cosx close vertical bar space equals space straight c apostrophe therefore space space space log space open vertical bar straight y space cosx close vertical bar space equals space straight c apostrophe therefore space space space space open vertical bar straight y space cosx close vertical bar space equals space straight e to the power of straight c apostrophe end exponent therefore space space space straight y space cosx space equals space plus-or-minus straight e to the power of straight c space space space or space space straight y space cosx space equals space straight c

Now y = 1 when x = 0

therefore space space space 1 space space cos space 0 space equals space straight c space space space space rightwards double arrow space space space space 1 cross times 1 space equals space straight c space space space space space rightwards double arrow space space space straight c space equals space 1 therefore space space from space left parenthesis 1 right parenthesis comma space space space straight y space cosx space equals space 1 comma space space which space is space required space solution.

Question 27

Wired Faculty App

Solve: $straight x squared dy over dx space equals straight x squared plus 5 xy plus 4 straight y squared.$

Solution

The given differential equation is

straight x squared dy over dx space equals space straight x squared plus 5 xy plus 4 straight y squared

dy over dx space equals space fraction numerator straight x squared plus 5 xy plus 4 straight y squared over denominator straight x squared end fraction

Put y = v x so that

dy over dx equals straight v plus straight x dv over dx

therefore space space space space space from space left parenthesis 1 right parenthesis comma space space straight v plus straight x dv over dx space equals fraction numerator straight x squared plus 5 straight x. space straight v space straight x space plus space 4 space straight v squared space straight x squared over denominator straight x squared end fraction

straight v plus straight x dv over dx space equals space 1 plus 5 straight v plus 4 straight v squared

straight x dv over dx space equals space 4 space straight v squared plus 4 space straight v space plus space 1

Separating the variables, we get,

fraction numerator 1 over denominator 4 space straight v squared plus 4 space straight v space plus space 1 end fraction dv space equals space 1 over straight x dx

Integrating

integral fraction numerator 1 over denominator 4 straight v squared plus 4 straight v plus 1 end fraction straight d space straight v space equals space integral 1 over straight x dx

integral fraction numerator 1 over denominator left parenthesis 2 space straight v space plus space 1 right parenthesis squared end fraction dv space equals space integral 1 over straight x dx space space space space space or space space space space integral left parenthesis 2 straight v plus 1 right parenthesis to the power of negative 2 end exponent space dv space equals space integral 1 over straight x dx

therefore space space space space fraction numerator left parenthesis 2 space straight v space plus space 1 right parenthesis to the power of negative 1 end exponent over denominator left parenthesis 2 right parenthesis space left parenthesis negative 1 right parenthesis end fraction space equals space log space open vertical bar straight x close vertical bar space plus straight c space space space space space or space space space space minus fraction numerator 1 over denominator 2 space left parenthesis 2 space straight v space plus 1 right parenthesis end fraction log space open vertical bar straight x close vertical bar plus straight c

negative fraction numerator 1 over denominator 2 open parentheses 2 begin display style straight y over straight x end style plus 1 close parentheses end fraction space equals space log space open vertical bar straight x close vertical bar plus straight c space space space space space space space space or space space space space space space space space space minus fraction numerator straight x over denominator 2 space left parenthesis straight x plus 2 space straight y right parenthesis end fraction space equals space log space open vertical bar straight x close vertical bar plus straight c

fraction numerator straight x over denominator 2 space left parenthesis straight x plus 2 space straight y right parenthesis end fraction plus log space open vertical bar straight x close vertical bar plus straight c space equals space 0

is the required solution.

Question 28

Wired Faculty App

Solve the following differential equation:
$straight x squared dy over dx space equals space 2 xy plus straight y squared$

Solution

The given differential equation is

straight x squared dy over dx space equals space 2 xy plus straight y squared

dy over dx space equals space fraction numerator 2 xy plus straight y squared over denominator straight x squared end fraction

...(1)
Put

straight y space equals space straight v space straight x space space space space so space that space space space dy over dx equals straight v plus dv over dx

therefore space space space from space left parenthesis 1 right parenthesis comma space space straight v plus straight x dv over dx space equals space fraction numerator 2 straight x. space vx plus straight v squared straight x squared over denominator straight x squared end fraction

straight v plus straight x space dv over dx space equals space 2 space straight v space plus space straight v squared

straight x dv over dx space equals straight v squared plus straight v

Separating the variables, we get,

fraction numerator 1 over denominator straight v squared plus straight v end fraction dv space equals space 1 over straight x dx

Integrating,

integral fraction numerator 1 over denominator straight v squared plus straight v end fraction dv space equals space integral 1 over straight x dx

therefore space space space integral fraction numerator 1 over denominator straight v left parenthesis straight v plus 1 right parenthesis end fraction dv space equals space integral 1 over straight x dx

space therefore space space space space integral open square brackets fraction numerator 1 over denominator straight v left parenthesis 0 plus 1 right parenthesis end fraction plus fraction numerator 1 over denominator left parenthesis negative 1 right parenthesis thin space left parenthesis straight v plus 1 right parenthesis end fraction close square brackets space dv space equals space integral space 1 over straight x dx therefore space space space space integral open parentheses 1 over straight v minus fraction numerator 1 over denominator straight v plus 1 end fraction close parentheses space dv space equals space integral 1 over straight x dx therefore space space space log space open vertical bar straight v close vertical bar space minus space log space open vertical bar straight v plus 1 close vertical bar space equals space log space open vertical bar straight x close vertical bar space plus space log space straight c subscript 1 therefore space space log space open vertical bar fraction numerator straight v over denominator straight v plus 1 end fraction close vertical bar space equals space log space left parenthesis straight c subscript 1 space open vertical bar straight x close vertical bar right parenthesis therefore space space open vertical bar fraction numerator straight v over denominator straight v plus 1 end fraction close vertical bar space equals space straight c subscript 1 open vertical bar straight x close vertical bar therefore space space space space space fraction numerator begin display style straight y over straight x end style over denominator begin display style straight y over straight x plus 1 end style end fraction equals space space cx therefore space space space space fraction numerator straight y over denominator straight x plus straight y end fraction space equals space straight c space straight x comma space which space is space required space solution. space

Question 29

Wired Faculty App

Solve the differential equation:
$straight x squared dy over dx space equals straight y left parenthesis straight x plus straight y right parenthesis.$

Solution

The given differential equation is

straight x squared dy over dx space equals space straight y left parenthesis straight x plus straight y right parenthesis space space space or space space space space dy over dx space equals fraction numerator straight x space straight y space plus space straight y squared over denominator straight x squared end fraction space space space space space space space space... left parenthesis 1 right parenthesis

Put y = v x so that

dy over dx space equals space straight v plus dv over dx

therefore space space space from space left parenthesis 1 right parenthesis comma space space straight v plus straight x dv over dx space equals space fraction numerator straight x. space straight v space straight x space plus space straight v squared space straight x squared over denominator straight x squared end fraction

straight v plus straight x dv over dx space equals space straight v plus straight v squared

straight x dv over dx space equals space straight v squared

Separating the variables, we get,

1 over straight v squared dv space equals space 1 over straight x dx space space space space space or space space space straight v to the power of negative 2 end exponent dv space equals space 1 over straight x dx

Integrating,

integral straight v to the power of negative 2 end exponent space dv space equals space integral 1 over straight x dx

therefore space space space space space space fraction numerator straight v to the power of negative 1 end exponent over denominator negative 1 end fraction space equals space log space open vertical bar straight x close vertical bar space plus straight c space space space space space space or space space space space minus 1 over straight v space equals space log space open vertical bar straight x close vertical bar space plus space straight c therefore space space space space space space fraction numerator negative straight x over denominator straight y end fraction space equals space log space open vertical bar straight x close vertical bar space plus straight c comma space space which space is space required space solution. space

Question 30

Wired Faculty App

Solve the differential equation:
$left parenthesis straight y plus straight x right parenthesis space dy over dx space equals space straight y minus straight x.$

Solution

The given differential equation is

left parenthesis straight y plus straight x right parenthesis space dy over dx space equals space straight y minus straight x

therefore space space space space space space space dy over dx space equals space fraction numerator straight y minus straight x over denominator straight y plus straight x end fraction

...(1)
Put y = v x so that

dy over dx space equals space straight v plus straight x dv over dx

therefore

from (1),

straight v plus straight x dv over dx space equals space fraction numerator vx minus straight x over denominator vx plus straight x end fraction

therefore space space space space straight v plus straight x dv over dx space equals space fraction numerator straight v minus 1 over denominator straight v plus 1 end fraction space space space space space space space space space space space space space or space space space space space space space straight x dv over dx equals fraction numerator straight v minus 1 over denominator straight v plus 1 end fraction minus straight v

straight x dv over dx space equals fraction numerator straight v minus 1 minus straight v squared minus straight v over denominator straight v plus 1 end fraction space space space space or space space space space straight x dv over dx space equals space minus fraction numerator 1 plus straight v squared over denominator 1 plus straight v end fraction

Separating the variables, we get,

fraction numerator 1 plus straight v over denominator 1 plus straight v squared end fraction dv space equals space minus 1 over straight x dx

Integrating,

integral fraction numerator 1 plus straight v over denominator 1 plus straight v squared end fraction dv equals space minus integral 1 over straight x dx

therefore space space space space integral fraction numerator 1 over denominator 1 plus straight v squared end fraction dv plus 1 half integral fraction numerator 2 straight v over denominator 1 plus straight v squared end fraction dv space equals space minus integral 1 over straight x dx

therefore space space space space tan to the power of negative 1 end exponent straight v plus 1 half log space left parenthesis 1 plus straight v squared right parenthesis space equals space minus log space open vertical bar straight x close vertical bar space plus space straight c subscript 1 therefore space space space space space tan to the power of negative 1 end exponent open parentheses straight y over straight x close parentheses plus 1 half log open parentheses 1 plus straight y squared over straight x squared close parentheses space equals space minus log space open vertical bar straight x close vertical bar plus straight c subscript 1 therefore space space space space space 2 space tan to the power of negative 1 end exponent open parentheses straight y over straight x close parentheses plus log space open parentheses fraction numerator straight x squared plus straight y squared over denominator straight x squared end fraction close parentheses space equals space minus log open vertical bar straight x close vertical bar squared plus 2 straight c subscript 1 therefore space space space space space 2 space tan to the power of negative 1 end exponent open parentheses straight y over straight x close parentheses plus log space open parentheses fraction numerator straight x squared plus straight y squared over denominator straight x squared end fraction close parentheses space equals space minus log space straight x squared plus straight c therefore space space space 2 space tan to the power of negative 1 end exponent open parentheses straight y over straight x close parentheses plus log space open parentheses fraction numerator straight x squared plus straight y squared over denominator straight x squared end fraction close parentheses plus log space straight x squared space equals space straight c therefore space space space 2 space tan to the power of negative 1 end exponent open parentheses straight y over straight x close parentheses plus log space open parentheses fraction numerator straight x squared plus straight y squared over denominator straight x squared end fraction cross times space straight x squared close parentheses space equals space straight c therefore space space space 2 space tan to the power of negative 1 end exponent open parentheses straight y over straight x close parentheses plus log space open parentheses straight x squared plus straight y squared close parentheses space equals space straight c

which is required solution.

Question 31

Wired Faculty App

Solve the following differential equation:
(y² – x²) dy = 3 x y dx

Solution

The given differential equation is

open parentheses straight y squared minus straight x squared close parentheses space dy space equals space 3 xy space dx

dy over dx space equals space fraction numerator 3 space straight x space straight y over denominator straight y squared minus straight x squared end fraction

...(1)
Put y = v x so that

dy over dx space equals straight v plus straight x dv over dx.

therefore space space space space space space space left parenthesis 1 right parenthesis space becomes comma space space straight v plus straight x dv over dx space equals space fraction numerator 3 straight x. space straight v space straight x over denominator straight v squared straight x squared minus straight x squared end fraction

straight v plus straight x dv over dx space equals space fraction numerator 3 straight v over denominator straight v squared minus 1 end fraction

therefore space space space space space space straight x space dv over dx space equals space fraction numerator 3 straight v over denominator straight v squared minus 1 end fraction minus straight v space space space space or space space space space straight x space dv over dx space equals space fraction numerator 4 straight v space minus straight v cubed over denominator straight v squared minus 1 end fraction

Separating the variables, we get,

fraction numerator straight v squared minus 1 over denominator 4 space straight v minus straight v cubed end fraction dv space equals space 1 over straight x dx space space space or space space space fraction numerator straight v squared minus 1 over denominator straight v left parenthesis 4 minus straight v squared right parenthesis end fraction space dv space equals space 1 over straight x dx

Integrating,

integral fraction numerator straight v squared minus 1 over denominator straight v space left parenthesis 2 minus straight v right parenthesis thin space left parenthesis 2 plus straight v right parenthesis end fraction dv space equals space integral 1 over straight x dx

therefore space space space space integral open square brackets fraction numerator 0 minus 1 over denominator straight v space left parenthesis 2 minus 0 right parenthesis thin space left parenthesis 2 plus 0 right parenthesis end fraction minus fraction numerator 4 minus 1 over denominator left parenthesis 2 right parenthesis thin space left parenthesis 2 minus straight v right parenthesis thin space left parenthesis 2 plus 2 right parenthesis end fraction plus fraction numerator 4 minus 1 over denominator left parenthesis negative 2 right parenthesis thin space left parenthesis 2 plus 2 right parenthesis thin space left parenthesis 2 plus straight v right parenthesis end fraction close square brackets dv space equals space integral 1 over straight x dx

therefore space space integral open square brackets negative fraction numerator 1 over denominator 4 straight v end fraction plus fraction numerator 3 over denominator 8 left parenthesis 2 minus straight v right parenthesis end fraction minus fraction numerator 3 over denominator 8 left parenthesis 2 plus straight v right parenthesis end fraction close square brackets space dv space equals space integral 1 over straight x dx therefore space space space space minus space 1 fourth open vertical bar straight v close vertical bar plus space 3 over 8 fraction numerator log space open vertical bar 2 minus straight v close vertical bar over denominator negative 1 end fraction space minus space 3 over 8 space log space open vertical bar 2 plus straight v close vertical bar space equals space log space straight x space plus straight c therefore space space space space space minus 1 fourth space log space open vertical bar straight y over straight x close vertical bar space minus space 3 over 8 space log space open vertical bar 2 minus straight y over straight x close vertical bar space minus space 3 over 8 log space open vertical bar 2 plus straight y over straight x close vertical bar space equals space logx plus straight c

which is required solution.

Question 32

Wired Faculty App

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
(x - y) y' = x + 2 y

Solution

The given differential equation is

left parenthesis straight x minus straight y right parenthesis space dy over dx space equals space straight x plus 2 straight y

dy over dx equals fraction numerator straight x plus 2 straight y over denominator straight x minus straight y end fraction

...(1)
Put y = vx so that

dy over dx equals straight v plus straight x dv over dx

therefore space space space from space left parenthesis 1 right parenthesis comma space space straight v plus straight x dv over dx equals fraction numerator straight x plus 2 space straight v space straight x over denominator 1 minus straight v space straight x end fraction space space space space space space or space space space space space space straight v plus straight x dv over dx equals fraction numerator 1 plus 2 straight v over denominator 1 minus straight v end fraction therefore space space space space space space space space space straight x dv over dx equals fraction numerator 1 plus 2 straight v over denominator 1 minus straight v end fraction minus straight v space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space straight x space dv over dx space equals fraction numerator 1 plus 2 space straight v minus straight v plus straight v squared over denominator 1 minus straight v end fraction or space space space space space space space space straight x dv over dx equals fraction numerator 1 plus straight v plus straight v squared over denominator 1 minus straight v end fraction

Separating the variables,

fraction numerator 1 minus straight v over denominator 1 plus straight v plus straight v squared end fraction dv space equals space 1 over straight x dx

Integrating,

integral fraction numerator 1 minus straight v over denominator 1 plus straight v plus straight v squared end fraction dv space equals space integral 1 over straight x dx

therefore space space space space integral fraction numerator negative begin display style 1 half end style left parenthesis 2 straight v plus 1 right parenthesis plus begin display style 3 over 2 end style over denominator 1 plus straight v plus straight v squared end fraction dv space equals space integral 1 over straight x dv

therefore space space space minus 1 half integral fraction numerator 2 straight v plus 1 over denominator 1 plus straight v plus straight v squared end fraction dv plus 3 over 2 integral fraction numerator 1 over denominator 1 plus straight v plus straight v squared end fraction dv space equals space integral 1 over straight x dx

therefore space space minus 1 half integral fraction numerator 2 straight v plus 1 over denominator 1 plus straight v plus straight v squared end fraction dv plus 3 over 2 integral fraction numerator 1 over denominator open parentheses straight v plus begin display style 1 half end style close parentheses squared plus open parentheses begin display style fraction numerator square root of 3 over denominator 2 end fraction end style close parentheses squared end fraction dv space equals space integral 1 over straight x dx

therefore space space space space minus 1 half log space open vertical bar 1 plus straight v plus straight v squared close vertical bar plus 3 over 2. space fraction numerator 1 over denominator begin display style fraction numerator square root of 3 over denominator 2 end fraction end style end fraction space tan to the power of negative 1 end exponent open parentheses fraction numerator straight v plus begin display style 1 half end style over denominator begin display style fraction numerator square root of 3 over denominator 2 end fraction end style end fraction close parentheses space equals space log space open vertical bar straight x close vertical bar plus straight c subscript 1

therefore space space space minus 1 half log space open vertical bar 1 plus straight y over straight x plus straight y squared over straight x squared close vertical bar plus square root of 3 space tan to the power of negative 1 end exponent space open parentheses fraction numerator 2 straight v plus 1 over denominator square root of 3 end fraction close parentheses space equals space log space open vertical bar straight x close vertical bar space plus space straight c subscript 1 therefore space space space space minus 1 half log space open vertical bar fraction numerator straight x squared plus straight x space straight y plus straight y squared over denominator straight x squared end fraction close vertical bar plus square root of 3 space tan to the power of negative 1 end exponent space open parentheses fraction numerator 2 begin display style straight y over straight x end style plus 1 over denominator square root of 3 end fraction close parentheses space equals space log space open vertical bar straight x close vertical bar space plus space straight c subscript 1 therefore space space space space 2 square root of 3 space tan to the power of negative 1 end exponent open parentheses fraction numerator 2 straight y plus straight x over denominator straight x square root of 3 end fraction close parentheses space minus space log space open vertical bar straight x squared plus xy plus straight y squared close vertical bar plus log space straight x squared space equals space log space straight x squared plus straight c therefore space space space 2 square root of 3 space tan to the power of negative 1 end exponent space open parentheses fraction numerator straight x plus 2 straight y over denominator straight x square root of 3 end fraction close parentheses space minus space log space open vertical bar straight x squared plus straight x space straight y space plus straight y squared close vertical bar space equals space straight c

Which is required solution.

Question 33

Wired Faculty App

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
(x² + y²) y' = 8 x² - 3 x y + 2 y²

Solution

The given differential equation is

left parenthesis straight x squared plus straight y squared right parenthesis space dy over dx space equals space 8 straight x squared minus 3 xy plus 2 straight y squared

therefore

dy over dx space equals space fraction numerator 8 straight x squared minus 3 xy plus 2 straight y squared over denominator straight x squared plus straight y squared end fraction

Put

straight y equals vx space so space that space dy over dx space equals space straight v plus straight x dv over dx

therefore space space space space space space space straight v plus straight x dv over dx space equals space fraction numerator 8 straight x squared minus 3 straight x cubed straight v plus 2 straight x squared straight v squared over denominator straight x squared plus straight v squared straight x squared end fraction therefore space space space space space straight v plus straight x dv over dx space equals space fraction numerator 8 minus 3 straight v plus 2 straight v squared over denominator 1 plus straight v squared end fraction therefore space space space space space space space space space space space straight x dv over dx space equals space fraction numerator 8 minus 3 straight v plus 2 straight v squared over denominator 1 plus straight v squared end fraction minus straight v therefore space space space space space space space space straight x dv over dx space equals space fraction numerator 8 minus 3 straight v plus 2 straight v squared minus straight v minus straight v cubed over denominator 1 plus straight v squared end fraction therefore space space space space space straight x dv over dx space equals space fraction numerator 8 minus 4 straight v plus 2 straight v squared minus straight v cubed over denominator 1 plus straight v squared end fraction

Separating the variables, we get,

fraction numerator 1 plus straight v squared over denominator 8 minus 4 straight v plus 2 straight v squared minus straight v cubed end fraction dv space equals space 1 over straight x dx

Integrating,

integral fraction numerator 1 plus straight v squared over denominator 8 minus 4 straight v plus 2 straight v squared minus straight v cubed end fraction dv space equals space integral 1 over straight x dx

therefore space space space space integral fraction numerator 1 plus straight v squared over denominator left parenthesis 2 minus straight v right parenthesis thin space left parenthesis 4 plus straight v squared right parenthesis end fraction dv space equals space integral 1 over straight x dx

Put

fraction numerator 1 plus straight v squared over denominator left parenthesis 2 minus straight v right parenthesis thin space left parenthesis 4 plus straight v squared right parenthesis end fraction space equals space fraction numerator straight A over denominator 2 minus straight v end fraction plus fraction numerator Bv plus straight c over denominator 4 plus straight v squared end fraction

therefore

1 plus straight v squared space space identical to space space straight A space left parenthesis 4 plus straight v squared right parenthesis space plus space straight B space straight v thin space left parenthesis 2 minus straight v right parenthesis space plus space straight C left parenthesis 2 minus straight v right parenthesis

...(1)
Put

2 minus straight v space equals space 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space or space space space space space space straight v space equals space 2 space in space left parenthesis 1 right parenthesis

therefore space space space space space space space space space space space 1 plus 4 space equals space straight A left parenthesis 4 plus 4 right parenthesis plus 0 plus 0 space space space space space space space space space space space space space space space rightwards double arrow space space 5 space equals space 8 straight A space space space space space space space rightwards double arrow space space straight A space equals space 5 over 8

(1) can be written as

1 plus straight v squared identical to space straight A left parenthesis 4 plus straight v squared right parenthesis space plus space straight B left parenthesis 2 straight v minus straight v squared right parenthesis space plus straight C left parenthesis 2 minus straight v right parenthesis

...(2)
Equating coefficients in (2) of
V²) 1 =A - B

rightwards double arrow space space space space 1 space equals space 5 over 8 minus straight B space space space space space space space space rightwards double arrow space space space space straight B space equals space minus 3 over 8

v) 0 = 2 B - C

rightwards double arrow space 0 space equals space 3 over 4 minus straight C space space space space space space rightwards double arrow space space space space straight C space equals space minus 3 over 4

therefore space space space space fraction numerator 1 plus straight v squared over denominator left parenthesis 2 minus straight v right parenthesis thin space left parenthesis 4 plus straight v squared right parenthesis end fraction space identical to space space fraction numerator 5 over denominator 8 space left parenthesis 2 minus straight v right parenthesis end fraction plus fraction numerator negative begin display style 3 over 8 end style straight v minus begin display style 3 over 4 end style over denominator 4 plus straight v end fraction therefore space space from space left parenthesis 1 right parenthesis comma space space integral open square brackets fraction numerator 5 over denominator 8 left parenthesis 2 minus straight v right parenthesis end fraction plus fraction numerator negative begin display style 3 over 8 end style straight v minus begin display style 3 over 4 end style over denominator straight v squared plus 4 end fraction close square brackets space dv space equals space integral 1 over straight x dx therefore space space space 5 over 8 integral fraction numerator 1 over denominator 2 minus straight v end fraction dv space minus 3 over 16 integral fraction numerator 2 straight v over denominator straight v squared plus 4 end fraction dv minus 3 over 4 integral fraction numerator 1 over denominator straight v squared plus left parenthesis 2 right parenthesis squared end fraction dv space equals space integral 1 over straight x dx therefore space space space 5 over 8 fraction numerator log space open vertical bar 2 minus straight v close vertical bar over denominator negative 1 end fraction minus 3 over 16 log space left parenthesis straight v squared plus 4 right parenthesis minus space 3 over 4 space 1 half space tan to the power of negative 1 end exponent straight v over 2 space equals space log space open vertical bar straight x close vertical bar space space plus straight c therefore space space space minus 5 over 8 log space open vertical bar 2 minus straight y over straight x close vertical bar minus 3 over 16 log space open parentheses straight y squared over straight x squared plus 4 close parentheses space minus space 3 over 8 tan to the power of negative 1 end exponent open parentheses fraction numerator straight y over denominator 2 straight x end fraction close parentheses space equals space log space open vertical bar straight x close vertical bar plus straight c

which is required primitive.

Question 34

Wired Faculty App

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
(3 x y + y²) dx = (x² + x y) dy

Solution

The given differential equation is
(3 x y + y²) dx = (x² + x y) dy
or

dy over dx equals fraction numerator 3 xy plus straight y squared over denominator straight x squared plus xy end fraction

Put y = v x so that

dy over dx equals straight v plus straight x dv over dx

therefore space space space space space space straight v plus straight x dv over dx equals fraction numerator 3 straight x squared straight v plus straight v squared straight x squared over denominator straight x squared plus straight x squared straight v end fraction therefore space space space straight v plus straight x dv over dx space equals space fraction numerator 3 straight v plus straight v squared over denominator 1 plus straight v end fraction therefore space space space space space space straight x dv over dx space equals space fraction numerator 3 straight v plus straight v squared over denominator 1 plus straight v end fraction minus straight v space space space space space or space space space straight x dv over dx space equals space fraction numerator 3 straight v plus straight v squared minus straight v minus straight v squared over denominator 1 minus straight v end fraction therefore space space space space space space straight x dv over dx equals fraction numerator 2 straight v over denominator 1 plus straight v end fraction

Separating the variables,

fraction numerator 1 plus straight v over denominator straight v end fraction dv space equals space 2 over straight x dx

Integrating,

integral open parentheses 1 over straight v plus 1 close parentheses space dv space equals space 2 integral 1 over straight x dx

therefore space space space space space log space open vertical bar straight v close vertical bar plus straight v space equals space 2 space log space open vertical bar straight x close vertical bar space plus space straight c therefore space space space space log space open vertical bar straight y over straight x close vertical bar plus straight y over straight x space equals space 2 space log space open vertical bar straight x close vertical bar plus straight c

is the required solution.

Question 35

Wired Faculty App

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
2 x y dx + (x² + 2 y²) dy = 0

Solution

The given differential equation is
2 x y dx + (x² + 2 y²) dy = 0 or

left parenthesis straight x squared plus 2 straight y squared right parenthesis space dy space equals space minus 2 xy space dx

therefore space space space space dy over dx space equals space minus fraction numerator 2 xy over denominator straight x squared plus 2 straight y squared end fraction

Put y = v x so that

dy over dx equals straight v plus straight x dv over dx

therefore space space space space space straight v plus straight x dv over dx equals negative fraction numerator 2. straight x. space straight v space straight x over denominator straight x squared plus 2 straight v squared straight x squared end fraction space space space or space space space straight v plus straight x dv over dx space equals space minus fraction numerator 2 straight v over denominator 1 plus 2 straight v squared end fraction therefore space space space space space space straight x dv over dx equals negative fraction numerator 2 straight v over denominator 1 plus 2 straight v squared end fraction minus straight v space space space space space or space space space space space straight x dv over dx equals space fraction numerator negative 2 straight v minus straight v minus 2 straight v cubed over denominator 1 plus 2 straight v squared end fraction therefore space space space space space straight x dv over dx space equals fraction numerator negative 3 straight v minus 2 straight v cubed over denominator 1 plus 2 straight v squared end fraction space space space space space space space rightwards double arrow space space space space space space fraction numerator 1 plus 2 straight v squared over denominator 3 straight v plus 2 straight v cubed end fraction dv space equals space minus 1 over straight x dx

Integrating,

1 third integral fraction numerator 3 plus 6 straight v squared over denominator 3 straight v plus 2 straight v cubed end fraction space equals space minus integral 1 over straight x dx

therefore space space space 1 third space log space open vertical bar 3 straight v space plus space 2 straight v cubed close vertical bar space equals space minus 3 space log space open vertical bar straight x close vertical bar space plus space 3 space log space open vertical bar straight c subscript 1 close vertical bar therefore space space space log space open vertical bar 3 straight v plus 2 straight v cubed close vertical bar space plus space log space open vertical bar straight x close vertical bar cubed space equals space log space left parenthesis straight c right parenthesis therefore space space space space open square brackets left parenthesis 3 space straight v plus space 2 straight v cubed right parenthesis space left parenthesis straight x right parenthesis close square brackets cubed space equals space straight c therefore space space space open parentheses 3 straight y over straight x plus 2 straight y cubed over straight x cubed close parentheses space left parenthesis straight x cubed right parenthesis space equals straight c or space space space 3 straight x squared straight y plus 2 straight y cubed space equals space straight c comma space which space is space required space solution. space

Question 36

Wired Faculty App

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
$left parenthesis 2 straight x squared straight y plus straight y cubed right parenthesis space dx plus space left parenthesis xy squared minus 3 straight x cubed right parenthesis space dy space equals space 0$

Solution

The given differential equation is
$left parenthesis 2 straight x squared straight y plus straight y cubed right parenthesis space dx space plus space left parenthesis straight x space straight y squared minus space 3 space straight x cubed right parenthesis space dy space equals space 0$
or $dy over dx space equals space minus fraction numerator 2 straight x squared straight y plus straight y cubed over denominator xy squared minus 3 straight x cubed end fraction$
Put y = v x so that $dy over dx equals straight v plus straight x dv over dx$
$therefore space space space space space straight v plus straight x dv over dx space equals space minus fraction numerator 2 straight x cubed straight v plus straight v cubed straight x cubed over denominator straight x cubed straight v squared minus 3 straight x cubed end fraction therefore space space space space straight v plus straight x dv over dx space equals space fraction numerator 2 straight v plus straight v cubed over denominator 3 minus straight v squared end fraction therefore space space space space space space straight x dv over dx space equals space fraction numerator 2 straight v cubed minus straight v over denominator 3 minus straight v squared end fraction therefore space space space space fraction numerator 3 minus straight v squared over denominator 2 straight v cubed minus straight v end fraction dv space space equals space 1 over straight x dx space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space.. left parenthesis 1 right parenthesis$
$therefore space space integral fraction numerator 3 minus straight v squared over denominator straight v left parenthesis 2 straight v squared minus 1 right parenthesis end fraction dv space equals space integral 1 over straight x dx$
Put $fraction numerator 3 minus straight v squared over denominator straight v left parenthesis 2 straight v squared minus 1 right parenthesis end fraction identical to space straight A over straight v plus fraction numerator Bv plus straight c over denominator 2 straight v squared minus 1 end fraction$
$therefore space space space space space space space 3 minus straight v squared space identical to space straight A space left parenthesis 2 straight v squared minus 1 right parenthesis space plus space Bv squared plus straight C space straight v$ ...(2)
Put v = 0 in (2)
$therefore space space space space space space space 3 minus 0 space equals space straight A left parenthesis 0 minus 1 right parenthesis plus 0 plus 0 space space space space space space rightwards double arrow space space space 3 equals space space minus straight A space space space space space space rightwards double arrow space space space straight A space equals space minus 3$
Equating coefficients in (2) of
$straight v squared right parenthesis$ -1 = 2 A + B $rightwards double arrow space space minus 1 space equals space minus 6 plus straight B space space space space space space space space space space rightwards double arrow space space space straight B space equals space 5$
$straight v right parenthesis space space space space space space space space space space 0 space equals space straight c space space space space space space space space space space rightwards double arrow space space space space straight C space equals space 0$
$therefore space space fraction numerator 3 minus straight v squared over denominator straight v left parenthesis 2 straight v squared minus 1 right parenthesis end fraction identical to fraction numerator negative 3 over denominator straight v end fraction plus fraction numerator 5 straight v over denominator 2 straight v squared minus 1 end fraction therefore space space space from space left parenthesis 1 right parenthesis space we space get comma therefore space space space space minus 3 space integral 1 over straight v plus 5 over 4 integral fraction numerator 4 straight v over denominator 2 straight v squared minus 1 end fraction dv space equals space integral 1 over straight x dx therefore space space space space minus 3 space log space open vertical bar straight v close vertical bar space plus space 5 over 4 log space open vertical bar 2 space straight v squared minus 1 close vertical bar space equals space log space open vertical bar straight x close vertical bar space plus space log space straight c subscript 1 space space space space space space space space space space minus 12 space log space open vertical bar straight v close vertical bar space plus space 5 space log space open vertical bar 2 straight v squared minus 1 close vertical bar space equals space 4 space log space open vertical bar straight x close vertical bar space space plus space 4 space log space straight c subscript 1 therefore space space space minus 12 space log space open vertical bar straight y over straight x close vertical bar space plus space 5 space log space open vertical bar fraction numerator 2 straight y squared over denominator straight x squared end fraction minus 1 close vertical bar space equals space 4 space log space open vertical bar straight x close vertical bar space plus space space 4 space log space straight c subscript 1 therefore space space space minus 12 space log space open vertical bar straight y close vertical bar space plus space 12 space log space open vertical bar straight x close vertical bar space plus space 5 space log space open vertical bar 2 straight y squared minus straight x squared close vertical bar space minus space 10 space log space open vertical bar straight x close vertical bar space minus space 4 space log space open vertical bar straight x close vertical bar space equals space 4 space log space straight c subscript 1 therefore space space space space minus 12 space log space open vertical bar straight y close vertical bar minus 2 space log space open vertical bar straight x close vertical bar space plus space 5 space log space open vertical bar 2 straight y squared minus straight x squared close vertical bar space equals space log space straight c therefore space space space log space open vertical bar fraction numerator 2 straight y squared minus straight x squared over denominator straight y to the power of 12 straight x squared end fraction close vertical bar space equals space log space straight c therefore space space space space space space space space fraction numerator 2 straight y squared minus straight x squared over denominator straight x squared space straight y to the power of 12 end fraction space equals space straight c therefore space space space space space space space 2 space straight y squared minus straight x squared space equals space straight c space straight x squared space straight y to the power of 12 space is space the space required space solution. space$

Question 37

Wired Faculty App

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
$straight x space straight y apostrophe space minus space straight y space plus space straight x space sin space space open parentheses straight y over straight x close parentheses space equals 0$

Solution

The given differential equation is

xy apostrophe space minus space straight y plus space straight x space sin space open parentheses straight y over straight x close parentheses space equals space 0

therefore space space space space space space space space straight x dy over dx space equals space straight y minus straight x space sin straight y over straight x therefore space space space space space space space space dy over dx space equals space straight y over straight x minus sin straight y over straight x

Put

straight y space equals space straight v space straight x space so space that space space dy over dx space equals space straight v plus straight x dv over dx

therefore space space space space space space straight v plus straight x dv over dx space equals space straight v minus sin space straight v comma space space space space space space space space space space space space space space space space therefore space space space straight x space dv over dx space equals space minus sin space straight v

fraction numerator 1 over denominator sin space straight v end fraction dv space equals space minus 1 over straight x dx

integral space cosec space straight v space dv space equals space minus integral 1 over straight x dx

therefore space space space space space space log space space open vertical bar tan space straight v over 2 close vertical bar space equals space minus log space open vertical bar straight x close vertical bar space plus space log space open vertical bar straight c close vertical bar

log space open vertical bar straight x space tan space straight v over 2 close vertical bar space equals space log space open vertical bar straight c close vertical bar space space space space space space space space space space space space or space space space space straight x space tan straight v over 2 space equals straight c

straight x space tan space open parentheses fraction numerator straight y over denominator 2 straight x end fraction close parentheses space equals space straight c space is space the space required space solution. space

Question 38

Wired Faculty App

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
$left parenthesis straight x plus 2 straight y right parenthesis space dx space minus space left parenthesis 2 straight x minus straight y right parenthesis space dy space equals space 0$

Solution

The given differential equation is
   $left parenthesis straight x plus 2 straight y right parenthesis space dx space minus space left parenthesis 2 straight x minus straight y right parenthesis space dy space equals space 0$
or    $dy over dx space equals space fraction numerator straight x plus 2 straight y over denominator 2 straight x minus straight y end fraction$
Put y = v x so that    $dy over dx space equals space straight v plus straight x dv over dx$
$therefore space space space straight v plus straight x dv over dx space equals fraction numerator straight x plus 2 vx over denominator 2 straight x minus vx end fraction therefore space space space space straight v plus straight x dv over dx space equals space fraction numerator 1 plus 2 straight v over denominator 2 minus straight v end fraction therefore space space space space space straight x dv over dx space equals space fraction numerator 1 plus 2 straight v over denominator 2 minus straight v end fraction minus straight v therefore space space space space straight x dv over dx space equals space fraction numerator 1 plus 2 straight v minus 2 straight v plus straight v squared over denominator 2 minus straight v end fraction therefore space space space space space space straight x dv over dx space equals space fraction numerator 1 plus straight v squared over denominator 2 minus straight v end fraction therefore space space space fraction numerator 2 minus straight v over denominator 1 plus straight v squared end fraction dv space equals space 1 over straight x dx therefore space space space space integral fraction numerator 2 minus straight v over denominator 1 plus straight v squared end fraction dv space equals space integral 1 over straight x dx therefore space space space 2 space integral fraction numerator 1 over denominator 1 plus straight v squared end fraction dv minus 1 half integral fraction numerator 2 straight v over denominator 1 plus straight v squared end fraction dv space equals space integral 1 over straight x dx therefore space 2 space tan to the power of negative 1 end exponent straight v space minus space 1 half log space left parenthesis 1 plus straight v squared right parenthesis space equals space log space open vertical bar straight x close vertical bar plus space log space straight c subscript 1 therefore space space 4 space tan to the power of negative 1 end exponent straight v space minus space log space left parenthesis 1 plus straight v squared right parenthesis space equals space 2 space log space open vertical bar straight x close vertical bar space plus space 2 space log space straight c subscript 1 therefore space space 4 space tan to the power of negative 1 end exponent straight y over straight x minus log space open parentheses 1 plus straight y squared over straight x squared close parentheses space equals space log space straight x squared plus space log space straight c therefore space space space 4 space tan to the power of negative 1 end exponent straight y over straight x minus log space left parenthesis straight x squared plus straight y squared right parenthesis space plus space log space straight x squared space equals space log space straight x squared plus space log space straight c therefore space space space 4 space tan to the power of negative 1 end exponent straight y over straight x minus log space left parenthesis straight x squared plus straight y squared right parenthesis space equals space log space straight c$
which is the required solution.

Question 39

Wired Faculty App

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
$1 over straight x cos straight y over straight x dx minus open parentheses straight x over straight y sin straight y over straight x plus cos straight y over straight x close parentheses dy space equals 0$

Solution

The given differential equation is

straight y over straight x cos straight y over straight x dx minus open parentheses straight x over straight y sin straight y over straight x plus cos straight y over straight x close parentheses dy space equals space 0

dy over dx space equals space fraction numerator begin display style straight y over straight x end style cos begin display style straight y over straight x end style over denominator begin display style straight x over straight y end style sin begin display style straight y over straight x end style plus cos begin display style straight y over straight x end style end fraction

Put

straight y space equals space space straight v space straight x space so space that space dy over dx space equals space straight v plus straight x dv over dx

therefore space space space space space space space space space space straight v plus straight x dv over dx space equals space fraction numerator straight v space cos space straight v over denominator begin display style 1 over straight v end style sinv space plus cos space straight v end fraction space space space rightwards double arrow space space space space straight v plus straight x dv over dx space equals fraction numerator straight v squared space cosv over denominator sin space straight v plus straight v space cos space straight v end fraction therefore space space space space space space space straight x dv over dx space equals space fraction numerator straight v squared space cos space straight v over denominator sin space straight v plus straight v space cos space straight v end fraction minus straight v therefore space space space space space straight x dv over dx space equals space fraction numerator straight v squared space cos space straight v minus space straight v space sin space straight v space minus space straight v squared space cos space straight v over denominator sin space straight v plus straight v space cos space straight v end fraction therefore space space space space straight x dv over dx space equals negative fraction numerator straight v space sin space straight v over denominator sin space straight v plus straight v space cos space straight v end fraction

Separating the variables and integrating ,we get,

integral fraction numerator sin space straight v plus space straight v space cosv over denominator straight v space sinv end fraction dv space equals negative integral 1 over straight x dx

therefore space space log space open vertical bar straight v space sinv close vertical bar space equals space minus space log space open vertical bar straight x close vertical bar space plus space log space open vertical bar straight A close vertical bar rightwards double arrow space space space log space open vertical bar straight y over straight x sin space straight y over straight x close vertical bar plus space log space open vertical bar straight x close vertical bar space equals space log space open vertical bar straight A close vertical bar rightwards double arrow space space log space open vertical bar straight y space sin straight y over straight x close vertical bar space minus space log space open vertical bar straight x close vertical bar space plus space log space open vertical bar straight x close vertical bar space equals space log space open vertical bar straight A close vertical bar rightwards double arrow space space log space open vertical bar straight y space sin straight y over straight x close vertical bar space equals space log space open vertical bar straight A close vertical bar rightwards double arrow space space space space space space straight y space sin space straight y over straight x space equals space straight A comma space space space which space is space required space solution. space

Sponsor Area

Question 40

Wired Faculty App

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
$2 space straight y space straight e to the power of straight x over 4 end exponent space dx plus open parentheses straight y minus 2 space straight x space straight e to the power of straight x over straight y end exponent close parentheses space dy space equals space 0$

Solution

The given differential equation is

2 ye to the power of straight x over straight y end exponent dx plus open parentheses straight y minus 2 xe to the power of straight x over straight y end exponent close parentheses dy space equals space 0

therefore space space space space space space space space open parentheses straight y minus 2 xe to the power of straight x over straight y end exponent close parentheses space dy space equals space minus 2 space straight y space straight e to the power of straight x over straight y end exponent dx therefore space space space space space space dx over dy space equals negative fraction numerator straight y minus 2 xe to the power of begin display style straight x over straight y end style end exponent over denominator 2 ye to the power of begin display style straight x over straight y end style end exponent end fraction therefore space space space space space space space dx over dy space equals space fraction numerator straight y open parentheses 2 begin display style straight x over straight y end style straight e to the power of begin display style straight x over straight y end style end exponent minus 1 close parentheses over denominator 2 straight y space. straight e to the power of begin display style straight x over straight y end style end exponent end fraction space space space space space space space or space space space space space space space space dx over dy space equals space fraction numerator 2 begin display style straight x over straight y end style straight e to the power of begin display style straight x over straight y end style end exponent minus 1 over denominator 2 straight e to the power of begin display style straight x over straight y end style end exponent end fraction

Put x = v y so that

dx over dy space equals space straight v plus straight y dv over dy

therefore space space space straight v plus straight y dv over dy space equals fraction numerator 2 ve to the power of straight v minus 1 over denominator 2 straight e to the power of straight v end fraction. space space space space space space therefore space space straight y space dv over dy space equals space fraction numerator 2 straight v space straight e to the power of straight v minus 1 over denominator 2 straight e to the power of straight v end fraction minus straight v therefore space space space space space straight y dv over dy space equals space fraction numerator 2 ve to the power of straight v minus 1 minus 2 ve to the power of straight v over denominator 2 straight e to the power of straight v end fraction therefore space space space straight y dv over dy space equals space minus fraction numerator 1 over denominator 2 straight e to the power of straight v end fraction therefore space space space space space straight e to the power of straight v space dv space equals space minus fraction numerator 1 over denominator 2 space straight y end fraction dy therefore space space space 2 space integral straight e to the power of straight v space dv space equals space minus integral 1 over straight y dy therefore space space space space space space space space 2 space straight e to the power of straight v space equals space minus log space open vertical bar straight y close vertical bar space plus space straight c therefore space space space space space space 2 space straight e to the power of straight x over straight y end exponent space equals space minus log space open vertical bar straight y close vertical bar plus straight c space is space the space required space solution.

Question 41

Wired Faculty App

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
$straight y space dx space plus space straight x space open parentheses log space straight y over straight x close parentheses space dy space minus space 2 space straight x space dy space equals space 0$

Solution

The given differential equation is
$straight y space dx plus straight x space log space open parentheses straight y over straight x close parentheses space dy space minus 2 straight x space dy space equals space 0$
$therefore space space straight x space open parentheses 2 minus log space straight y over straight x close parentheses space dy space equals space straight y space dx therefore space space space space space space space space space space space space space space space space space space space space space dy over dx equals fraction numerator begin display style straight y over straight x end style over denominator 2 minus log space begin display style straight y over straight x end style end fraction$
Put y = v x so that $dy over dx equals straight v plus dv over dx$
$therefore space space space space space space space space space straight v plus straight x dv over dx equals fraction numerator straight v over denominator 2 minus log space straight v end fraction therefore space space space space space space space space straight x space dv over dx space equals space fraction numerator straight v over denominator 2 minus log space straight v end fraction minus straight v therefore space space space space straight x dv over dx equals fraction numerator straight v minus 2 straight v plus straight v space log space straight v over denominator 2 minus log space straight v end fraction therefore space space space space space straight x dv over dx space equals space fraction numerator negative straight v plus straight v space log space straight v over denominator 2 minus space log space straight v end fraction$
$therefore space space space space fraction numerator 2 minus space log space straight v over denominator negative straight v plus straight v space log space straight v end fraction dv space equals space 1 over straight x dx therefore space space space space integral fraction numerator 1 minus log space left parenthesis straight v minus 1 right parenthesis over denominator straight v left parenthesis log space straight v minus 1 right parenthesis space end fraction dv space equals space integral 1 over straight x dx therefore space space space space integral fraction numerator 1 over denominator straight v left parenthesis log space straight v minus 1 right parenthesis end fraction dv space minus space integral 1 over straight v dv space equals space integral 1 over straight x dx therefore space space space integral fraction numerator begin display style 1 over straight v end style over denominator log space straight v minus 1 end fraction dv space minus space integral 1 over straight v dv space equals space integral 1 over straight x dx therefore space space space log space open vertical bar log space space straight v minus 1 close vertical bar space minus space log space open vertical bar straight v close vertical bar space equals space log space open vertical bar straight x close vertical bar space plus space log space open vertical bar straight c close vertical bar therefore space space space log space open vertical bar fraction numerator log space straight v minus 1 over denominator straight v end fraction close vertical bar space equals space log space left parenthesis open vertical bar cx close vertical bar right parenthesis therefore space space space space log space open vertical bar fraction numerator log space begin display style straight y over straight x end style minus 1 over denominator begin display style straight y over straight x end style end fraction close vertical bar space equals space log space left parenthesis open vertical bar straight c space straight x close vertical bar right parenthesis therefore space space space space space space fraction numerator log space begin display style straight y over straight x end style minus 1 over denominator begin display style straight y over straight x end style end fraction space equals straight c space straight x therefore space space space log space straight y over straight x minus 1 space equals space straight c space straight y space is space the space required space solution.$

Question 42

Wired Faculty App

Show that the given differential equation is homogeneous and solve it:
$left parenthesis straight x squared plus xy right parenthesis space dy space equals space left parenthesis straight x squared plus straight y squared right parenthesis space dx$

Solution

The given differential equation is

open parentheses straight x squared plus straight x space straight y close parentheses dy space equals space left parenthesis straight x squared plus straight y squared right parenthesis space dx

dy over dx space equals space fraction numerator straight x squared plus straight y squared over denominator straight x squared plus xy end fraction

...(1)
It is a differential equation of the form

dy over dx space equals space straight F left parenthesis straight x comma space straight y right parenthesis

Here,

straight F left parenthesis straight x comma space straight y right parenthesis space equals space fraction numerator straight x squared plus straight y squared over denominator straight x squared plus straight x space straight y end fraction

Replacing x by

λx space and space straight y space by space λy comma

we get

straight F left parenthesis λx comma space λy right parenthesis space equals space fraction numerator straight lambda squared straight x squared plus straight lambda squared straight y squared over denominator straight lambda squared straight x squared plus straight lambda squared xy end fraction space equals space fraction numerator straight lambda squared left parenthesis straight x squared plus straight y squared right parenthesis over denominator straight lambda squared left parenthesis straight x squared plus xy right parenthesis end fraction space equals space straight lambda degree space left square bracket straight F space left parenthesis straight x comma space straight y right parenthesis right square bracket

therefore space space space straight F left parenthesis straight x comma space straight y right parenthesis

is a homogeneous function of degree zero.

therefore space space space

given differential equation is a homogeneous differential equation.
Put y = v x so that

dy over dx space equals space straight v plus straight x dv over dx

therefore space space space from space left parenthesis 1 right parenthesis comma space space straight v plus straight x dv over dx space equals space fraction numerator straight x squared plus straight v squared straight x squared over denominator straight x squared plus vx squared end fraction rightwards double arrow space space space straight v plus straight x dv over dx space equals space fraction numerator 1 plus straight v squared over denominator 1 plus straight v end fraction space space space space rightwards double arrow space space space straight x dv over straight d space equals space fraction numerator 1 plus straight v squared over denominator 1 plus straight v end fraction minus straight v therefore space space space space straight x dv over dx space equals space fraction numerator 1 minus straight v over denominator 1 plus straight v end fraction space space space space rightwards double arrow space space space fraction numerator 1 plus straight v over denominator 1 minus straight v end fraction dv space equals space 1 over straight x dx space space space rightwards double arrow space space space space integral open parentheses negative 1 plus fraction numerator 2 over denominator 1 minus straight v end fraction close parentheses dv space equals space integral 1 over straight x dx therefore space space space space space minus straight v minus 2 space log space left parenthesis 1 minus straight v right parenthesis space equals space log space straight x space plus straight c rightwards double arrow space space space minus straight y over straight x minus 2 log space open parentheses 1 minus straight y over straight x close parentheses space equals space log space straight x space plus space straight c rightwards double arrow space space space space minus straight y minus 2 straight x space log space open parentheses fraction numerator straight x minus straight y over denominator straight x end fraction close parentheses space equals space straight x space log space straight x space plus space straight c space straight x

which is required solution.

Question 43

Wired Faculty App

Show that the given differential equation is homogeneous and solve it:
$straight y apostrophe space equals space fraction numerator straight x plus straight y over denominator straight x end fraction$

Solution

The given differential equation is

straight y apostrophe space equals space fraction numerator straight x plus straight y over denominator straight x end fraction space space or space space space dy over dx space equals space fraction numerator straight x plus straight y over denominator straight x end fraction

...(1)
It is a differential equation of the form

dy over dx space equals space straight F left parenthesis straight x comma space straight y right parenthesis

Here

straight F left parenthesis straight x comma space straight y right parenthesis space equals space fraction numerator straight x plus straight y over denominator straight x end fraction

Replacing x by

λx space and space straight y space by space λy comma

we get,

straight F left parenthesis λx comma space λy right parenthesis space equals space fraction numerator λx space plus space λy over denominator λx end fraction space equals space fraction numerator straight lambda left parenthesis straight x plus straight y right parenthesis over denominator λx end fraction space equals space straight lambda degree space left square bracket straight F left parenthesis straight x comma space straight y right parenthesis right square bracket

therefore space space space straight F left parenthesis straight x comma space straight y right parenthesis space is space straight a space

homogeneous function of degree zero.

therefore space space given

differential equation is a homogeneous differential equation.
Put y = v x,

therefore space space space dy over dx space equals space straight v. space 1 space plus space straight x dv over dx

therefore space space from space left parenthesis 1 right parenthesis comma space straight v plus straight x dv over dx space equals space fraction numerator straight x plus straight v space straight x over denominator straight x end fraction

rightwards double arrow space space straight v plus straight x dv over dx space equals space 1 plus space straight v space space space space space rightwards double arrow space space space straight x dv over dx space equals space 1 space space space space space space space space space space space rightwards double arrow space space space space dv space equals space space 1 over straight x dx

Integrating,

integral space dv space equals space integral 1 over straight x dx comma space space or space space space straight v space equals space log space open vertical bar straight x close vertical bar space plus space straight c

straight y over straight x space equals space space log space open vertical bar straight x close vertical bar space plus space straight c comma space space which space is space required space solution space. space

Question 44

Wired Faculty App

Show that the given differential equation is homogeneous and solve it:
(x-y) dy - (x+y) dx = 0

Solution

The given differential equation is
(x-y) dy - (x+y) dx = 0 or (x-y) dy = (x+y) dy
or

dy over dx space equals space fraction numerator straight x plus straight y over denominator straight x minus straight y end fraction

...(1)
It is a differential equation of the form

dy over dx space equals space straight F left parenthesis straight x comma space straight y right parenthesis

Here,

straight F left parenthesis straight x comma space straight y right parenthesis space equals space fraction numerator straight x plus straight y over denominator straight x minus straight y end fraction

Replacing x by

λx

and y by

λx comma space we space get comma

straight F left parenthesis λx comma space λy right parenthesis space equals space fraction numerator λx plus λy over denominator λx minus λy end fraction space equals space fraction numerator straight lambda left parenthesis straight x plus straight y right parenthesis over denominator straight lambda left parenthesis straight x minus straight y right parenthesis end fraction space equals space straight lambda degree space space left square bracket straight F left parenthesis straight x comma space straight y right parenthesis right square bracket

∴ F(x, y) is a homogeneous function of degree zero.
∴ given differential equation is a homogeneous differential equation.
Put y = v x so that $dy over dx space equals space straight v plus straight x dv over dx$
$therefore space space space from space left parenthesis 1 right parenthesis comma space space space straight v plus straight x dv over dx space equals space fraction numerator straight x plus vx over denominator straight x minus vx end fraction space space or space space space straight x dv over dx equals fraction numerator 1 plus straight v over denominator 1 minus straight v end fraction minus straight v therefore space space space straight x dv over dx space equals fraction numerator 1 plus straight v minus straight v plus straight v squared over denominator 1 minus straight v end fraction space space or space space space straight x dv over dx space equals space fraction numerator 1 plus straight v squared over denominator 1 minus straight v end fraction$
Separating the variables,    $fraction numerator 1 minus straight v over denominator 1 plus straight v squared end fraction dv space equals space 1 over straight x dx$
Integrating , $integral fraction numerator 1 minus straight v over denominator 1 plus straight v squared end fraction dv space equals space integral 1 over straight x dx$
or    $integral fraction numerator 1 over denominator 1 plus straight v squared end fraction dv space minus space 1 half integral fraction numerator 2 straight v over denominator 1 plus straight v squared end fraction dv space equals space integral 1 over straight x dx space space or space space tan to the power of negative 1 end exponent straight v minus 1 half log space left parenthesis 1 plus straight v squared right parenthesis space equals space log space open vertical bar straight x close vertical bar plus straight c$
or $tan to the power of negative 1 end exponent straight y over straight x minus 1 half log space open parentheses 1 plus straight y squared over straight x squared close parentheses space equals space log space open vertical bar straight x close vertical bar plus straight c$
or    $tan to the power of negative 1 end exponent straight y over straight x minus 1 half log space open parentheses fraction numerator straight x squared plus straight y squared over denominator straight x squared end fraction close parentheses space equals space log space open vertical bar straight x close vertical bar space plus space straight c space is space the space required space solution. space$

Question 45

Wired Faculty App

Show that the given differential equation is homogeneous and solve it:
$open parentheses straight x squared minus straight y squared close parentheses space dx space plus space 2 xy space dy space equals space 0$

Solution

The given differential equation is
$open parentheses straight x squared minus straight y squared close parentheses dx plus 2 xy space dy space equals space 0 space space or space space space 2 xy space dy space equals space left parenthesis straight y squared minus straight x squared right parenthesis space dx$
or $dy over dx space equals fraction numerator straight y squared minus straight x squared over denominator 2 xy end fraction$
It is a differential equation of the form $dy over dx space equals straight F left parenthesis straight x comma space straight y right parenthesis$
Here, $straight F left parenthesis straight x comma space straight y right parenthesis space equals space fraction numerator straight y squared minus straight x squared over denominator 2 space straight x space straight y end fraction$
Replacing x by $λx$ and y by $λy comma$ we get,
$straight F left parenthesis λx comma space λy right parenthesis space equals space fraction numerator straight lambda squared straight y squared minus straight lambda squared straight x squared over denominator 2 space straight lambda squared space straight x space straight y end fraction space equals space fraction numerator straight lambda squared space left parenthesis straight y squared minus straight x squared right parenthesis over denominator straight lambda squared space left parenthesis 2 xy right parenthesis end fraction space equals space straight lambda degree space space left square bracket straight F left parenthesis straight x comma space straight y right parenthesis right square bracket$
∴ F(x, y) is a homogeneous function of degree zero.
∴ given differential equation is a homogeneous differential equation.
Put y = vx so that $dy over dx equals straight v plus straight x dv over dx$
$therefore space space space space space straight v plus straight x dv over dx equals fraction numerator straight v squared straight x squared minus straight x squared over denominator 2 space straight v space straight x squared end fraction space space or space space space straight v plus straight x dv over dx space equals space fraction numerator straight v squared minus 1 over denominator 2 straight v end fraction therefore space space space space space space space space space straight x dv over dx space equals space fraction numerator straight v squared minus 1 over denominator 2 straight v end fraction minus straight v space space space or space space space straight x dv over dx space equals fraction numerator straight v squared minus 1 minus 2 straight v squared over denominator 2 straight v end fraction therefore space space space space space space straight x dv over dx space equals fraction numerator negative 1 minus straight v squared over denominator 2 straight v end fraction space space space space rightwards double arrow space space space space space space fraction numerator 2 straight v over denominator 1 plus straight v squared end fraction dv space equals space minus 1 over straight x space dx therefore space space space space space integral fraction numerator 2 straight v over denominator 1 plus straight v squared end fraction dv space equals space minus integral 1 over straight x dx therefore space space space space space log space open vertical bar 1 plus straight v squared close vertical bar space equals space minus log space open vertical bar straight x close vertical bar plus straight c apostrophe therefore space space space space log space open vertical bar 1 plus straight v squared close vertical bar space plus space log space open vertical bar straight x close vertical bar space equals space space straight c apostrophe therefore space space log space open vertical bar left parenthesis 1 plus straight v squared right parenthesis thin space left parenthesis straight x right parenthesis close vertical bar space equals space straight c apostrophe therefore space space space space space space space space straight x left parenthesis 1 plus straight v squared right parenthesis space equals space straight e to the power of straight c apostrophe end exponent space space space space space space space space rightwards double arrow space space space space space space space space straight x open parentheses 1 plus straight y squared over straight x squared close parentheses space equals space straight c therefore space space space space straight x squared plus straight y squared space equals space straight c space straight x space$
is the required solution.

Question 46

Wired Faculty App

Show that the given differential equation is homogeneous and solve it:
$straight x squared dy over dx space equals space straight x squared minus 2 straight y squared plus straight x space straight y$

Solution

The given differential equation is

space straight x squared dy over dx space equals space straight x squared minus 2 straight y squared plus xy

dy over dx space equals fraction numerator straight x squared minus 2 straight y squared plus straight x space straight y over denominator straight x squared end fraction

...(1)
It is a differential equation of the form

dy over dx space equals space straight F left parenthesis straight x comma space straight y right parenthesis

Here,

straight F left parenthesis straight x comma space straight y right parenthesis space equals space fraction numerator straight x squared minus 2 straight y squared plus straight x space straight y over denominator straight x squared end fraction

Replacing x by

λx

and y by

λy

, we get.

straight F left parenthesis λx comma space λy right parenthesis space equals space fraction numerator straight lambda squared straight x squared minus 2 straight lambda squared straight y squared plus straight lambda squared straight x space straight y over denominator straight lambda squared straight x squared end fraction space equals space fraction numerator straight lambda squared left parenthesis straight x squared minus 2 straight y squared plus xy right parenthesis over denominator straight lambda squared straight x squared end fraction space equals space straight lambda degree space left square bracket straight F left parenthesis straight x comma space straight y right parenthesis right square bracket

∴ F(x, y) is a homogeneous function of degree zero.
∴ given differential equation is a homogeneous differential equation.
Put y = vx $rightwards double arrow space space dy over dx space equals space straight v. space 1 space space plus straight x space dv over dx$
Substituting these values of y and $dy over dx$ in the given equation, we get
$straight v plus straight x dv over dx space equals space fraction numerator straight x squared minus 2 straight v squared straight x squared plus vx squared over denominator straight x squared end fraction$
or $straight v plus straight x dv over dx equals 1 minus 2 straight v squared plus straight v space space space space space rightwards double arrow space space space space space straight x dv over dx equals 1 minus 2 straight v squared$
$rightwards double arrow space space space space space space space space 1 over straight x dx space equals space fraction numerator 1 over denominator 1 minus 2 straight v squared end fraction dv$
Integrating, $integral 1 over straight x dx space equals space integral fraction numerator 1 over denominator 1 minus 2 straight v squared end fraction dv$
$therefore space space space space space integral 1 over straight x dx space equals space 1 half integral fraction numerator dv over denominator open parentheses begin display style fraction numerator 1 over denominator square root of 2 end fraction end style close parentheses squared minus straight v squared end fraction therefore space space space space space log space open vertical bar straight x close vertical bar space space equals 1 half. space fraction numerator 1 over denominator 2. space begin display style fraction numerator 1 over denominator square root of 2 end fraction end style end fraction log space open vertical bar fraction numerator begin display style fraction numerator 1 over denominator square root of 2 end fraction end style plus straight v over denominator begin display style fraction numerator 1 over denominator square root of 2 end fraction end style minus straight v end fraction close vertical bar plus straight c rightwards double arrow space space space space space space space log space open vertical bar straight x close vertical bar space equals space fraction numerator 1 over denominator 2 square root of 2 end fraction space log space space open vertical bar fraction numerator 1 plus square root of 2 space straight v over denominator 1 minus square root of 2 straight v end fraction close vertical bar plus straight c rightwards double arrow space space space log space open vertical bar straight x close vertical bar space equals space fraction numerator 1 over denominator 2 square root of 2 end fraction log space open vertical bar fraction numerator 1 plus square root of 2. space begin display style straight y over straight x end style over denominator 1 minus square root of 2. begin display style straight y over straight x end style end fraction close vertical bar plus straight c rightwards double arrow space space space space log space open vertical bar straight x close vertical bar space equals space fraction numerator 1 over denominator 2 square root of 2 end fraction log space open vertical bar fraction numerator straight x plus square root of 2 straight y over denominator straight x minus square root of 2 straight y end fraction close vertical bar plus straight c$

Question 47

Wired Faculty App

Show that the given differential equation is homogeneous and solve it:
$open curly brackets straight x space cos space open parentheses straight y over straight x close parentheses plus straight y space sin space open parentheses straight y over straight x close parentheses close curly brackets straight y space dx space equals space open curly brackets straight y space sin space open parentheses straight y over straight x close parentheses minus straight x space cos space open parentheses straight y over straight x close parentheses close curly brackets space straight x space dy$

Solution

The given differential equation is
$open curly brackets straight x space cos space open parentheses straight y over straight x close parentheses space plus space straight y space sin space open parentheses straight y over straight x close parentheses close curly brackets straight y space dy space equals space open curly brackets straight y space sin space open parentheses straight y over straight x close parentheses minus straight x space cos space open parentheses straight y over straight x close parentheses close curly brackets straight x space dy$
or $dy over dx space equals space fraction numerator straight x space straight y space open curly brackets cos space begin display style straight y over straight x end style plus begin display style straight y over straight x end style sin space begin display style straight y over straight x end style close curly brackets over denominator straight x squared open curly brackets begin display style straight y over straight x end style sin begin display style straight y over straight x end style minus cos space begin display style straight y over straight x end style close curly brackets end fraction$
or    $dy over dx space equals space fraction numerator begin display style straight y over straight x end style open curly brackets cos space begin display style straight y over straight x end style plus begin display style straight y over straight x end style sin begin display style straight y over straight x end style close curly brackets over denominator begin display style straight y over straight x end style sin space begin display style straight y over straight x end style minus cos begin display style straight y over straight x end style end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis$
Since R.H.S. of (1) is of the form $straight F open parentheses straight y over straight x close parentheses comma$ therefore, it is a homogeneous differential equation.
Put y= v x so that $dy over dx space equals space straight v plus straight x space dv over dx$
$therefore space space space space left parenthesis 1 right parenthesis space space becomes comma space space space straight v plus straight x dv over dx space equals space fraction numerator straight v left parenthesis cos space straight v space plus space straight v space sin space straight v right parenthesis space over denominator straight v space sin space straight v space minus space cos space straight v end fraction$
or    $straight x dv over dx space equals space fraction numerator straight v space cosv plus straight v squared space sin space straight v over denominator straight v space sin space straight v space minus space cos space straight v end fraction minus straight v$
or $straight x dv over dx space equals space fraction numerator 2 space straight v space cos space straight v over denominator straight v space sin space straight v space minus space cos space straight v end fraction$
Separating the variables,
   $open parentheses fraction numerator straight v space sin space straight v space minus space cos space straight v over denominator straight v space cos space straight v end fraction close parentheses space dv space equals space 2 over straight x dx space space or space space space open parentheses tan space straight v minus space 1 over straight v close parentheses dv space equals space 2 over straight x dx$
Integrating, $integral open parentheses tan space straight v minus 1 over straight v close parentheses space dv space equals space 2 space integral 1 over straight x dx$
$therefore space space space minus log space open vertical bar cos space straight v close vertical bar minus log space open vertical bar straight v close vertical bar space equals space 2 space log space open vertical bar straight x close vertical bar plus straight c rightwards double arrow space space space space log space open vertical bar straight v space cos space straight v close vertical bar space plus space 2 space log open vertical bar straight x close vertical bar space equals space minus straight c rightwards double arrow space space space log space left parenthesis open vertical bar straight v space cos space straight v close vertical bar space open vertical bar straight x squared close vertical bar right parenthesis space equals space minus straight c rightwards double arrow space space space space space open vertical bar straight v space cos space straight v close vertical bar space open vertical bar straight x close vertical bar squared space equals space straight e to the power of negative straight c end exponent rightwards double arrow space space straight x squared straight v space cos space straight v space equals space plus-or-minus straight e to the power of negative straight c end exponent rightwards double arrow space space space space straight x squared straight y over straight x cos space open parentheses straight y over straight x close parentheses space equals straight A comma space space where space straight A space space equals space plus-or-minus space straight e to the power of negative straight c end exponent rightwards double arrow space space space straight x space straight y space cos space straight y over straight x space equals straight A comma space which space is space the space required space general space solution. space$

Question 48

Wired Faculty App

Solve $straight x space dy space minus space straight y space dx space equals space square root of straight x squared plus straight y squared end root space dx.$

Solution

The given differential equation is

straight x space dy minus space straight y space dx space equals space square root of straight x squared plus straight y squared end root space dx

straight x space dy space equals space open parentheses straight y plus square root of straight x squared plus straight y squared end root close parentheses dx

dy over dx space equals space fraction numerator straight y plus square root of straight x squared plus straight y squared end root over denominator straight x end fraction

Put y = v x so that

dy over dx space equals space straight v plus straight x dv over dx

therefore space space space space left parenthesis 1 right parenthesis space becomes comma space space space space space straight v plus straight x dv over dx space equals space fraction numerator vx plus square root of straight x squared plus straight v squared straight x squared end root over denominator straight x end fraction

straight x dv over dx space equals straight v plus square root of 1 plus straight v squared end root minus straight v comma space space space or space space space space straight x dv over dx space equals space square root of 1 plus straight v squared end root

Separating the variables,

fraction numerator 1 over denominator square root of 1 plus straight v squared end root end fraction dv space equals space 1 over straight x dx

Integrating

integral fraction numerator 1 over denominator square root of 1 plus straight v squared end root end fraction dv space equals space integral 1 over straight x dx space space space space space space space space space space space space space space space space rightwards double arrow space space space space log space open vertical bar straight v plus square root of 1 plus straight v squared end root close vertical bar space equals space log space open vertical bar straight x close vertical bar space plus space log space straight c

open vertical bar straight v plus square root of 1 plus straight v squared end root close vertical bar space equals space straight c open vertical bar straight x close vertical bar space space space space space space space space space space space space space space space space space space space space space space space space space space space or space space space space space open vertical bar straight y over straight x plus square root of 1 plus straight y squared over straight x squared end root close vertical bar space equals space straight c open vertical bar straight x close vertical bar

open vertical bar straight y plus square root of straight x squared plus straight y squared end root space close vertical bar space equals space straight c open vertical bar straight x close vertical bar squared comma space space or space space space space open vertical bar straight y plus square root of straight x squared plus straight y squared end root close vertical bar space equals space straight c space straight x squared

therefore space space straight y plus square root of straight x squared plus straight y squared end root space equals space plus-or-minus space straight c space straight x squared space space space or space space space straight y plus square root of straight x squared plus straight y squared end root space equals space straight A space straight x squared space space where space straight A space equals space plus-or-minus straight c

which is required solution.

Question 49

Wired Faculty App

Find a one parameter family of solutions of each of the following differential equation:
y² + x² y' = x y y'

Solution

The given differential equation is

straight y squared plus straight x squared dy over dx equals space straight x. space straight y space dy over dx

left parenthesis straight x space straight y space minus space straight x squared right parenthesis space dy over dx space equals straight y squared

therefore space space space space space space space space space dy over dx space equals space fraction numerator straight y squared over denominator xy minus straight y squared end fraction

Put

straight y space equals space straight v space straight x space so space that space dy over dx space equals straight v plus straight x dv over dx

therefore space space space space straight v plus straight x dv over dx equals space fraction numerator straight v squared minus straight x squared over denominator straight x squared straight v minus straight x squared end fraction therefore space space straight v plus straight x dv over dx space equals space fraction numerator straight v squared over denominator straight v minus 1 end fraction therefore space space space space space space space space straight x dv over dx space equals fraction numerator straight v squared over denominator straight v minus 1 end fraction minus straight v therefore space space space space space space straight x dv over dx space equals space fraction numerator straight v over denominator straight v minus 1 end fraction therefore space space space space space fraction numerator straight v minus 1 over denominator straight v end fraction dv space equals space 1 over straight x dx therefore space space space space integral open parentheses 1 minus 1 over straight v close parentheses space dv space equals space integral 1 over straight x dx therefore space space space space space straight v minus log space straight v space equals space logx plus log space straight c therefore space space space space straight y over straight x minus log space straight y over straight x space equals space log space straight x space plus space log space straight c therefore space space space space straight y over straight x minus log space straight y space equals space log space straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space space straight y over straight x space equals space log space straight c space straight y therefore space space space space space space space space space straight c space straight y space equals space straight e to the power of straight y over straight x end exponent space is space the space required space solution. space

Question 50

Wired Faculty App

Find a one parameter family of solutions of each of the following differential equation:
(y² – 2xy) dx = (x² – 2xy) dy

Solution

The given differential equation is
(y² – 2xy) dx = (x² – 2xy) dy
or

dy over dx space equals fraction numerator straight y squared minus 2 xy over denominator straight x squared minus 2 xy end fraction

Put y = v x so that

dy over dx space equals space straight v plus straight x dv over dx

therefore space space space space space space space space straight v plus straight x dv over dx space equals space fraction numerator straight v squared straight x squared minus 2 vx squared over denominator straight x squared minus 2 vx squared end fraction therefore space space space space space space space straight v plus straight x dv over dx space equals fraction numerator straight v squared minus 2 straight v over denominator 1 minus 2 straight v end fraction therefore space space space space space space space space space space straight x dv over dx space equals space fraction numerator straight v squared minus 2 straight v over denominator 1 minus 2 straight v end fraction minus straight v space equals space fraction numerator straight v squared minus 2 straight v minus straight v plus 2 straight v squared over denominator 1 minus 2 straight v end fraction therefore space space space space space space space straight x dv over dx space equals space fraction numerator 3 straight v squared minus 3 straight v over denominator 1 minus 2 straight v end fraction

Separating the variables and integrating, we get,

integral fraction numerator 1 minus 2 straight v over denominator 3 straight v squared minus 3 straight v end fraction equals space space integral 1 over straight x dx space space or space space 1 third integral fraction numerator 1 minus 2 straight v over denominator straight v squared minus straight v end fraction dv space equals integral 1 over straight x dx

therefore space space space space minus 1 third space log space left parenthesis straight v squared minus straight v right parenthesis space equals space log space straight x plus space log space straight c therefore space space space space log space left parenthesis straight v squared minus straight v right parenthesis to the power of fraction numerator negative 1 over denominator 3 end fraction end exponent space equals space log space cx space therefore space space space space space space space space space space space left parenthesis straight v squared minus straight v right parenthesis to the power of negative 1 third end exponent space equals space straight c space straight x therefore space space space space space space space open parentheses straight y squared over straight x squared minus straight y over straight x close parentheses to the power of negative 1 third end exponent space equals space space straight c space straight x therefore space space space space space space open parentheses straight y squared minus straight x space straight y close parentheses to the power of negative 1 third end exponent space equals space cx space cross times space straight x to the power of fraction numerator negative 2 over denominator 3 end fraction end exponent therefore space space space space space space space space left parenthesis straight y squared minus straight x space straight y right parenthesis to the power of fraction numerator negative 1 over denominator 3 end fraction end exponent space equals space straight c space straight x to the power of 1 third end exponent therefore space space space space space space space space space left parenthesis straight y squared minus straight x space straight y right parenthesis to the power of negative 1 end exponent space equals space Ax therefore space space space space space space fraction numerator 1 over denominator straight y squared minus xy end fraction space equals space Ax or space space space xy squared minus straight x squared straight y space equals space straight c comma space which space is space required space solution. space

Question 51

Wired Faculty App

Find a one parameter family of solutions of each of the following differential equation:
y² dx + (x² – x y + y²) dy = 0

Solution

The given differential equation is
y² dx + (x² – x y + y²) dy = 0 or (x² – x y + y²) dy = - y² dx

therefore space space space space space space space space space space space space space space space space space space space space dy over dx equals negative fraction numerator straight y squared over denominator straight x squared minus xy plus straight y squared end fraction

Put y = vx so that

dy over dx equals straight v plus straight x dv over dx

therefore space space space space space space straight v plus straight x dv over dx equals negative fraction numerator straight v squared straight x squared over denominator straight x squared minus vx squared plus straight v squared straight x squared end fraction therefore space space space space space straight v plus straight x dv over dx equals negative fraction numerator straight v squared over denominator 1 minus straight v plus straight v squared end fraction therefore space space space space space space straight x dv over dx space equals space minus fraction numerator straight v squared over denominator 1 minus straight v plus straight v squared end fraction minus straight v therefore space space space space space straight x dv over dx equals fraction numerator negative straight v squared minus straight v plus straight v squared minus straight v cubed over denominator 1 minus straight v plus straight v squared end fraction therefore space space space space space space straight x dv over dx equals fraction numerator negative straight v minus straight v cubed over denominator 1 minus straight v plus straight v squared end fraction therefore space space fraction numerator 1 minus straight v plus straight v squared over denominator straight v plus straight v cubed end fraction dv space equals space minus 1 over straight x dx therefore space space space space space integral fraction numerator 1 minus straight v plus straight v squared over denominator straight v left parenthesis 1 plus straight v squared right parenthesis end fraction space identical to space straight A over straight v plus fraction numerator Bv plus straight c over denominator 1 plus straight v squared end fraction

Multiplying both sides by v (1 + v²), we get.

1 minus straight v plus straight v squared space equals space straight A thin space left parenthesis 1 plus straight v squared right parenthesis space plus space Bv squared plus Cv

...(1)
Put v = 0 in (1)

therefore space space space space space space 1 minus 0 plus 0 space equals space straight A space left parenthesis 1 plus 0 right parenthesis space plus space 0 space plus space 0. space space space space space space rightwards double arrow space space space straight A space equals space 1

Equating coefficients in (1) of

straight v squared right parenthesis

1 = A + B

rightwards double arrow 1 space equals 1 space plus space straight B space space space space space space space space space space space space space space space space rightwards double arrow space space space straight B space equals space 0

v) -1 = C

rightwards double arrow space straight C space equals space minus 1

therefore space space space space space fraction numerator 1 minus straight v plus straight v squared over denominator straight v left parenthesis 1 plus straight v squared right parenthesis end fraction identical to 1 over straight v plus fraction numerator negative 1 over denominator 1 plus straight v squared end fraction

therefore space space space space space from space left parenthesis 1 right parenthesis comma space we space get

integral open parentheses 1 over straight v minus fraction numerator 1 over denominator 1 plus straight v squared end fraction close parentheses space dv space equals space minus integral 1 over straight x dx

therefore space space log space open vertical bar straight v close vertical bar space minus space tan to the power of negative 1 end exponent straight v space equals space space minus space log space open vertical bar straight x close vertical bar space plus space straight c therefore space space space log space open vertical bar straight y over straight x close vertical bar space minus space tan to the power of negative 1 end exponent straight y over straight x space equals space minus log space open vertical bar straight x close vertical bar space plus space log space straight A therefore space space space log space open vertical bar straight y close vertical bar space minus space log space open vertical bar straight x close vertical bar space minus space tan to the power of negative 1 end exponent straight y over straight x space equals space minus log space open vertical bar straight x close vertical bar plus space space log space straight A therefore space space space log space open vertical bar straight y close vertical bar minus log space straight A space equals space space tan to the power of negative 1 end exponent straight y over straight x therefore space space space log space open vertical bar fraction numerator open vertical bar straight y close vertical bar over denominator straight A end fraction close vertical bar space equals tan to the power of negative 1 end exponent straight y over straight x space space space space rightwards double arrow space space space space fraction numerator open vertical bar straight y close vertical bar over denominator straight A end fraction space equals space straight e to the power of tan to the power of negative 1 end exponent straight y over straight x end exponent therefore space space space space space space space space space space space space space space space space space space open vertical bar straight y close vertical bar space equals space Ae to the power of tan to the power of negative 1 end exponent straight y over straight x end exponent therefore space space space space space space space space space space space space space space space space space space straight y squared space equals space space straight c space straight e space to the power of 2 space tan to the power of negative 1 end exponent straight y over straight x end exponent

is the required solution.

Question 52

Wired Faculty App

Find a one parameter family of solutions of each of the following differential equation:
(x² + x y) dy = (x² + y²) dx

Solution

The given differential equation is
(x² + x y) dy = (x² + y²) dx
or

dy over dx space equals space fraction numerator straight x squared plus straight y squared over denominator straight x squared plus xy end fraction

...(1)
Put y = v x so that

dy over dx equals straight v plus straight x dv over dx

therefore space space space from space left parenthesis 1 right parenthesis comma space space space space straight v plus straight x dv over dx equals fraction numerator straight x squared plus straight v squared space straight x squared over denominator straight x squared plus straight v space straight x squared end fraction rightwards double arrow space space space space straight v plus straight x dv over dx equals fraction numerator 1 plus straight v squared over denominator 1 plus straight v end fraction space space space space space space space rightwards double arrow space space space straight x dv over dx space equals fraction numerator 1 plus straight v squared over denominator 1 plus straight v end fraction minus straight v therefore space space space space straight x dv over dx space equals space fraction numerator 1 minus straight v over denominator 1 plus straight v end fraction space space space space rightwards double arrow space space space space fraction numerator 1 plus straight v over denominator 1 minus straight v end fraction dv space equals space 1 over straight x dx space space space space rightwards double arrow space space integral open parentheses negative 1 plus fraction numerator 2 over denominator 1 minus straight v end fraction close parentheses dv space equals integral 1 over straight x dx therefore space space minus straight v minus 2 space log space left parenthesis 1 minus straight v right parenthesis space equals space log space straight x space plus straight c space space space space rightwards double arrow space space space space space minus straight y over straight x minus 2 space log space open parentheses 1 minus straight y over straight x close parentheses space equals space log space straight x plus straight c therefore space space space space space straight y minus 2 straight x space log space open parentheses fraction numerator straight x minus straight y over denominator straight x end fraction close parentheses space equals space straight x space logx plus space cx

which is required solution.

Question 53

Wired Faculty App

Solve the following differential equation:
$open parentheses 1 plus straight e to the power of straight x over straight y end exponent close parentheses space dx space plus space straight e to the power of straight x over straight y end exponent space open parentheses 1 minus straight x over straight y close parentheses dy space equals space 0$ .

Solution

The given differential equation is

open parentheses 1 plus straight e to the power of straight x over straight y end exponent close parentheses space dx space plus space straight e to the power of straight x over straight y end exponent open parentheses 1 minus straight x over straight y close parentheses dy space equals space 0 space space space space space or space space space dy over dx equals fraction numerator 1 plus straight e to the power of begin display style straight x over straight y end style end exponent over denominator straight e to the power of begin display style straight x over straight y end style end exponent open parentheses begin display style straight x over straight y minus 1 end style close parentheses end fraction space space space space space space space... left parenthesis 1 right parenthesis

Put

straight x space equals space straight v space straight y space space space space or space space dx over dy equals straight v plus straight y dv over dy

therefore space space from space left parenthesis 1 right parenthesis comma space space fraction numerator 1 over denominator straight v plus straight y begin display style dv over dy end style end fraction equals space fraction numerator 1 plus straight e to the power of straight v over denominator straight e to the power of straight v left parenthesis straight v minus 1 right parenthesis end fraction space space space or space space space space straight v plus straight y dv over dy space equals fraction numerator straight e to the power of straight v left parenthesis straight v minus 1 right parenthesis over denominator 1 plus straight e to the power of straight v end fraction

straight y dv over dy space equals fraction numerator straight e to the power of straight v left parenthesis straight v minus 1 right parenthesis over denominator 1 plus straight e to the power of straight v end fraction minus straight v space space equals space fraction numerator ve to the power of straight v minus straight e to the power of straight v minus straight v minus straight v space straight e to the power of straight v over denominator 1 plus straight e to the power of straight v end fraction

straight y dv over dy equals space fraction numerator negative left parenthesis straight e to the power of straight v plus straight v right parenthesis over denominator 1 plus straight e to the power of straight v end fraction space space space or space space space fraction numerator 1 plus straight e to the power of straight v over denominator straight v plus straight e to the power of straight v end fraction dv space equals space minus dy over straight y

integral fraction numerator 1 plus straight e to the power of straight v over denominator straight v plus straight e to the power of straight v end fraction dv space equals space minus integral 1 over straight y dy space space space space space or space space space space log space open vertical bar straight v plus straight e to the power of straight v close vertical bar space equals space minus log space open vertical bar straight y close vertical bar space plus space log space open vertical bar straight c close vertical bar

therefore space space log space open vertical bar straight v plus straight e to the power of straight v close vertical bar space equals space log space open vertical bar straight c over straight y close vertical bar space space space or space space left parenthesis straight v plus straight e to the power of straight v right parenthesis space straight y space equals space straight c space space space space or space space space space open parentheses straight x over straight y plus straight e to the power of straight x over straight y end exponent close parentheses straight y space equals space straight c

straight x plus straight y space straight e to the power of straight x over straight y end exponent equals space straight c space is space the space required space solution. space

Question 54

Wired Faculty App

Solve $open parentheses straight x space sin space straight y over straight x close parentheses dy space equals open parentheses straight y space sin space straight y over straight x minus straight x close parentheses dx$

Solution

The given differential equation is

open parentheses straight x space sin space straight y over straight x close parentheses dy space equals space open parentheses ysin straight y over straight x minus straight x close parentheses dx space space space space space space or space space space space dy over dx equals fraction numerator ysin begin display style straight y over straight x end style minus straight x over denominator straight x space sin begin display style straight y over straight x end style end fraction space space space space space... left parenthesis 1 right parenthesis

Put y = v x so that

dy over dx equals straight v plus straight x dv over dx

therefore space space left parenthesis 1 right parenthesis space becomes comma space space straight v plus straight x dv over dx equals fraction numerator straight v space straight x space sin space straight v space minus space straight x over denominator straight x space sin space straight v end fraction

straight v plus straight x dv over dx equals fraction numerator straight v space sinv minus 1 over denominator sin space straight v end fraction space space space or space space space straight x dv over dx equals fraction numerator straight v space sinv minus 1 over denominator sin space straight v end fraction minus straight v space space space or space space straight x dv over dx equals negative fraction numerator 1 over denominator sin space straight v end fraction

Separating the variables,

sin space straight v space dv space equals space minus 1 over straight x dx

Integrating,

negative cos space straight v space equals space minus log space open vertical bar straight x close vertical bar space minus space straight c space space space or space cos space open parentheses straight y over straight x close parentheses space equals space log space open vertical bar straight x close vertical bar space plus straight c

which is required solution.

Question 55

Wired Faculty App

Show that the differential equation $straight x space cos space open parentheses straight y over straight x close parentheses dy over dx space equals straight y space cos space open parentheses straight y over straight x close parentheses plus straight x$ is homogeneous and solve it.

Solution

The given differential equation is

space straight x space cos space open parentheses straight y over straight x close parentheses space dy over dx space equals space straight y space cos space open parentheses straight y over straight x close parentheses plus space straight x space space space or space space space space dy over dx space equals fraction numerator straight y space cos space open parentheses begin display style straight y over straight x end style close parentheses plus straight x over denominator straight x space cos space open parentheses begin display style straight y over straight x end style close parentheses end fraction

...(1)
It is differential equation of the form

dy over dx space equals space straight F left parenthesis straight x comma space straight y right parenthesis.

Here,

straight F left parenthesis straight x comma space straight y right parenthesis space equals space fraction numerator straight y space cos space open parentheses begin display style straight y over straight x end style close parentheses plus straight x over denominator straight x space cos space open parentheses begin display style straight y over straight x end style close parentheses end fraction

Replacing x by

straight lambda space straight x

and y by

λy comma

we get

space straight F left parenthesis λx comma space λy right parenthesis space equals space fraction numerator straight lambda space open square brackets straight y space cos space open parentheses begin display style straight y over straight x end style close parentheses plus straight x close square brackets over denominator straight lambda space open parentheses straight x space cos space begin display style straight y over straight x end style close parentheses end fraction space equals space straight lambda degree space space left square bracket straight F space left parenthesis straight x comma space straight y right parenthesis right square bracket

therefore space space space straight F left parenthesis straight x comma space straight y right parenthesis

is a homogeneous function of degree zero.

therefore space

the given differential equation is a homogeneous differential equation
Put y = vx that

dy over dx equals straight v plus straight x dv over dx

therefore space space space from space left parenthesis 1 right parenthesis comma space space straight v plus straight x dv over dx space equals space fraction numerator straight v space straight x space cos space open parentheses begin display style vx over straight x end style close parentheses plus straight x over denominator straight x space cos space open parentheses begin display style vx over straight x end style close parentheses end fraction

therefore space space space straight v plus straight x dv over dx space equals space fraction numerator straight v space straight x space cos space straight v space plus space straight x over denominator straight x space cos space straight v end fraction

straight v plus straight x dv over dx space equals fraction numerator straight v space cos space straight v plus 1 over denominator cos space straight v end fraction space space or space space straight x dv over dx space equals space fraction numerator straight v space cos space straight v space plus space 1 over denominator cos space straight v end fraction minus straight v

therefore space space space space space space space space straight x dv over dx space equals space fraction numerator straight v space cos space straight v plus 1 minus straight v space cos space straight v over denominator cos space straight v end fraction therefore space space space space straight x dv over dx equals fraction numerator 1 over denominator cos space straight v end fraction

Separating the variables, we get,

cosv space dv space equals space 1 over straight x dx

Integrating,

integral space cos space straight v space dv space equals space integral 1 over straight x dx

therefore

sin space straight v space equals space log space open vertical bar straight x close vertical bar space plus space log space open vertical bar straight c close vertical bar

sin space straight y over straight x space equals space log space open vertical bar straight c space straight x close vertical bar space which space is space required space solution. space

Question 56

Wired Faculty App

Show that the differential equation:
$left parenthesis straight x space dy space minus space straight y space dx right parenthesis space straight y space sin space open parentheses straight y over straight x close parentheses space equals space left parenthesis straight y space dx space plus space straight x space dy right parenthesis space straight x space cos space open parentheses straight y over straight x close parentheses$

Solution

The given differential equation is

left parenthesis straight x space dy space minus space straight y space dx right parenthesis space straight y space sin space straight y over straight x space equals space left parenthesis straight y space dx plus space straight x space dy right parenthesis straight x space cos straight y over straight x

straight x space straight y space sin space straight y over straight x space dy space minus space straight y squared space sin space straight y over straight x dx space equals space xy space sin space straight y over straight x dx space plus space straight x squared space cos space straight y over straight x dy

open square brackets straight x space straight y space sin space open parentheses straight y over straight x close parentheses minus space straight x squared space cos space open parentheses straight y over straight x close parentheses close square brackets dy space equals space open square brackets straight x space straight y space cos space open parentheses straight y over straight x close parentheses plus straight y squared space sin space open parentheses straight y over straight x close parentheses close square brackets dx

dy over dx space equals fraction numerator straight x space straight y space cos space open parentheses begin display style straight y over straight x end style close parentheses plus straight y squared space sin space open parentheses begin display style straight y over straight x end style close parentheses over denominator space straight x space straight y space sin space open parentheses begin display style straight y over straight x end style close parentheses minus straight x squared space cos space open parentheses begin display style straight y over straight x end style close parentheses end fraction

Dividing numerator and denominator on R.H.S. by

straight x squared comma

we get

dy over dx space equals space fraction numerator begin display style straight y over straight x end style cos space open parentheses begin display style straight y over straight x end style close parentheses plus open parentheses begin display style straight y squared over straight x squared end style close parentheses space sin space open parentheses begin display style straight y over straight x end style close parentheses over denominator begin display style straight y over straight x end style sin space open parentheses begin display style straight y over straight x end style close parentheses minus cos space open parentheses begin display style straight y over straight x end style close parentheses end fraction space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis

which is a homogeneous differential equation of the form

dy over dx space equals space straight F open parentheses straight y over straight x close parentheses.

Put y = v x so that

dy over dx space equals space straight v plus space straight x dv over dx

therefore space space space from space left parenthesis 1 right parenthesis comma space space straight v plus straight x dv over dx space equals space fraction numerator straight v space cos space straight v plus space straight v squared space sin space straight v over denominator straight v space sin space straight v minus cos space straight v end fraction or space space space space space space space straight x dv over dx equals space fraction numerator straight v space cos space straight v plus space straight v squared space sinx over denominator straight v space sin space straight v space minus space cos space straight v end fraction minus straight v or space space space space space straight x dv over dx space equals space fraction numerator straight v space cos space straight v plus space straight v squared space sin space straight v space minus straight v squared space sin space space straight v space plus space straight v space cos space straight v over denominator straight v space sin space straight v minus space cos space straight v end fraction or space space space space straight x dv over dx space equals space fraction numerator 2 straight v space cos space straight v over denominator straight v space sin space straight v space minus space cos space straight v end fraction

Separating the variables, we get,

open parentheses fraction numerator straight v space sin space straight v space minus space cos space straight v over denominator straight v space cos space straight v end fraction close parentheses dv space equals space fraction numerator 2 dx over denominator straight x end fraction

Integrating,

integral open parentheses fraction numerator straight v space sin space straight v space minus space cos space straight v over denominator straight v space cos space straight v end fraction close parentheses space equals space 2 space integral 1 over straight x dx

integral tan space straight v space dv space minus space integral 1 over straight v dv space equals space 2 space integral 1 over straight x dx

log space open vertical bar sec space straight v close vertical bar space minus space log space open vertical bar straight v close vertical bar space equals space 2 space log space open vertical bar straight x close vertical bar plus log space open vertical bar straight c subscript 1 close vertical bar

log space open vertical bar sec space straight v close vertical bar space minus space log space open vertical bar straight v close vertical bar space minus space log space open vertical bar straight x close vertical bar squared space equals space log space open vertical bar straight c subscript 1 close vertical bar

log space open vertical bar fraction numerator sec space straight v over denominator straight v space straight x squared end fraction close vertical bar space equals space log space open vertical bar straight c subscript 1 close vertical bar

fraction numerator sec space straight v over denominator straight v space straight x squared end fraction space equals plus-or-minus straight c subscript 1

...(2)
Now,

straight y space equals space straight v space straight x space space space or space space straight v space equals space straight y over straight x

therefore space space space space space space space fraction numerator sec open parentheses begin display style straight y over straight x end style close parentheses over denominator open parentheses begin display style straight y over straight x end style close parentheses space left parenthesis straight x squared right parenthesis end fraction space equals space straight c comma space space space space where space straight c space equals space plus-or-minus space straight c subscript 1

sec space open parentheses straight y over straight x close parentheses space equals space straight c space straight x space straight y

which is the general solution of the given differential equation.

Question 57

Wired Faculty App

Solve:
$straight y space straight e to the power of straight x over straight y end exponent dx space equals open parentheses xe to the power of straight x over straight y end exponent plus straight y squared close parentheses space space dy comma space space space straight y not equal to 0$

Solution

The given differential equation is
   $straight y space straight e to the power of straight x over straight y end exponent space dx space equals space open parentheses xe to the power of straight x over straight y end exponent plus straight y squared close parentheses space dy$
or $dx over dy space equals space fraction numerator xe to the power of begin display style straight x over straight y end style end exponent plus straight y squared over denominator ye to the power of begin display style straight x over straight y end style end exponent end fraction$
or $dx over dy space equals fraction numerator xe to the power of begin display style straight x over straight y end style end exponent plus straight y squared over denominator ye to the power of begin display style straight x over straight y end style end exponent end fraction$
or $dx over dy space equals space straight x over straight y plus straight y over straight e to the power of begin display style straight x over straight y end style end exponent$
Put y = v y so that $dx over dy equals straight v plus straight y dv over dx$
$therefore space space space space space space space space space straight v plus straight y dv over dy space equals space straight v plus ye to the power of negative straight v end exponent therefore space space space space space space space space straight y dv over dy space equals space straight y space straight e to the power of negative straight v end exponent space space space space space space space space space space space or space space space space space space dv over dy space equals straight e to the power of negative straight v end exponent$
Separting the variables and integration, we get,
   $integral straight e to the power of straight v dv space equals space integral dy space space space space space space space space space or space space space space space straight e to the power of straight v space equals space straight y plus straight c$
or $straight e to the power of straight x over straight y end exponent space equals space straight y plus straight c$

Question 58

Wired Faculty App

Solve:
$straight x dy over dx minus straight y plus straight x space tan straight y over straight x equals 0$

Solution

The given differential equation is

straight x dy over dx minus straight y plus straight x space tan straight y over straight x space equals space 0

straight x dy over dx space equals space straight y minus straight x space tan straight y over straight x

dy over dx space equals space straight y over straight x minus tan straight y over straight x

Put y = v x so that

dy over dx space equals straight v plus straight x dv over dx

therefore space space space space space space space space straight v plus straight x dv over dx space equals straight v minus tan space straight v therefore space space space space space straight x dv over dx space equals space minus tan space straight v or space space space space space space space space space space space space space space fraction numerator 1 over denominator tan space straight v end fraction dv space equals space minus 1 over straight x dx therefore space space space space space space space space space space integral fraction numerator 1 over denominator tan space straight v end fraction dv space equals space minus integral 1 over straight x dx therefore space space space space space space space space space space log space open vertical bar sin space straight v close vertical bar space equals space minus log space open vertical bar straight x close vertical bar space plus space log space open vertical bar straight c close vertical bar or space space space space space space space space space space space space log space open vertical bar straight x space space sinv close vertical bar space equals space log space open vertical bar straight c close vertical bar therefore space space space space space space space space space space space space space straight x space space sin space straight v space equals space straight c or space space space space space space space space space space space space space space space space straight x space sin space straight y over straight x space equals space straight c space is space the space required space solution. space

Question 59

Wired Faculty App

Solve:
$straight y apostrophe space equals space straight y over straight x plus sin straight y over straight x$

Solution

The given differential equation is

dy over dx space equals space straight y over straight x plus sin space straight y over straight x

Put y = v x so that

dy over dx equals straight v plus straight x dv over dx

therefore space space space space from space left parenthesis 1 right parenthesis comma space space straight v plus straight x dv over dx equals straight v plus sin space straight v space space space or space space space straight x dv over dx space equals space sin space straight v

fraction numerator 1 over denominator sin space straight v end fraction dv space equals space dx over straight x

integral cosec space straight v space dv space equals space integral dx over straight x

log space open vertical bar tan space straight v over 2 close vertical bar space equals space log space open vertical bar straight x close vertical bar space plus space log space open vertical bar straight c close vertical bar space space or space space space log space open vertical bar tan space straight v over 2 close vertical bar space equals space log space open vertical bar cx close vertical bar space space or space space tan space straight v over 2 space equals space cx

tan space fraction numerator straight y over denominator 2 straight x end fraction space equals space cx space space is space the space required space solution.

Question 60

Wired Faculty App

Solve:
$dy over dx space equals straight y over straight x plus tan straight y over straight x open parentheses 0 less than straight y over straight x less than straight pi over 2 close parentheses$

Solution

The given differential equation is
$dy over dx space equals space straight y over straight x plus tan straight y over straight x space space space space space space space space space space space space open parentheses 0 less than straight y over straight x less than straight pi over 2 close parentheses$ ...(1)
Put y = v x so that $dy over dx equals straight v plus straight x dv over dx$
$therefore space space space space from space left parenthesis 1 right parenthesis comma space space straight v plus straight x dv over dx space equals space straight v plus tan space straight v space space or space space space straight x dv over dx space equals space tan space straight v space space or space space space fraction numerator 1 over denominator tan space straight v end fraction dv space equals space dx over straight x$
or $integral cot space straight v space dv space equals space integral dx over straight x$
or $log space open vertical bar sin space straight v close vertical bar space equals space log space open vertical bar straight x close vertical bar space plus space log space open vertical bar straight c close vertical bar space space or space space space log space open vertical bar sin space straight v close vertical bar space equals space log space open vertical bar cx close vertical bar$
or $sin space straight v space equals space cx space space space space or space space space space space sin straight y over straight x space equals space space straight c space straight x space is space the space required space solution. space$

Question 61

Wired Faculty App

For the given differential equations, find the particular solution satisfying the given condition:
$left parenthesis straight x plus straight y right parenthesis space dy space plus space left parenthesis straight x minus straight y right parenthesis dx space equals space 0 space semicolon space space straight y space equals space 1 space space space when space straight x space equals space 1$

Solution

The given differential equation is

left parenthesis straight x plus straight y right parenthesis space dy space plus space left parenthesis straight x minus straight y right parenthesis space dx space equals space 0 comma space space space or space space space left parenthesis straight x plus straight y right parenthesis space dy space equals space minus left parenthesis straight x minus straight y right parenthesis space dx

therefore space space space space space dy over dx space equals space minus fraction numerator straight x minus straight y over denominator straight x plus straight y end fraction space space space or space space space dy over dx space equals space fraction numerator straight y minus straight x over denominator straight y plus straight x end fraction

...(1)
Put y = v x, so that

dy over dx space equals straight v plus straight x dv over dx

therefore space space space space space from space left parenthesis 1 right parenthesis comma space space space space straight v plus straight x dv over dx space equals space fraction numerator straight v space straight x space minus straight x over denominator vx plus straight x end fraction space space or space space straight v plus straight x dv over dx space equals space fraction numerator straight v minus 1 over denominator straight v plus 1 end fraction

straight x dy over dx space equals fraction numerator straight v minus 1 over denominator straight v plus 1 end fraction minus straight v space equals space fraction numerator negative 1 minus straight v squared over denominator straight v plus 1 end fraction space equals space fraction numerator negative left parenthesis 1 plus straight v squared right parenthesis over denominator 1 plus straight v end fraction space space or space space fraction numerator 1 plus straight v over denominator 1 plus straight v squared end fraction dv space equals negative dx over straight x

integral fraction numerator 1 plus straight v over denominator 1 plus straight v squared end fraction dv space equals space minus integral dx over straight x

integral fraction numerator 1 over denominator 1 plus straight v squared end fraction dv plus 1 half integral fraction numerator 2 straight v over denominator 1 plus straight v squared end fraction dv space equals space minus integral 1 over straight x dx

tan to the power of negative 1 end exponent straight v space plus space 1 half log space open vertical bar 1 plus straight v squared close vertical bar plus log space open vertical bar straight x close vertical bar space equals space straight c

tan to the power of negative 1 end exponent straight v plus 1 half left square bracket log space open vertical bar 1 plus straight v squared close vertical bar space plus space 2 space log space open vertical bar straight x close vertical bar right square bracket space equals space straight c

tan to the power of negative 1 end exponent straight v plus 1 half log space open vertical bar straight x squared close vertical bar space space open vertical bar 1 plus straight v squared close vertical bar space equals space straight c

or space space tan space to the power of negative 1 end exponent straight y over straight x space plus space 1 half space log space open vertical bar straight x squared plus open parentheses 1 plus straight y squared over straight x squared close parentheses close vertical bar space equals space straight c or space space space tan space to the power of negative 1 end exponent straight y over straight x plus 1 half log space open vertical bar straight x squared plus straight y squared close vertical bar space equals space straight c therefore space space space space tan to the power of negative 1 end exponent straight y over straight x plus 1 half log space left parenthesis straight x squared plus straight y squared right parenthesis space equals space straight c space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis

Now y = 1 when x = 1

therefore space space space space space space tan to the power of negative 1 end exponent 1 plus 1 half log 2 space equals space straight c space space space space space rightwards double arrow space space space space space straight c space equals space straight pi over 4 plus 1 half log space 2 therefore space space space space from space left parenthesis 2 right parenthesis comma space space tan to the power of negative 1 end exponent straight y over straight x plus 1 half log space left parenthesis straight x squared plus straight y squared right parenthesis space equals space straight pi over 4 plus 1 half log space 2

is the required solution.

Question 62

Wired Faculty App

For the given differential equation, find the particular solution satisfying the given condition:
$straight x squared dy plus left parenthesis xy plus straight y squared right parenthesis dx space equals space 0 semicolon space space space straight y space equals space 1 space when space straight x space equals space 1$

Solution

The given differential equation is

straight x squared dy space plus straight y left parenthesis straight x plus straight y right parenthesis space dx space equals space 0 comma space space space or space space space straight x squared space dy space equals space minus straight y left parenthesis straight x plus straight y right parenthesis space dx

therefore space space space space space dy over dx space equals space minus fraction numerator straight y left parenthesis straight x plus straight y right parenthesis over denominator straight x squared end fraction

...(1)
Put y = v x so that

dy over dx equals straight v plus straight x dv over dx

therefore space space space space from space left parenthesis 1 right parenthesis comma space space space straight v plus straight x dv over dx space equals space minus fraction numerator straight v space straight x left parenthesis straight x plus vx right parenthesis over denominator straight x squared end fraction space space space space space rightwards double arrow space space space space straight v plus straight x dv over dx equals negative straight v minus straight v squared

rightwards double arrow space space space space straight x space dv over dx space equals space straight v squared minus 2 straight v

therefore space space space space space fraction numerator 1 over denominator straight v squared plus 2 straight v end fraction dv space equals space minus 1 over straight x dx space space space space space rightwards double arrow space space space integral fraction numerator 1 over denominator straight v left parenthesis straight v plus 2 right parenthesis end fraction dv space equals space minus integral 1 over straight x dx rightwards double arrow space space space space integral open square brackets fraction numerator 1 over denominator straight v left parenthesis 0 plus 2 right parenthesis end fraction plus fraction numerator 1 over denominator left parenthesis negative 2 right parenthesis thin space left parenthesis straight v plus 2 right parenthesis end fraction close square brackets dv space equals space minus integral 1 over straight x dx rightwards double arrow space space space space 1 half integral open parentheses 1 over straight v minus fraction numerator 1 over denominator straight v plus 2 end fraction close parentheses space dv space equals space minus integral 1 over straight x dx therefore space space space space space 1 half open square brackets log space straight v space minus space log space left parenthesis straight v plus 2 right parenthesis close square brackets space equals space minus log space straight x space plus space log space straight c rightwards double arrow space space space space space 1 half log space open parentheses fraction numerator straight v over denominator straight v plus 2 end fraction close parentheses space equals space log space open parentheses straight c over straight x close parentheses therefore space space space space space fraction numerator begin display style straight y over straight x end style over denominator begin display style straight y over straight x end style plus 2 end fraction space equals space straight c squared over straight x squared space space space space rightwards double arrow space space space space space fraction numerator straight y over denominator 2 straight x plus straight y end fraction space equals space straight c squared over straight x squared rightwards double arrow space space space space space space space straight x squared straight y space equals space straight k space left parenthesis 2 straight x plus straight y right parenthesis comma space where space straight k space equals space straight c squared space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis space space space space space space space space space space space space space space space space space Now space straight y space equals space 1 space when space straight x space equals space 1 space space space space space space rightwards double arrow space space space space space 1 space equals space straight k left parenthesis 2 plus 1 right parenthesis space space space rightwards double arrow space space space straight k space equals space 1 third

therefore space space space space from space left parenthesis 2 right parenthesis comma space space straight x squared straight y space equals space 1 third left parenthesis 2 straight x plus straight y right parenthesis space which space is space required space solution. space

Question 63

Wired Faculty App

For the given differential equation, find the particular solution satisfying the given condition:
$open square brackets straight x space sin squared open parentheses straight y over straight x close parentheses minus straight y close square brackets space dx plus space straight x space dy space equals space 0 colon space space space straight y space equals straight pi over 4 space space when space straight x space equals space 1$

Solution

The given differential equation is

open square brackets straight x space sin squared space open parentheses straight y over straight x close parentheses minus straight y close square brackets space dx plus straight x space dy space equals space 0

sin squared open parentheses straight y over straight x close parentheses minus straight y over straight x plus dy over dx space equals 0

dy over dx space equals straight y over straight x minus sin squared space open parentheses straight y over straight x close parentheses

...(1)
Put y = v x so that

dy over dx equals straight v plus straight x dv over dx

therefore space space space space space space left parenthesis 1 right parenthesis space becomes comma space space space straight v plus straight x dv over dx equals straight v minus sin squared straight v or space space space space space space space space space straight x dv over dx equals negative sin squared straight v

Separating the variables,

fraction numerator 1 over denominator sin squared straight v end fraction dv space equals space minus dx over straight x

Integrating,

integral cosec squared straight v space dv space equals space minus integral dx over straight x

rightwards double arrow space space space log space open vertical bar straight x close vertical bar minus cot space straight v space equals space straight c rightwards double arrow space space space log space open vertical bar straight x close vertical bar minus cot space open parentheses straight y over straight x close parentheses space equals space straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis

Now, x = 1, when

straight y space equals space straight pi over 4

log space open vertical bar 1 close vertical bar space minus space cot space open parentheses straight pi over 4 close parentheses space equals straight c

rightwards double arrow space space space space straight c space equals space 0 minus 1 space equals space minus 1

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log space open vertical bar straight x close vertical bar space minus space cot space straight y over straight x space equals space space minus space log space straight e

log space open vertical bar straight e space straight x close vertical bar space equals space cot space open parentheses straight y over straight x close parentheses

.
which is the required particular solution.

Question 64

Wired Faculty App

For the given differential equation, find the particular solution satisfying the given condition:
$dy over dx minus straight y over straight x plus cosec space open parentheses straight y over straight x close parentheses space equals space 0 space semicolon space space space straight y space equals space 0 space space space when space straight x space equals space 1$

Solution

The given differential equation is

dy over dx space equals space straight y over straight x minus cosec space straight y over straight x

Put y = v x so that

dy over dx space equals space straight v plus straight x dv over dx

therefore space space space space space space space space straight v plus straight x dv over dx space equals space straight v minus cosec space straight v therefore space space space space space space space space space space space space space straight x dv over dx space equals space minus cosec space straight v comma space space space space space space space space space space space space space space space space therefore space space space space minus fraction numerator 1 over denominator cosec space straight v end fraction dv space equals space 1 over straight x dx therefore space space space space space minus integral sin space straight v space dv space equals space integral 1 over straight x dx therefore space space space space space space space cos space straight v space space equals space log space open vertical bar straight x close vertical bar plus straight c therefore space space space space space cos space straight y over straight x space equals space log space open vertical bar straight x close vertical bar plus straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis space space space space Now space straight y space equals space 0 space space space space when space straight x space space equals space 1 therefore space space space space space space space cos space 0 space equals space log space open vertical bar 1 close vertical bar space plus space straight c space space space space space space rightwards double arrow space space space space space 1 space equals space 0 plus space straight c space rightwards double arrow space space space straight c space space equals 1 therefore space space from space left parenthesis 1 right parenthesis comma space space cos space straight y over straight x space equals space log space open vertical bar straight x close vertical bar plus 1 space is space the space required space solution. space

Question 65

Wired Faculty App

For the given differential equation, find the particular solution satisfying the given condition:
$2 xy plus straight y squared minus 2 straight x squared dy over dx equals 0 semicolon space space space straight y space equals 2 space space when space straight x space equals space 1.$

Solution

The given differential equation is

2 xy plus straight y minus 2 straight x squared dy over dx space equals space 0 space space space space space space space space space or space space space space space space 2 straight x squared dy over dx space equals space 2 xy plus straight y squared

therefore space space space space space space space space space space dy over dx space space equals fraction numerator 2 xy plus straight y squared over denominator 2 straight x squared end fraction

Put

straight y equals space straight v space straight x space space so space that space dy over dx space equals space straight v plus straight x dv over dx

therefore

straight v plus straight x dv over dx equals fraction numerator 2 straight x squared straight v plus straight v squared straight x squared over denominator 2 straight x squared end fraction

therefore space space space space space space straight v plus straight x dv over dx space equals fraction numerator 2 straight v plus straight v squared over denominator 2 end fraction therefore space space space space space space space space space space straight x dv over dx space equals space fraction numerator 2 straight v plus straight v squared over denominator 2 end fraction minus straight v space space space space or space space space straight x dv over dx equals straight v squared over 2 therefore space space space space space space 2 space 1 over straight v squared dv space equals space 1 over straight x dx

Integrating,

2 integral straight v to the power of negative 2 end exponent space dv space equals space integral 1 over straight x dx

therefore space space space space space 2 fraction numerator straight v to the power of negative 1 end exponent over denominator negative 1 end fraction space equals space log space open vertical bar straight x close vertical bar plus space straight c space space space space space or space space space space fraction numerator negative 2 over denominator straight v end fraction space equals space log space open vertical bar straight x close vertical bar plus straight c therefore space space space space space minus 2 straight x over straight y space equals log space open vertical bar straight x close vertical bar space plus space straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis

Now v = 2 when x = 1

therefore space space space space space space space space space space space space space space minus 2. space 1 half minus log space 1 space plus space straight c space space space space space space space rightwards double arrow space space space minus 1 space equals space 0 plus straight c space space space space rightwards double arrow space space space straight c space equals space minus 1 therefore space space space space space from space left parenthesis 1 right parenthesis comma space space space minus fraction numerator 2 straight x over denominator straight y end fraction space equals space log space open vertical bar straight x close vertical bar minus 1 space space space space space or space space space space straight y space equals space fraction numerator 2 straight x over denominator 1 minus log space open vertical bar straight x close vertical bar end fraction is space the space required space solution. space

Question 66

Wired Faculty App

For the given differential equation, find the particular solution satisfying the given condition:
$2 straight x squared straight y apostrophe space minus space 2 xy plus straight y squared space equals space 0 comma space space space space straight y left parenthesis straight e right parenthesis space equals space straight e$

Solution

The given differential equation is
$2 straight x squared straight y apostrophe space minus space 2 xy plus straight y squared space equals space 0$
or $2 straight x squared dy over dx space equals space 2 xy minus straight y squared space space space space space or space space space space dy over dx space equals space fraction numerator 2 xy minus straight y squared over denominator 2 straight x squared end fraction$
Put y = v x so that $dy over dx space equals straight v plus straight x dv over dx$
$therefore space space space space space straight v plus straight x dv over dx space equals space fraction numerator 2 straight x squared straight v minus straight v squared straight x squared over denominator 2 straight x squared end fraction$
$therefore space space space space straight v plus straight x dv over dx equals fraction numerator 2 straight v minus straight v squared over denominator 2 end fraction therefore space space space space space space straight x dv over dx space equals space fraction numerator 2 straight v minus straight v squared over denominator 2 end fraction minus straight v space space space space space or space space space straight x dv over dx space equals negative straight v squared over 2 therefore space space space space space space minus 2 over straight v squared dv space equals space 1 over straight x dx$
Integrating $2 integral straight v to the power of negative 2 end exponent dv space equals space integral 1 over straight x dx$
$therefore space space space minus 2 fraction numerator straight v to the power of negative 1 end exponent over denominator negative 1 end fraction space equals space log space open vertical bar straight x close vertical bar space plus space straight c space space space space space or space space space space fraction numerator negative 2 over denominator straight v end fraction equals log space open vertical bar straight x close vertical bar plus straight c therefore space space space space space space fraction numerator 2 straight x over denominator straight y end fraction space equals space log space open vertical bar straight x close vertical bar space plus space straight c$
Now $straight y left parenthesis straight e right parenthesis space equals straight e$ $rightwards double arrow space space space straight y space equals space straight e space space space when space straight x space equals space straight e$
$therefore space space space 2 straight e over straight e equals log space straight e plus straight c space space space space space space space space space space rightwards double arrow space space space space 2 space equals space 1 plus straight c space space space space space space space rightwards double arrow space space space straight c space equals space 1$
$therefore space space space solution space is space space fraction numerator 2 straight x over denominator straight y end fraction space equals space log space open vertical bar straight x close vertical bar plus 1.$

Question 67

Wired Faculty App

For the given differential equation, find the particular solution satisfying the given condition:
$2 xy plus straight y squared minus 2 straight x squared straight y apostrophe space equals space 0 space space space space space space space space space space space space space straight y left parenthesis 1 right parenthesis space equals space 2$

Solution

The given differential equation is

2 xy plus straight y squared minus 2 straight x squared straight y apostrophe space equals space 0 space space space space space space space or space space space space space 2 straight x squared straight y apostrophe space equals space 2 xy plus straight y squared

therefore space space space space space space space space space dy over dx space equals space fraction numerator 2 xy plus straight y squared over denominator 2 straight x squared end fraction

Put y - v x so that

dy over dx equals straight v plus straight x dv over dx

therefore space space space space space space straight v plus straight x dv over dx equals fraction numerator 2 straight x squared straight v plus straight v squared straight x squared over denominator 2 straight x squared end fraction therefore space space space space space straight v plus straight x dv over dx equals fraction numerator 2 straight v plus straight v squared over denominator 2 end fraction therefore space space space space space space space space straight x dv over dx equals fraction numerator 2 straight v plus straight v squared over denominator 2 end fraction minus straight v space space space space space or space space space straight x dv over dx equals straight v squared over 2 therefore space space space space space 2 1 over straight v squared dv space equals space 1 over straight x dx

Integration,

2 space integral straight v to the power of negative 2 end exponent space dv space equals space integral 1 over straight x dx

therefore space space space space space 2 fraction numerator straight v to the power of negative 1 end exponent over denominator negative 1 end fraction space equals space log space open vertical bar straight x close vertical bar plus straight c space space space space space or space space space space fraction numerator negative 2 over denominator straight v end fraction equals log space open vertical bar straight x close vertical bar plus straight c therefore space space space space minus 2 straight x over straight y space equals space log space open vertical bar straight x close vertical bar plus straight c

Now,

straight y left parenthesis 1 right parenthesis space equals space 2 space space space space space space space space space space space space space space rightwards double arrow space space space space straight y space equals space 2 space space space when space straight x space equals space 1

therefore space space space space minus 2. space 1 half space equals space log space 1 plus straight c space space space space space space space space space space rightwards double arrow space space space space minus 1 space equals space 0 space plus straight c space space space space rightwards double arrow space space space space straight c space equals space minus 1 therefore space space solution space is space minus fraction numerator 2 straight x over denominator straight y end fraction space equals space log space open vertical bar straight x close vertical bar minus 1 space space space space space or space space space straight y space equals space fraction numerator 2 straight x over denominator 1 minus log space open vertical bar straight x close vertical bar end fraction.

Question 68

Wired Faculty App

Find a particular solution of the differential equation
(x – y) (dx + dy) = dx – dy. given that y = – 1, when x = 0.

Solution

The given differential equation is
(x – y) (dx + dy) = dx – dy ...(1)
or

dx plus dy space equals space fraction numerator dx minus dy over denominator straight x minus straight y end fraction

Integrating, we get

integral left parenthesis dx plus dy right parenthesis space equals space integral fraction numerator straight d space left parenthesis straight x minus straight y right parenthesis over denominator straight x minus straight y end fraction plus straight c

rightwards double arrow space space space space space space space space space space space space straight x plus straight y space equals space log space open vertical bar straight x minus straight y close vertical bar plus straight c

....(2)
Now x = 0, y = -1

therefore space space space space space space space 0 plus left parenthesis negative 1 right parenthesis space equals space log space open vertical bar 0 plus 1 close vertical bar space plus space straight c rightwards double arrow space space space space space space space space space space space straight c space equals space minus 1 therefore space space space from space left parenthesis 2 right parenthesis comma space space space straight x plus straight y space equals space log space open vertical bar straight x minus straight y close vertical bar space minus space 1

which is required solution.

Question 69

Wired Faculty App

Solve the following differential equation:
$dy over dx equals fraction numerator straight x left parenthesis 2 straight y minus straight x right parenthesis over denominator straight x left parenthesis 2 straight y plus straight x right parenthesis end fraction comma space space if space space straight y space equals space 1 space space when space straight x space equals space 1.$

Solution

The given differential equation is
$dy over dx equals fraction numerator straight x left parenthesis 2 straight y minus straight x right parenthesis over denominator straight x left parenthesis 2 straight y plus straight x right parenthesis end fraction space space space or space space space dy over dx equals fraction numerator 2 straight y minus straight x over denominator 2 straight y plus straight x end fraction$
Put $straight y space equals space straight v space straight x space space so space that space space dy over dx space equals space straight v plus straight x dv over dx$
$therefore space space space space space straight v plus straight x dv over dx space equals space fraction numerator 2 space straight v space straight x space minus straight x over denominator 2 space straight v space straight x space plus 1 end fraction space space space or space space straight v plus straight x dv over dx equals fraction numerator 2 straight v minus 1 over denominator 2 straight v plus 1 end fraction therefore space space space space space space space space space straight x dv over dx equals fraction numerator 2 straight v minus 1 over denominator 2 straight v plus 1 end fraction minus straight v space space space or space space space straight x dv over dx equals fraction numerator 2 straight v minus 1 minus 2 straight v squared minus straight v over denominator 2 straight v plus 1 end fraction therefore space space space space straight x dv over dx space equals space fraction numerator negative 2 straight v squared plus straight v minus 1 over denominator 2 straight v plus 1 end fraction$
Separating the variables, we get,
$fraction numerator 2 straight v plus 1 over denominator negative 2 straight v squared plus straight v minus 1 end fraction dv space equals space 1 over straight x dx$
Integrating, $integral fraction numerator 2 straight v plus 1 over denominator negative 2 straight v squared plus straight v minus 1 end fraction dv space equals space 1 over straight x dx$
$therefore space space space space space space space integral fraction numerator begin display style 1 half left parenthesis 4 straight v minus 1 right parenthesis plus 3 over 2 end style over denominator 2 straight v squared minus straight v plus 1 end fraction dv space equals space minus integral 1 over straight x dx therefore space space space 1 half integral fraction numerator 4 straight v minus 1 over denominator 2 straight v squared minus straight v plus 1 end fraction dv plus 3 over 2 integral fraction numerator 1 over denominator 2 straight v squared minus straight v plus 1 end fraction dv space equals space minus integral 1 over straight x dx therefore space space space space 1 half integral fraction numerator 4 straight v minus 1 over denominator 2 straight v squared minus straight v plus 1 end fraction dv plus 3 over 4 integral fraction numerator 1 over denominator straight v squared minus begin display style 1 half end style straight v plus begin display style 1 half end style end fraction dv space equals space minus integral 1 over straight x dx therefore space space space space space 1 half space integral fraction numerator 4 straight v minus 1 over denominator 2 straight v squared minus straight v plus 1 end fraction dv plus 3 over 4 integral fraction numerator 1 over denominator open parentheses straight v squared minus begin display style 1 half end style straight v plus begin display style 1 over 16 end style close parentheses plus begin display style 7 over 16 end style end fraction dv space equals space minus integral 1 over straight x dx therefore space space space space space space 1 half integral fraction numerator 4 straight v minus 1 over denominator 2 straight v squared minus straight v plus 1 end fraction dv plus 3 over 4 integral fraction numerator 1 over denominator open parentheses straight v squared minus begin display style 1 fourth end style close parentheses squared plus open parentheses begin display style fraction numerator square root of 7 over denominator 4 end fraction end style close parentheses squared end fraction dv space equals space minus integral 1 over straight x dx therefore space space space space space space 1 half log space open vertical bar 2 straight v squared minus straight v plus 1 close vertical bar plus 3 over 4. fraction numerator 1 over denominator begin display style fraction numerator square root of 7 over denominator 4 end fraction end style end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator straight v minus begin display style 1 fourth end style over denominator begin display style fraction numerator square root of 7 over denominator 4 end fraction end style end fraction close parentheses equals negative log space open vertical bar straight x close vertical bar plus straight c therefore space space 1 half log space open vertical bar 2 straight v squared minus straight v plus 1 close vertical bar plus fraction numerator 3 over denominator square root of 7 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator 4 straight v minus 1 over denominator square root of 7 end fraction close parentheses space equals space minus log space open vertical bar straight x close vertical bar plus straight c therefore space space 1 half log space open vertical bar fraction numerator 2 straight y squared over denominator straight x squared end fraction minus straight y over straight x plus 1 close vertical bar plus space fraction numerator 3 over denominator square root of 7 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style fraction numerator 4 straight y over denominator straight x end fraction minus 1 end style over denominator square root of 7 end fraction close parentheses space equals space minus log space open vertical bar straight x close vertical bar plus straight c space space space therefore space space space 1 half log space open vertical bar fraction numerator 2 straight y squared minus xy plus straight x squared over denominator straight x squared end fraction close vertical bar plus fraction numerator 3 over denominator square root of 7 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator 4 straight y minus straight x over denominator square root of 7 straight x end fraction close parentheses space equals space minus log space open vertical bar straight x close vertical bar plus straight c space space space space... left parenthesis 1 right parenthesis$
Now y = 1, when x = 1
$therefore space space space space 1 half log space open vertical bar fraction numerator 2 minus 1 plus 1 over denominator 1 end fraction close vertical bar plus fraction numerator 3 over denominator square root of 7 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator 4 minus 1 over denominator square root of 7 end fraction close parentheses space equals space log space open vertical bar 1 close vertical bar plus straight c therefore space space space 1 half log space 2 space plus fraction numerator 3 over denominator square root of 7 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator 3 over denominator square root of 7 end fraction close parentheses space equals space straight c$
Putting this value of c in (1), we get,
$space space space space space space space space space 1 half log space open vertical bar fraction numerator 2 straight y squared minus xy plus straight x squared over denominator straight x squared end fraction close vertical bar plus fraction numerator 3 over denominator square root of 7 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator 4 straight y minus straight x over denominator square root of 7 straight x end fraction close parentheses space equals space minus log open vertical bar straight x close vertical bar plus 1 half log 2 plus fraction numerator 3 over denominator square root of 7 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator 3 over denominator square root of 7 end fraction close parentheses or space space space 1 half log space open vertical bar 2 straight y squared minus xy plus straight x squared close vertical bar minus 1 half log open vertical bar straight x squared close vertical bar plus fraction numerator 3 over denominator square root of 7 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator 4 straight y minus straight x over denominator square root of 7 straight x end fraction close parentheses space equals space minus log space open vertical bar straight x close vertical bar space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space plus space 1 half log space 2 plus fraction numerator 3 over denominator square root of 7 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator 3 over denominator square root of 7 end fraction close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space therefore space space space space 1 half log space open vertical bar 2 straight y squared minus xy plus straight x squared close vertical bar minus log space open vertical bar straight x close vertical bar plus fraction numerator 3 over denominator square root of 7 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator 4 straight y minus straight x over denominator square root of 7 straight x end fraction close parentheses space equals space minus log space open vertical bar straight x close vertical bar space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space plus 1 half log space 2 plus fraction numerator 3 over denominator square root of 7 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator 3 over denominator square root of 7 end fraction close parentheses therefore space space space space 1 half log space open vertical bar 2 straight y squared minus xy plus straight x squared close vertical bar plus fraction numerator 3 over denominator square root of 7 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator 4 straight y minus straight x over denominator square root of 7 straight x end fraction close parentheses space equals 1 half log space 2 plus fraction numerator 3 over denominator square root of 7 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator 3 over denominator square root of 7 end fraction close parentheses$
which is required solution.

Question 70

Wired Faculty App

Solve the following initial value problem:
$dy over dx space equals space fraction numerator straight y left parenthesis straight x plus 2 straight y right parenthesis over denominator straight x left parenthesis 2 straight x plus straight y right parenthesis end fraction. space space space straight y left parenthesis 1 right parenthesis space equals space 2$

Solution

The given differential equation is

dy over dx space equals space fraction numerator straight y left parenthesis straight x plus 2 straight y right parenthesis over denominator straight x left parenthesis 2 straight x plus straight y right parenthesis end fraction

Put y = vx so that

dy over dx space equals space straight v plus straight x dv over dx

therefore space space space space space space space straight v plus straight x dv over dx space equals space fraction numerator vx left parenthesis straight x plus 2 vx right parenthesis over denominator straight x left parenthesis 2 straight x plus vx right parenthesis end fraction

therefore space space space space straight v plus straight x dv over dx space equals space fraction numerator straight v left parenthesis 1 plus 2 straight v right parenthesis over denominator 2 plus straight v end fraction therefore space space space space space straight x dv over dx space equals space fraction numerator straight v left parenthesis 1 plus 2 straight v right parenthesis over denominator 2 plus straight v end fraction minus straight v space equals space space fraction numerator straight v plus 2 straight v squared minus 2 straight v minus straight v squared over denominator 2 plus straight v end fraction therefore space space space space space straight x dv over dx space equals space fraction numerator straight v squared minus straight v over denominator straight v plus 2 end fraction

Separating the variables and integrating, we get,

integral fraction numerator straight v plus 2 over denominator straight v squared minus straight v end fraction dv space equals space integral 1 over straight x dx

therefore space space space integral open square brackets fraction numerator straight v plus 2 over denominator straight v left parenthesis straight v minus 1 right parenthesis end fraction close square brackets dv space equals space integral 1 over straight x dx therefore space space space space integral open square brackets fraction numerator 0 plus 2 over denominator straight v left parenthesis 0 minus 1 right parenthesis end fraction plus fraction numerator 1 plus 2 over denominator left parenthesis 1 right parenthesis thin space left parenthesis straight v minus 1 right parenthesis end fraction close square brackets dv space equals space integral 1 over straight x dx therefore space space space minus integral 2 over straight v dv plus integral fraction numerator 3 over denominator straight v minus 1 end fraction dv space equals space integral 1 over straight x dx therefore space space space minus 2 space logv plus space 3 space log space left parenthesis straight v minus 1 right parenthesis space equals space log space straight x space plus straight c therefore space space space minus 2 log space straight y over straight x plus 3 space log space open parentheses straight y over straight x minus 1 close parentheses space equals space logx plus space straight c

Now,

straight y left parenthesis 1 right parenthesis space equals space 2 space space space space space space space space rightwards double arrow space space space space straight y space equals space 2 space space space space space space when space space space space straight x space equals space 1

therefore space space space space space space minus 2 space log space 2 over 1 plus 3 space log space open parentheses 2 over 1 minus 1 close parentheses space equals space log space 1 space plus space straight c comma space space space space space space therefore space space space minus 2 space log space 2 space plus space 3 space log space 1 space equals space log space 1 space plus straight c therefore space space space space minus 2 space log space 2 space plus space 0 space equals space 0 plus space straight c space space space space space space space space space space space rightwards double arrow space space space space straight c space equals space minus 2 space log space 2 therefore space space space solution space is space space minus 2 space log space straight y over straight x plus 3 space log space open parentheses fraction numerator straight y minus straight x over denominator straight x end fraction close parentheses space equals space log space straight x space plus space 2 space log space 2

Question 71

Wired Faculty App

Show that the family of curves for which the slope of the tangent at any point (x, y) on it is $fraction numerator straight x squared plus straight y squared over denominator 2 xy end fraction comma$ is given by $straight x squared minus straight y squared space equals space straight c space straight x.$

Solution

We know that $dy over dx$ is slope of tangent to the curve at point (x, y).
$therefore space space space space space space space space space dy over dx space equals space fraction numerator straight x squared plus straight y squared over denominator 2 xy end fraction$
Put y = v x so that $dy over dx equals straight v plus straight x dv over dx$
$therefore space space space space straight v plus straight x dv over dx equals fraction numerator straight x squared plus straight v squared straight x squared over denominator 2 vx squared end fraction therefore space space space space straight v plus straight x dv over dx space equals fraction numerator 1 plus straight v squared over denominator 2 straight v end fraction therefore space space space space space space space space space straight x dv over dx space equals space fraction numerator 1 plus straight v squared over denominator 2 straight v end fraction minus straight v space space space space space space or space space space space space straight x dv over dx equals space fraction numerator 1 minus straight v squared over denominator 2 straight v end fraction$
Separating the variables and integrating, we get,
$integral fraction numerator 2 straight v over denominator 1 minus straight v squared end fraction dv space equals space integral 1 over straight x dx space space space space or space space space integral fraction numerator 2 straight v over denominator straight v squared minus 1 end fraction dv space equals space minus integral 1 over straight x dx$
$therefore space space space log space open vertical bar straight v squared minus 1 close vertical bar space equals space minus log space open vertical bar straight x close vertical bar plus log space open vertical bar straight c subscript 1 close vertical bar space space space or space space space log space open vertical bar left parenthesis straight v squared minus 1 right parenthesis thin space left parenthesis straight x right parenthesis close vertical bar space equals space log space open vertical bar straight c subscript 1 close vertical bar or space space space space space space space left parenthesis straight v squared minus 1 right parenthesis space straight x space equals space plus-or-minus space space straight c subscript 1$
Replacing v by $straight y over straight x comma space$ we get
$open parentheses straight y squared over straight x squared minus 1 close parentheses space straight x space equals space plus-or-minus space straight c subscript 1 space space space or space space space space left parenthesis straight y squared minus straight x squared right parenthesis space equals space space plus-or-minus space straight c subscript 1 straight x space space space or space space space straight x squared minus straight y squared space equals space cx.$

Question 72

Wired Faculty App

The general solution of the differential equation $fraction numerator straight y space dx space minus space straight x space dy over denominator straight y end fraction space equals space 0$ is

xy = C

x = Cy²
y = Cx
y = Cx²

Solution

y = Cx The given differential equation is

fraction numerator straight y space dx minus space straight x space dy over denominator straight y end fraction space equals space 0

fraction numerator straight y space dx space minus space straight x space dy over denominator straight x squared end fraction space equals space 0 space space space space space or space space straight d open parentheses straight x over straight y close parentheses space equals space 0

therefore space space space space space space space space space space straight x over straight y space equals space constant. therefore space space space space space space space space space straight y over straight x space equals space straight C space space or space space space space straight y space equals space Cx therefore space space space space left parenthesis straight C right parenthesis space is space correct space answer.

Question 73

Wired Faculty App

A homogeneous differential equation of the from $dx over dy space equals space straight h open parentheses straight x over straight y close parentheses$ can be solved by making the substitution.

y = vx
v = yx

x = vy

x = v

Solution

x = vy

A homogeneous differential equation of the from

dy over dx space equals space straight h space open parentheses straight x over straight y close parentheses

can be solved by making the substitution

straight v space equals straight x over straight y space space straight i. straight e. comma space space space straight x space equals space vy

∴ (C) is correct answer.

Question 74

Wired Faculty App

Which of the following is a homogeneous differential equation?

(4x + 6y + 5) dy – (3y + 2x + 4) dx = 0

(xy) dx – (x³ + y³) dy = 0

(x³ + 2 y²) dx + 2xy dy = 0

y² dx + (x² – x y – y²) dy =0

Solution

y² dx + (x² – x y – y²) dy =0

The equation y²dx + (x² – xy – y²) dy = 0. which can be written as

dy over dx equals negative fraction numerator straight y squared over denominator straight x squared minus xy minus straight y squared end fraction space equals space minus fraction numerator open parentheses begin display style straight y over straight x end style close parentheses squared over denominator 1 minus open parentheses begin display style straight y over straight x end style close parentheses minus open parentheses begin display style straight y over straight x end style close parentheses squared end fraction comma space is space homogeneous space as space straight R. straight H. straight S. space is space of space form space straight F open parentheses straight y over straight x close parentheses

as R.H.S. is of form

straight F open parentheses straight y over straight x close parentheses

therefore space space space left parenthesis straight D right parenthesis space is space correct space answer.

Question 75

Wired Faculty App

Solve : $dy over dx plus straight y space equals space sin space straight x comma space space left parenthesis straight x space element of space straight R right parenthesis$

Solution

The given differential equation is $dy over dx plus straight y space equals space sin space straight x$
Comparing it with $dy over dx plus straight P space straight y space equals space straight Q comma space space space we space get comma space space space straight P space equals space 1 comma space space straight Q space equals space sin space straight x$
$therefore space space space integral straight P space dx space equals space integral 1 space dx space equals space straight x. space space space straight I. straight F. space equals space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of straight x$
$therefore space space space$ solution of differential equation is
$straight y space. straight e to the power of straight x space equals space integral space sinx. space straight e to the power of straight x space dx space plus space straight c space or space space straight y space. straight e to the power of straight x space equals space fraction numerator 1 over denominator square root of 1 plus 1 end root end fraction straight e to the power of straight x space sin space open parentheses straight x minus tan to the power of negative 1 end exponent 1 over 1 close parentheses plus straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space space integral straight e to the power of ax space sin space bx space equals space fraction numerator 1 over denominator square root of straight a squared plus straight b squared end root end fraction straight e to the power of ax space sin space open parentheses bx minus tan to the power of negative 1 end exponent straight b over straight a close parentheses close square brackets or space space space straight y space equals space fraction numerator 1 over denominator square root of 2 end fraction sin space open parentheses straight x minus straight pi over 4 close parentheses plus space straight c space straight e to the power of negative straight x end exponent$

Question 76

Wired Faculty App

Solve the differential equation:
$straight x dy over dx minus straight y minus 2 straight x cubed space equals space 0.$

Solution

The given differential equation is

straight x dy over dx minus straight y minus 2 straight x cubed space equals space 0

straight x dy over dx minus straight y space equals space 2 straight x cubed

dy over dx minus 1 over straight x. space straight y space equals space 2 straight x squared

Comparing it with

dy over dx plus Py space equals straight Q comma space space we space get comma space straight P space equals space minus 1 over straight x comma space space straight Q space equals space 2 straight x squared

integral Pdx space equals space minus integral 1 over straight x dx space equals space minus log space straight x space equals space log space straight x to the power of negative 1 end exponent

therefore space space space straight e to the power of integral Pdx end exponent space equals space straight e to the power of log space straight x to the power of negative 1 end exponent end exponent space equals space straight x to the power of negative 1 end exponent space equals space 1 over straight x

therefore

solution of differential equation is

straight y. space straight e to the power of integral straight P space dx end exponent space equals space integral straight Q. space straight e to the power of integral straight P space dx end exponent dx plus straight c

straight y.1 over straight x space equals space integral 2 space straight x squared. space 1 over straight x dx space plus space straight c space space space space space space or space space space straight y over straight x space equals space 2 integral straight x space dx space plus space straight c

straight y over straight x space equals space straight x squared plus straight c

straight y space equals space straight x cubed plus cx

Question 77

Wired Faculty App

Solve the differential equation:
$dy over dx minus 2 straight y space equals space 3 straight x.$

Solution

The given differential equation is

dy over dx minus 2 straight y space equals space 3 straight x

Comparing it with

dy over dx plus space straight P space straight y space equals space straight Q comma space space we space get comma

P = -2, Q = 3x

integral straight P space dx space equals space minus 2 space integral space dx space equals space minus 2 straight x space space space straight I. straight F. space equals space straight e to the power of integral Pdx end exponent space equals space straight e to the power of negative 2 straight x end exponent

Solution of given differential equation is

straight y space straight e to the power of integral straight P space dx end exponent space equals space integral straight Q space straight e to the power of integral straight P space dx end exponent dx space plus space straight c

straight y space straight e to the power of negative 2 straight x end exponent space equals space integral space 3 space straight x space straight e to the power of negative 2 straight x end exponent space dx plus straight c

ye to the power of negative 2 straight x end exponent space equals space 3 space integral space xe to the power of negative 2 straight x end exponent dx plus straight c

ye to the power of negative 2 straight x end exponent space equals space 3 space open square brackets straight x fraction numerator straight e to the power of negative 2 straight x end exponent over denominator negative 2 end fraction minus integral 1. space fraction numerator straight e to the power of negative 2 straight x end exponent over denominator negative 2 end fraction dx close square brackets plus straight c

ye to the power of negative 2 straight x end exponent space equals negative 3 over 2 xe to the power of negative 2 straight x end exponent plus 3 over 2 integral straight e to the power of negative 2 end exponent dx plus straight c

ye to the power of negative 2 straight x end exponent space equals space minus 3 over 2 xe to the power of negative 2 straight x end exponent plus 3 over 2 fraction numerator straight e to the power of negative 2 straight x end exponent over denominator negative 2 end fraction plus straight c

straight y equals negative 3 over 2 straight x minus 3 over 4 plus ce to the power of 2 straight x end exponent

which is required solution.

Question 78

Wired Faculty App

Solve $dy over dx plus straight y over straight x space equals space straight e to the power of straight x comma space space left parenthesis straight x greater than 0 right parenthesis.$

Solution

The given differential equation is

dy over dx plus straight y over straight x space equals space straight e to the power of straight x

Comparing it with

dy over dx plus Py space equals space straight Q comma

we get,

straight P space equals space 1 over straight x comma space space straight Q space equals space straight e to the power of straight x

integral Pdx space equals integral 1 over straight x dx space equals space log space straight x comma space space space space space straight e to the power of integral straight p space dx end exponent space equals space straight e to the power of logx space equals space straight x

therefore space space space solution space of space differential space equation space is

straight y. space straight e to the power of integral straight P space dx end exponent space equals space integral straight Q. space straight e to the power of integral straight P space dx end exponent dx space plus space straight c space space space or space space space straight y. space straight x space equals space integral straight e to the power of straight x. space straight x space dx plus straight c

straight x space straight y space equals space straight x space straight e to the power of straight x space space minus space integral 1. space straight e to the power of straight x space dx plus space straight c space or space straight x space straight y space equals space straight x space straight e to the power of straight x minus straight e to the power of straight x plus straight c

straight y equals fraction numerator left parenthesis straight x minus 1 right parenthesis space straight e to the power of straight x over denominator straight x end fraction plus straight c over straight x

Question 79

Wired Faculty App

Solve the differential equation:
$straight x dy over dx plus 2 straight y space equals straight x squared.$

Solution

The given differential equation is

straight x dy over dx plus 2 straight y space equals straight x squared space space or space space dy over dx plus 2 over straight x straight y space equals space straight x

Comparing it with

dy over dx plus Py space equals space straight Q comma space space we space get comma space straight P space equals space 2 over straight x comma space space straight Q space equals space straight x

therefore space space space space integral straight P space dx space equals space integral 2 over straight x dx space equals space 2 space log space straight x space equals space log space straight x squared

therefore space space space space space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of log space straight x squared end exponent space equals space straight x squared space

open square brackets because space space straight e to the power of log space straight x end exponent space equals space straight x close square brackets

therefore space space space solution space of space given space differential space equation space is

straight y. space straight e to the power of integral straight P space dx end exponent space equals space integral space straight Q. space space straight e to the power of integral space straight P space dx end exponent dx space plus straight c

straight y. space straight x squared space equals space integral straight x. space straight x squared space dx plus straight c space space space space or space space space space straight x squared straight y space equals space integral straight x cubed space dx plus straight c

straight x squared straight y space equals space straight x to the power of 4 over 4 plus straight c space space or space space straight y space equals space 1 fourth straight x squared plus cx squared

Sponsor Area

Question 80

Wired Faculty App

Solve:
$straight v plus 2 straight y space equals 4 straight x$

Solution

The given differential equation is $dy over dx plus 2 straight y space equals space 4 straight x$
Comparing it with $dy over dx plus Py space equals space straight Q comma space we space get.$
P = 2, Q = 4 x
$therefore space space space integral straight P space dx space equals space integral 2 space dx space equals space 2 space straight x therefore space space space straight I. straight F. space equals space straight e to the power of integral straight P space dx end exponent space equals space straight e squared therefore space space space solution space of space differential space equation space is$
   $straight y. space straight e to the power of 2 straight x end exponent space space space integral space 4 straight x. space straight e to the power of 2 straight x end exponent space dx plus straight c space space or space space space straight y space straight e to the power of 2 straight x end exponent space equals space 4 space integral straight x space straight e to the power of 2 straight x end exponent space dx$
or    $straight y. straight e to the power of 2 straight x end exponent space equals space 4 open square brackets straight x straight e to the power of 2 straight x end exponent over 2 minus integral 1. space straight e to the power of 2 straight x end exponent over 2 dx close square brackets plus straight c$
or $straight y. straight e to the power of 2 straight x end exponent space equals space 2 space straight x space straight e to the power of 2 straight x end exponent minus 2 integral straight e to the power of 2 straight x end exponent dx plus straight c space space or space space straight y space straight e to the power of 2 straight x end exponent space equals space 2 straight x space straight e to the power of 2 straight x end exponent minus 2. straight e to the power of 2 straight x end exponent over 2 plus straight c$
or    $straight y. space straight e to the power of 2 straight x end exponent space equals space 2 space straight x space straight e to the power of 2 straight x end exponent space space space space comma space space space straight e to the power of 2 straight x end exponent plus straight c space space space or space space straight y space equals space 2 straight x minus 1 space plus space straight c space straight e to the power of 2 straight x end exponent$

Question 81

Wired Faculty App

Solve:
$dy over dx plus straight y over straight x equals straight x squared.$

Solution

The given differential equation is

dy over dx plus straight y over straight x equals straight x squared

Comparing it with

dy over dx plus straight P space straight y space equals space straight Q comma space space we space get comma space space straight P space equals 1 over straight x. space space straight Q space equals straight x squared

integral straight P space dx space equals space integral 1 over straight x dx space equals space log space straight x

therefore space space space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of logx space equals straight x

therefore space space space solution space of space given space differential space equation space is

straight y space straight e to the power of integral straight P space dx end exponent space equals space integral straight Q. space straight e to the power of integral straight P space dx end exponent space dx plus straight c space space space or space space space straight y space. straight x space equals space integral straight x squared. space straight x space dx plus straight c

straight x space straight y space equals space integral straight x cubed dx plus straight c space space space space or space space space straight x space straight y space equals space straight x to the power of 4 over 4 plus straight c space space space space space or space space space straight y space equals space straight x cubed over 4 plus straight c over straight x

Question 82

Wired Faculty App

Solve:
$dy over dx plus 4 straight y space equals space 5 straight x$

Solution

The given differential equation is

dy over dx plus 4 straight y space equals space 5 straight x

Comparing it with

dy over dx plus straight P space straight y space equals space straight Q comma

we get,
P = 4, Q = 5x

integral straight P space dx space equals space integral 4 space dx space equals space 4 straight x

straight I. straight F. space equals space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of 4 straight x end exponent

Solution of given differential equation is

ye to the power of integral straight P space dx end exponent space equals space integral space straight Q space straight e to the power of integral space straight P space dx end exponent dx space plus space straight c

ye to the power of 4 straight x end exponent space equals space integral 5 space straight x space straight e to the power of 4 straight x end exponent dx space plus straight c

or space space space space space straight y space straight e to the power of 4 straight x end exponent space equals space 5 space integral straight x space straight e to the power of 4 straight x end exponent dx plus straight c or space space space space straight y space straight e to the power of 4 straight x end exponent space equals space 5 space open square brackets straight x space straight e to the power of 4 straight x end exponent over 4 minus space integral 1. space straight e to the power of 4 straight x end exponent over 4 dx close square brackets plus straight c or space space space space straight y space straight e to the power of 4 straight x end exponent space equals space 5 over 4 straight x space straight e to the power of 4 straight x end exponent minus 5 over 4 integral straight e to the power of 4 straight x end exponent dx space plus space straight c or space space space straight y space straight e to the power of 4 straight x end exponent space equals space 5 over 4 straight x space straight e to the power of 4 straight x end exponent space minus space 5 over 4. space straight e to the power of 4 straight x end exponent over 4 plus straight c therefore space space space space space space space space straight y space equals space 5 over 4 straight x minus 5 over 16 plus ce to the power of negative 4 straight x end exponent

is required solution.

Question 83

Wired Faculty App

Solve:
$dy over dx plus straight y space equals space 1$

Solution

The given differential equation is $dy over dx plus straight y space equals space 1$
Comparing it with $dy over dx plus space straight P space straight y space equals space straight Q comma space we space get comma space space space straight P space equals space 1 comma space space straight Q space equals space 1$
$therefore space space space space integral straight P space dx space equals space integral 1. space dx space equals space straight x therefore space space space straight I. space straight F. space equals space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of straight x therefore space space space solution space of space differential space equation comma space is space straight y. space straight e to the power of straight x space equals space integral 1. space straight e to the power of straight x space dx space plus straight c or space space space straight y space straight e to the power of straight x space equals space straight e to the power of straight x plus straight c space space space or space space space straight y space equals space 1 plus straight c space straight e to the power of negative straight x end exponent$

Question 84

Wired Faculty App

Solve:
$fraction numerator dy apostrophe over denominator dx end fraction minus straight y over straight x equals space 2 straight x squared$

Solution

The given differential equation is $dy over dx minus straight y over straight x space equals space 2 straight x squared$
Comparing it with $dy over dx plus straight P space straight y space equals space straight Q comma space we space get comma space straight P space equals negative 1 over straight x comma space space straight Q space equals space 2 straight x squared.$
$integral straight P space dx space equals space minus integral 1 over straight x dx space equals space minus log space straight x space equals space log space straight x to the power of negative 1 end exponent$
$therefore space space space space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of logx to the power of negative 1 end exponent end exponent space equals space straight x to the power of negative 1 end exponent space equals space 1 over straight x$
$therefore$ solution of differential equation is,
$straight y space. straight e to the power of integral straight P space dx end exponent space equals space integral straight Q. space straight e to the power of integral straight P space dx end exponent dx space plus space straight c$
or $straight y.1 over straight x space equals space integral 2 space straight x squared. space 1 over straight x dx plus straight c space space space space or space space space straight y over straight x space equals space 2 integral straight x space dx plus straight c space space or space space straight y over straight x equals straight x squared plus straight c space space or space space straight y space equals space straight x cubed plus cx$

Question 85

Wired Faculty App

Solve:
$2 straight x dy over dx plus straight y space equals space 6 straight x cubed space space or space space space dy over dx plus fraction numerator straight y over denominator 2 straight x end fraction equals 3 straight x squared.$

Solution

The given differential equation is

2 straight x dy over dx plus straight y space equals space 6 straight x cubed

dy over dx plus fraction numerator 1 over denominator 2 straight x end fraction straight y space equals space 3 straight x squared

Comparing it with

dy over dx plus straight P space straight y space space equals space straight Q comma space we space get comma

straight P space equals space fraction numerator 1 over denominator 2 straight x end fraction comma space space space straight Q space equals space 3 straight x squared

integral straight P space dx space equals space 1 half integral 1 over straight x dx space equals space 1 half log space straight x space equals space log space square root of straight x therefore space space space straight e to the power of integral space straight P space dx end exponent space equals space straight e to the power of log space square root of straight x end exponent space equals space square root of straight x therefore space space space solution space is space space space space space space space space space straight y. space straight e to the power of integral straight P space dx end exponent space equals space integral space straight Q. space straight e to the power of integral straight P space dx end exponent dx space plus straight c

straight y. space square root of straight x space equals space integral 3 space straight x squared. space square root of straight x space dx space plus space straight c

therefore space space space square root of straight x space space straight y space equals space 3 space integral straight x to the power of 5 over 2 end exponent dx plus straight c space space space space or space space space straight y square root of straight x space equals space 3 fraction numerator straight x to the power of begin display style 7 over 2 end style end exponent over denominator begin display style 7 over 2 end style end fraction plus straight c

straight y square root of straight x space equals space 6 over 7 space straight x to the power of 7 over 2 end exponent plus straight c

which is required solution.

Question 86

Wired Faculty App

Solve $dy over dx plus fraction numerator 1 over denominator straight x space logx end fraction straight y space equals space 1 over straight x.$

Solution

The given differential equation is $dy over dx plus fraction numerator 1 over denominator straight x space logx end fraction. straight y space equals space 1 over straight x$
Comparing it with $dy over dx plus Py space equals space straight Q comma space we space get comma space space straight P space equals space fraction numerator 1 over denominator straight x space logx end fraction comma space straight Q space equals 1 over straight x$
$integral straight P space dx space equals space integral fraction numerator 1 over denominator straight x space logx end fraction dx space equals space integral fraction numerator begin display style 1 over straight x end style dx over denominator log space straight x end fraction space equals space log space left parenthesis log space straight x right parenthesis$
$therefore space space space space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of log space left parenthesis log space straight x right parenthesis end exponent space equals space log space straight x therefore space space space space solution space of space differential space equation space is space space space space space space space space space space space space space straight y space straight e to the power of integral straight P space dx end exponent space equals space integral straight Q. space straight e to the power of integral straight P space dx end exponent space dx space plus space straight c apostrophe space space or space straight y. space log space straight x space equals space integral 1 over straight x. space log space straight x space dx space plus space straight c apostrophe$
or $straight y space log space straight x space equals space integral left parenthesis log space straight x right parenthesis. space 1 over straight x dx plus straight c apostrophe space space or space space straight y space log space straight x space equals space fraction numerator left parenthesis log space straight x right parenthesis squared over denominator 2 end fraction plus straight c apostrophe$
or $2 space straight y space log space straight x space equals space left parenthesis log space straight x right parenthesis squared plus straight c$

Question 87

Wired Faculty App

Solve:
$straight x dy over dx plus straight y space equals space straight x space log space straight x$

Solution

The given differential equation is $straight x dy over dx plus straight y space equals space straight x space log space straight x$ .
or    $dy over dx plus 1 over straight x straight y space equals space log space straight x$
Comparing it with $dy over dx plus Py space equals space straight Q comma space space we space get comma space space straight P space equals space 1 over straight x comma space space straight Q space equals space log space straight x$
$integral straight P space dx space equals space integral 1 over straight x dx space equals space log space straight x comma space space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of logx space equals straight x$
$therefore space space space$ solution of differential equation is
   $straight y. space straight e to the power of integral straight P space dx end exponent space equals space integral space straight Q. space straight e to the power of integral straight P space dx end exponent space dx space plus straight c apostrophe space space or space space straight y. space straight x space equals space integral left parenthesis log space straight x right parenthesis. space straight x space dx space plus space straight c space$
or $straight y. straight x space equals left parenthesis log space straight x right parenthesis. space straight x squared over 2 minus integral 1 over straight x. straight x squared over 2 dx plus straight c apostrophe space space space or space space straight x space straight y space equals space straight x squared over 2 log space straight x space minus space 1 half integral straight x space dx plus straight c apostrophe$
or    $straight x space straight y space equals space straight x squared over 2 log space straight x space minus straight x squared over 4 plus straight c apostrophe space space or space space 4 xy space equals space 2 straight x squared logx space minus space straight x squared plus straight c$

Question 88

Wired Faculty App

Solve:
$straight x dy over dx plus 2 straight y space equals space straight x squared log space straight x$

Solution

The given differential equation is

straight x dy over dx plus 2 straight y space equals space straight x squared space logx

dy over dx plus 2 over straight x straight y space equals space straight x space logx

Comparing it with

dy over dx plus Py space equals space straight Q comma space space we space get comma space space straight P space equals space 2 over straight x comma space space straight Q space equals space straight x space log space straight x

therefore space space space integral straight P space dx space equals space 2 space integral 1 over straight x dx space equals space 2 space log space straight x. space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of 2 space log space straight x end exponent space equals space straight e to the power of log space straight x squared end exponent space equals space straight x squared

Solution of given differential equation is

straight y space straight e to the power of integral straight P space dx end exponent space equals space integral straight Q space straight e to the power of integral straight P space dx end exponent dx plus straight c

straight y. straight x squared space equals space integral straight x space log space straight x. space straight x squared space dx plus straight c

or space space straight x squared. straight y space equals space integral left parenthesis log space straight x right parenthesis. space straight x cubed space dx space plus straight c or space space straight x squared straight y space equals space left parenthesis log space straight x right parenthesis. space straight x to the power of 4 over 4 minus integral 1 over straight x. straight x to the power of 4 over 4 dx plus straight c or space space space straight x squared straight y space equals space straight x to the power of 4 over 4 log space straight x space minus space 1 fourth integral straight x cubed space dx plus straight c or space space space straight x squared straight y space equals space straight x to the power of 4 over 4 log space straight x minus space 1 fourth. space straight x to the power of 4 over 4 plus straight c or space space space straight y space equals straight x squared over 4 log space straight x space minus space 1 over 16 straight x squared plus straight c space straight x to the power of negative 2 end exponent space is space the space required space solution. space

Question 89

Wired Faculty App

Solve:
$straight x space log space straight x space dy over dx plus straight y space equals space 2 over straight x log space straight x$

Solution

The given differential equation is

straight x space log space straight x space dy over dx plus straight y space equals space 2 over straight x log space straight x space space space or space space space dy over dx plus space fraction numerator 1 over denominator straight x space log space straight x end fraction straight y space equals space 2 over straight x squared

Comparing it with

dy over dx plus Py space equals straight Q comma space we space get comma space space straight P space equals fraction numerator 1 over denominator straight x space log space straight x end fraction comma space straight Q space equals 2 over straight x squared

therefore space space space space space space integral straight P space dx space equals space integral fraction numerator 1 over denominator straight x space log space straight x end fraction dx space equals integral fraction numerator begin display style 1 over straight x end style over denominator log space straight x end fraction dx space equals space log space left parenthesis log space straight x right parenthesis therefore space space space space space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of log space left parenthesis log space straight x right parenthesis end exponent space equals space log space straight x.

therefore

solution of given differential equation is

straight y space straight e to the power of integral straight P space dx end exponent space equals space integral straight Q space straight e to the power of integral straight P space dx end exponent dx space plus straight c

straight y space log space straight x space equals space integral 2 over straight x squared log space straight x space dx space plus space straight c

straight y space log space straight x space equals space 2 space integral left parenthesis log space straight x right parenthesis. space straight x squared space dx space plus straight c

or space space space straight y space log space straight x space equals space 2 open square brackets left parenthesis log space straight x right parenthesis. space fraction numerator straight x to the power of negative 1 end exponent over denominator negative 1 end fraction minus integral 1 over straight x. fraction numerator straight x to the power of negative 1 end exponent over denominator negative 1 end fraction dx close square brackets plus straight c or space space straight y space log space straight x space equals space minus fraction numerator 2 space log space straight x over denominator straight x end fraction plus 2 space integral straight x to the power of negative 2 end exponent space dx plus straight c or space space straight y space logx space space equals space minus fraction numerator 2 space log space straight x over denominator straight x end fraction plus 2 fraction numerator straight x to the power of negative 1 end exponent over denominator negative 1 end fraction plus straight c or space space space straight y space logx space equals negative 2 over straight x left parenthesis 1 plus log space straight x right parenthesis space plus space straight c comma space which space is space required space solution.

Question 90

Wired Faculty App

Find one-parameter families of solution curves of the following differential equations:
$straight y apostrophe plus 3 straight y space equals straight e to the power of negative 2 straight x end exponent$

Solution

The given differential equation is

straight y apostrophe plus 3 straight y space equals space straight e to the power of negative 2 straight x end exponent

dy over dx plus 3 straight y space equals space straight e to the power of negative 2 straight x end exponent

Comparing it with

dy over dx plus Py space equals straight Q comma space we space get comma space straight P space equals space 3 comma space straight Q space equals space straight e to the power of negative 2 straight x end exponent

therefore space space space space space integral straight P space dx space equals space 3 space integral space 1 space dx space equals space 3 space straight x. space space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of 3 straight x end exponent

Solution of given differential equation is

straight y space straight e to the power of integral straight P space dx end exponent space equals space integral space straight Q space straight e to the power of integral straight P space dx end exponent dx plus space straight c

straight y space straight e to the power of 3 straight x end exponent space equals space integral space straight e to the power of negative 2 straight x end exponent. space straight e to the power of 3 straight x end exponent space dx plus straight c

or space space space straight y space straight e to the power of 3 straight x end exponent space equals space integral straight e to the power of straight x space dx plus straight c or space space space space straight y space straight e to the power of 3 straight x end exponent space equals space straight e to the power of straight x plus straight c or space space space space space space space straight y space space equals space straight e to the power of negative 2 straight x end exponent plus straight c space straight e to the power of negative 3 straight x end exponent space is space required space solution.

Question 91

Wired Faculty App

Find one-parameter families of solution curves of the following differential equation:
$straight y apostrophe space minus space straight y space equals space cos space straight x$

Solution

The given differential equation is

straight y apostrophe minus straight y space equals space cos space straight x space space space space space space or space space space space space dy over dx minus straight y space equals space cos space straight x

Comparing it with

dy over dx plus Py space equals space straight Q comma space space we space space get space comma space space straight P space equals space minus 1 comma space space space straight Q space equals space cos space straight x space

integral space thin space straight P space dx space equals space minus integral 1 space dx space equals space minus straight x comma space space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of negative straight x end exponent

solution is

space straight y space straight e to the power of integral straight P space dx end exponent space equals space integral straight Q space straight e to the power of integral space straight P space dx end exponent space dx plus straight c

straight y space straight e to the power of negative straight x end exponent space equals space integral space left parenthesis cosx right parenthesis. space straight e to the power of negative straight x end exponent dx plus straight c

therefore space space space space space space space space space space straight y space straight e to the power of negative straight x end exponent space equals space fraction numerator 1 over denominator 1 plus 1 end fraction straight e to the power of negative straight x end exponent space left parenthesis sin space straight x minus space cos space straight x right parenthesis space plus space straight c therefore space space space space space space space space space space space space space space straight y space space equals 1 half left parenthesis sin space straight x space minus space cos space straight x right parenthesis space plus space ce to the power of straight x

which is required solution.

Question 92

Wired Faculty App

Find one-parameter families of solution curves of the following differential equation:
y'+3 y = e^mx(m: a given real number)

Solution

The given differential equation is

dy over dx plus 3 straight y space equals space straight e to the power of mx

Comparing it with

dy over dx plus Py space equals space straight Q comma space space we space get space straight P space equals space 3 comma space space straight Q space equals space straight e to the power of mx

therefore space space space space space space space space space straight e to the power of integral space straight p space dx end exponent space equals straight e to the power of 3 space integral space dx end exponent space space equals space straight e to the power of 3 straight x end exponent therefore space space space space solution space is

straight y space straight e to the power of integral straight P space dx end exponent space equals integral straight Q space straight e to the power of integral straight P space dx end exponent dx plus straight c

straight y space straight e to the power of 3 straight x end exponent space equals space integral straight e to the power of mx. space straight e to the power of 3 straight x end exponent dx plus straight c

straight y space straight e to the power of 3 straight x end exponent space equals space integral straight e to the power of left parenthesis straight m plus 3 right parenthesis straight x end exponent dx space equals space straight c

straight y space straight e to the power of 3 straight x end exponent space equals space fraction numerator straight e to the power of left parenthesis straight m plus 3 right parenthesis straight x end exponent over denominator straight m plus 3 end fraction plus straight c comma space space space if space straight m space plus space 3 space not equal to space 0

straight m plus 3 space equals space 0 comma space space space space space straight i. straight e. comma space straight m space equals negative 3 comma space then space solution space is space

space space space straight y space straight e to the power of 3 straight x end exponent space equals space integral 1 space dx plus straight c

straight y space straight e to the power of 3 straight x end exponent space equals space straight x plus straight c

Question 93

Wired Faculty App

Find one-parameter families of solution curves of the following differential equation:
y' - y = cos 2x

Solution

The given differential equation is

straight y apostrophe minus straight y space equals space cos space 2 straight x space space space space space space space space space space space space space space space space space space space space space space space space space space or space space space dy over dx minus straight y space equals space cos space 2 straight x

Comparing it with

dy over dx plus straight P space straight y space equals straight Q comma space we space get comma space space straight P equals space minus 1 comma space space straight Q space equals space cos space 2 straight x

integral straight P space dx space equals space minus integral 1 space dx space equals space minus straight x comma space space space space space straight I. straight F. space equals space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of straight x

therefore space space space solution space of space differential space equation space is

straight y. space straight e to the power of straight x space equals space integral cos space 2 straight x. space straight e to the power of negative straight x end exponent space dx plus straight c space space or space space space straight y space straight e to the power of negative straight x end exponent space equals space integral straight e to the power of negative straight x end exponent space cos space 2 straight x space dx plus straight c

straight y space straight e to the power of straight x space equals space fraction numerator 1 over denominator square root of left parenthesis negative 1 right parenthesis squared plus left parenthesis 2 right parenthesis squared end root end fraction straight e to the power of negative straight x end exponent space cos space open square brackets 2 straight x minus tan to the power of negative 1 end exponent open parentheses fraction numerator 2 over denominator space space space space space space space space space minus 1 end fraction close parentheses close square brackets plus straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space space integral straight e to the power of ax cos space bx space dx space equals space fraction numerator 1 over denominator square root of straight a squared plus straight b squared end root end fraction straight e to the power of ax space cos space open parentheses bx minus tan to the power of negative 1 end exponent straight b over straight a close parentheses plus straight c close square brackets space space space space space space or space space space space space straight y space straight e to the power of negative straight x end exponent space equals space fraction numerator 1 over denominator square root of 5 end fraction straight e to the power of negative straight x end exponent space cos space left square bracket 2 straight x space minus space tan to the power of negative 1 end exponent left parenthesis negative 2 right parenthesis right square bracket space plus space straight c or space space space space space space space straight y space equals space fraction numerator 1 over denominator square root of 5 end fraction cos space left square bracket 2 straight x space minus tan to the power of negative 1 end exponent left parenthesis negative 2 right parenthesis right square bracket space plus space straight c space straight e to the power of straight x

Question 94

Wired Faculty App

Find one-parameter families of solution curves of the following differential equation:
x y' - y = (x+1) e^-x

Solution

The given differential equation is
x y' - y = (x+1) e^-xor

dy over dx minus 1 over straight x straight y space equals open parentheses 1 plus 1 over straight x close parentheses straight e to the power of negative straight x end exponent

Comparing with

dy over dx plus straight P space straight y space equals space straight Q comma space we space get space straight P space equals space minus 1 over straight x comma space space straight Q space equals space open parentheses 1 plus 1 over straight x close parentheses straight e to the power of negative straight x end exponent

integral straight P space dx space equals negative integral 1 over straight x dx space equals negative log space straight x comma space space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of log space straight x end exponent space equals space straight e to the power of log space straight x to the power of negative 1 end exponent end exponent space equals space straight x to the power of negative 1 end exponent space equals space 1 over straight x

Solution of differential equation is

straight y space straight e to the power of integral straight P space dx end exponent space equals space integral straight Q space straight e to the power of Pdx space end exponent dx plus straight c

straight y.1 over straight x space equals space integral open parentheses 1 plus 1 over straight x close parentheses straight e to the power of negative straight x end exponent. space 1 over straight x dx plus straight c

straight y over straight x equals integral open parentheses 1 over straight x plus 1 over straight x squared close parentheses space straight e to the power of negative straight x end exponent dx plus straight c

or space space straight y over straight x equals integral 1 over straight x. straight e to the power of negative straight x end exponent space dx plus integral 1 over straight x squared straight e to the power of negative straight x end exponent. space dx plus straight c or space space space space straight y over straight x equals 1 over straight x. space fraction numerator straight e to the power of negative straight x end exponent over denominator negative 1 end fraction minus space integral fraction numerator negative 1 over denominator straight x squared end fraction. space fraction numerator straight e to the power of negative straight x end exponent over denominator negative 1 end fraction dx plus integral 1 over straight x squared straight e to the power of negative straight x end exponent dx plus straight c or space space space space straight y over straight x equals negative 1 over xe to the power of straight x minus integral fraction numerator 1 over denominator straight x squared straight e to the power of straight x end fraction dx plus integral fraction numerator 1 over denominator straight x squared straight e to the power of straight x end fraction dx plus straight c or space space space space straight y over straight x equals negative 1 over xe to the power of straight x plus straight c or space space space space space straight y space equals negative straight e to the power of negative straight x end exponent plus cx

Question 95

Wired Faculty App

Find one-parameter families of solution curves of the following differential equation:
x y' + y = x⁴

Solution

The given differential equation is
x y' + y = x⁴or

dy over dx plus 1 over straight x straight y space equals space straight x cubed

Comparing it with

dy over dx plus Py space equals straight Q comma space we space get space straight P space equals 1 over straight x. straight Q space equals space straight x cubed

straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of integral 1 over straight x dx end exponent space equals space straight e to the power of log space straight x end exponent space equals space straight x

Solution of differential equation is

straight y space straight e to the power of integral space straight P space dx end exponent space equals space integral straight Q space straight e to the power of integral straight P space dx end exponent dx plus straight c

straight y. space straight e space equals space integral straight x cubed. space straight x space dx plus straight c space space space space space space or space space space straight x space straight y space equals space integral straight x to the power of 4 dx plus straight c

xy space equals straight x to the power of 5 over 5 plus straight c space space space space space space space space space space space space space space space space space space space or space space space space space straight y space equals space straight x to the power of 4 over 5 plus straight c over straight x

Question 96

Wired Faculty App

Find one-parameter families of solution curves of the following differential equation:
x y' log x + y = log x

Solution

The given differential equation is

left parenthesis straight x space logx right parenthesis dy over dx plus straight y space equals space log space straight x space space or space space dy over dx plus fraction numerator 1 over denominator straight x space log space straight x end fraction straight y space equals space 1 over straight x

Comparing it with

dy over dx plus straight P space straight y space equals space straight Q comma space we space get comma space straight P space equals space fraction numerator 1 over denominator straight x space logx end fraction comma space space straight Q space equals space 1 over straight x

integral straight P space dx space equals space integral fraction numerator 1 over denominator straight x space log space straight x end fraction dx space equals space integral fraction numerator begin display style 1 over straight x end style dx over denominator log space straight x end fraction space equals space log space left parenthesis log space straight x right parenthesis

therefore space space straight e to the power of integral straight P space dx end exponent space equals straight e to the power of log space left parenthesis log space straight x right parenthesis end exponent space equals space log space straight x

therefore

solution of differential equation is

straight y space straight e to the power of integral Pdx end exponent space equals space integral straight Q. space straight e to the power of integral straight P end exponent dx space plus space straight c apostrophe space space space or space space straight y. space log space straight x space equals space integral 1 over straight x. space log space straight x space dx plus space straight c apostrophe

straight y space log space straight x space equals integral left parenthesis log space straight x right parenthesis. space 1 over straight x dx plus straight c apostrophe space space or space space straight y space log space straight x space equals space fraction numerator left parenthesis log space straight x right parenthesis squared over denominator 2 end fraction plus straight c apostrophe

2 straight y space log space straight x space equals space left parenthesis log space straight x right parenthesis squared plus straight c

Question 97

Wired Faculty App

Find one-parameter families of solution curves of the following differential equation:
$dy over dx minus fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction comma space space straight y space equals space straight x squared plus 2$

Solution

The given differential equation is

dy over dx minus fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction straight y space equals space straight x squared plus 2

Comparing it with

dy over dx plus Py space equals space straight Q comma space we space get comma space straight P space equals negative fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction comma space space straight Q space equals straight x squared plus 2

therefore space space space space space integral straight P space dx space equals space minus integral fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction dx equals negative log space left parenthesis 1 plus straight x squared right parenthesis therefore space space space space space straight e to the power of integral space Pdx end exponent space equals space straight e to the power of negative log space left parenthesis 1 plus straight x squared right parenthesis end exponent space equals space straight e to the power of log space left parenthesis 1 plus straight x squared right parenthesis to the power of negative 1 end exponent end exponent space equals space fraction numerator 1 over denominator 1 plus straight x squared end fraction

therefore

solution of differential equation is

straight y space straight e to the power of integral straight P space dx end exponent space equals integral straight Q. space straight e to the power of integral straight P space dx end exponent dx space plus space straight c space space space or space space straight y. space fraction numerator 1 over denominator 1 plus straight x squared end fraction space equals space integral left parenthesis straight x squared plus 2 right parenthesis. space fraction numerator 1 over denominator straight x squared plus 1 end fraction dx plus straight c

fraction numerator straight y over denominator 1 plus straight x squared end fraction space equals integral fraction numerator straight x squared plus 2 over denominator straight x squared plus 1 end fraction dx plus straight c

fraction numerator straight y over denominator 1 plus straight x squared end fraction space equals space integral open parentheses 1 plus fraction numerator 1 over denominator straight x squared plus 1 end fraction close parentheses dx plus straight c

fraction numerator straight y over denominator 1 plus straight x squared end fraction space equals space straight x plus tan to the power of negative 1 end exponent straight x plus straight c

Question 98

Wired Faculty App

Find one-parameter families of solution curves of the following differential equation:
$straight y apostrophe plus straight y space cosx space equals space straight e to the power of sin space straight x end exponent space cos space straight x$

Solution

The given differential equation is

straight y apostrophe plus straight y space cosx space equals space straight e to the power of sinx space cos space straight x space space space space space space space or space space space space space dy over dx plus left parenthesis cos space straight x right parenthesis. space straight y space equals space straight e to the power of sin space straight x end exponent space space cosx space

Comparing it with

dy over dx plus straight P space straight y space equals space straight Q comma space we space get space straight P space equals cosx comma space straight Q space equals space straight e to the power of sinx. space cosx

therefore space space space space space space space space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of integral space cosx space dx end exponent space equals straight e to the power of sin space straight x end exponent

Solution of given differential equation is

straight y space. straight e to the power of integral straight P space dx end exponent space equals space integral straight Q. space straight e to the power of integral straight P space dx end exponent dx plus straight c

straight y space straight e to the power of sinx space equals space integral straight e to the power of sinx space cosx. space straight e to the power of sinx dx plus straight c

straight y space straight e to the power of sinx space equals space integral straight e to the power of 2 space sinx end exponent space cosx space dx plus straight c

...(1)
Let I =

integral straight e to the power of 2 sinx end exponent space cos space straight x space dx

Put sinx = t,

therefore space space space cos space straight x space dx space equals dt

therefore

I =

integral straight e to the power of 2 straight t end exponent dt space equals space straight e to the power of 2 straight t end exponent over 2 space equals 1 half straight e to the power of 2 sinx end exponent

therefore space space from space left parenthesis 1 right parenthesis comma space space space ye to the power of sinx space equals space 1 half straight e to the power of 2 sinx end exponent plus straight c comma space which space is space required space solution.

Question 99

Wired Faculty App

Find one-parameter families of solution curves of the following differential equation:
$left parenthesis 1 plus straight x squared right parenthesis space dy space plus space 2 xy space dx space equals space cot space straight x space dx space space space space left parenthesis straight x not equal to 0 right parenthesis$

Solution

The given differential equation is

left parenthesis 1 plus straight x squared right parenthesis space dy plus space 2 xy space dx space equals space cot space straight x space dx

open parentheses 1 plus straight x squared close parentheses dy over dx plus 2 space straight x space straight y space equals space space cot space straight x

dy over dx plus fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction. space straight y space equals space fraction numerator cot space straight x over denominator 1 plus straight x squared end fraction

Comparing it with

dy over dx plus Py space equals straight Q comma space we space get comma space straight P equals space fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction comma space space straight Q space equals space fraction numerator cot space straight x over denominator 1 plus straight x squared end fraction

integral straight P space dx space equals space integral fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction dx space equals space log space left parenthesis 1 plus straight x squared right parenthesis. space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of log space left parenthesis 1 plus straight x right parenthesis end exponent space equals space 1 plus straight x squared

Solution of given differential equation is

ye to the power of integral straight P space dx end exponent space equals space integral straight Q space straight e to the power of integral straight P space dx end exponent dx plus straight c

straight y left parenthesis 1 plus straight x squared right parenthesis space equals space integral open parentheses fraction numerator cot space straight x over denominator 1 plus straight x squared end fraction close parentheses. space left parenthesis 1 plus straight x squared right parenthesis space dx plus straight c

or space space space straight y left parenthesis 1 plus straight x squared right parenthesis space equals space integral cot space straight x space dx plus straight c or space space straight y left parenthesis 1 plus straight x squared right parenthesis space equals space log space open vertical bar sin space straight x close vertical bar space plus straight c space or space space space space space space space space space space straight y space equals space left parenthesis 1 plus straight x squared right parenthesis to the power of negative 1 end exponent space log space open vertical bar sin space straight x close vertical bar space plus space straight c space left parenthesis 1 plus straight x squared right parenthesis to the power of negative 1 end exponent which space is space required space solution. space

Question 100

Wired Faculty App

Find one-parameter families of solution curves of the following differential equation:
$straight x dy over dx plus straight y minus straight x plus straight x space straight y space cot space straight x space equals space 0 space space space left parenthesis straight x not equal to 0 right parenthesis$

Solution

The given differential equation is
   $straight x dy over dx plus straight y minus straight x plus straight x space straight y space cot space straight x space equals space 0$
or $straight x dy over dx plus left parenthesis 1 plus space straight x space cot space straight x right parenthesis space straight y space equals straight x$
or    $dy over dx plus open parentheses 1 over straight x plus cot space straight x close parentheses. straight y space equals space 1$
Comparing it with $dy over dx plus Py space equals straight Q comma space space we space get comma space space space straight P space equals space 1 over straight x plus cot space straight x comma space space straight Q space equals space 1$
$integral straight P space dx space equals space integral open parentheses 1 over straight x plus cot space straight x close parentheses space dx space equals space log space straight x plus space log space sinx space equals space log space left parenthesis straight x space sin space straight x right parenthesis$
$therefore space space space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of log space left parenthesis straight x space sinx right parenthesis end exponent space equals space space straight x space sin space straight x$
Solution of given differential equation is
$straight y space straight e to the power of integral straight P space dx end exponent space equals space integral straight Q space straight e to the power of integral straight P space dx end exponent dx plus straight c$
or $straight y. space left parenthesis straight x space sin space straight x right parenthesis space equals space integral space 1. space straight x space sinx space dx space plus straight c$
or    $straight x space straight y space sin space straight x space equals integral space straight x space sin space straight x space dx space plus space straight c$
or $straight x space straight y space sin space straight x space equals straight x. space left parenthesis negative cos space straight x right parenthesis minus space integral 1. space left parenthesis negative cos space straight x right parenthesis space dx plus space straight c$
or    $straight x space straight y space sin space straight x space equals space minus straight x space cos space straight x space space plus space integral space cosx space dx plus straight c$
or    $straight x space straight y space sin space straight x space equals space minus straight x space cos space straight x space plus space sin space straight x space plus straight c$
or    $straight y equals negative cot space straight x plus 1 over straight x plus fraction numerator straight c over denominator straight x space sin space straight x end fraction$
which is required solution.

Question 101

Wired Faculty App

Solve: $open parentheses fraction numerator straight e to the power of negative 2 square root of straight x end exponent over denominator square root of straight x end fraction minus fraction numerator straight y over denominator square root of straight x end fraction close parentheses dx over dy equals 1 comma space space space straight x not equal to space 0.$

Solution

The given differential equation is

open parentheses fraction numerator straight e to the power of negative 2 square root of straight x end exponent over denominator square root of straight x end fraction minus fraction numerator straight y over denominator square root of straight x end fraction close parentheses space dx over dy space equals 1 space space space space space space space space space or space space space space space dy over dx space equals fraction numerator straight e to the power of negative 2 square root of straight x end exponent over denominator square root of straight x end fraction minus fraction numerator straight y over denominator square root of straight x end fraction

dy over dx plus fraction numerator 1 over denominator square root of straight x end fraction straight y space equals space fraction numerator straight e to the power of negative 2 square root of straight x end exponent over denominator square root of straight x end fraction

Comparing it with

dy over dx plus straight P space straight y space equals space straight Q comma space space we space get comma

straight P space equals fraction numerator 1 over denominator square root of straight x end fraction space space straight Q space equals space fraction numerator straight e to the power of negative 2 square root of straight x end exponent over denominator square root of straight x end fraction

therefore space space space space space integral straight P space dx space equals space integral fraction numerator 1 over denominator square root of straight x end fraction dx space equals space integral straight x to the power of negative 1 half end exponent dx space equals space fraction numerator straight x to the power of begin display style 1 half end style end exponent over denominator begin display style 1 half end style end fraction space equals space 2 square root of straight x

therefore space space space space space space space straight I. straight F. space equals space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of 2 square root of straight x end exponent

therefore space space space space solution space of space differential space equation space is

straight y. space straight e to the power of integral Pdx end exponent space equals space integral straight Q space straight e to the power of integral straight P space dx end exponent dx plus straight c

straight y space straight e to the power of 2 square root of straight x end exponent space equals integral fraction numerator straight e to the power of negative 2 square root of straight x end exponent over denominator square root of straight x end fraction space straight e to the power of 2 square root of straight x end exponent dx plus straight c

straight y space straight e to the power of 2 square root of straight x end exponent space equals space integral straight x to the power of negative 1 half end exponent dx plus straight c

ye to the power of 2 square root of straight x end exponent space equals space fraction numerator straight x to the power of begin display style 1 half end style end exponent over denominator begin display style 1 half end style end fraction plus straight c space space space space space space space or space space space space straight y space straight e to the power of 2 square root of straight x end exponent space space equals space 2 square root of straight x plus straight c

Question 102

Wired Faculty App

Solve the following differential equation:
$dy over dx space plus 2 space straight y space tan space straight x space equals space sin space straight x$

Solution

The given differential equation is

dy over dx plus 2 space straight y space tanx space equals space sin space straight x

Comparing it with

dy over dx plus Py space equals space straight Q comma

we get,

straight P space equals space 2 space tanx space space space straight Q space equals space sin space straight x

therefore space space space space integral straight P space dx space equals space 2 space integral space tan space straight x space dx space equals space minus 2 space log space cosx space space equals space log space left parenthesis cosx right parenthesis squared

therefore space space space space space space space space space space straight I. straight F. space equals space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of log space left parenthesis cosx space right parenthesis to the power of negative 2 end exponent end exponent space equals space left parenthesis cos space straight x right parenthesis to the power of negative 2 end exponent space equals space fraction numerator 1 over denominator cos squared straight x end fraction

therefore space space space space space space solution space of space differential space equation space is

straight y. space fraction numerator 1 over denominator cos squared straight x end fraction space equals space integral sin space straight x. space space fraction numerator 1 over denominator cos squared straight x end fraction dx plus straight c

open square brackets therefore space space space straight y space straight e to the power of integral straight P space dx end exponent space equals space integral straight Q space straight e to the power of integral straight P space dx end exponent dx plus straight c close square brackets

fraction numerator straight y over denominator cos squared straight x end fraction space equals space integral space tanx space secx space dx plus straight c

fraction numerator straight y over denominator cos squared straight x end fraction space equals space secx plus straight c

straight y equals cosx plus straight c space cos squared straight x

which is required solution.

Question 103

Wired Faculty App

For the differential equation, find the general solution:
$dy over dx plus 2 straight y space equals sin space straight x$

Solution

The given different equation is

dy over dx plus 2 straight y space equals space sin space straight x

Comparing it with

dy over dx plus straight P space straight y space space equals space straight Q comma space we space get comma space

P = 2, Q = sin x

rightwards double arrow space

integral straight P space dx space equals space integral space 2 space dx space equals space 2 straight x

straight I. straight F. space equals space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of 2 straight x end exponent

therefore

solution of differential equation is

straight y. space straight e to the power of 2 straight x end exponent space equals space integral space sinx. space straight e to the power of 2 straight x end exponent space dx plus straight c space space space space space space space space space space open square brackets because space space space straight y space straight e to the power of integral Pdx end exponent space equals space integral straight Q space straight e to the power of integral straight P space dx end exponent dx plus straight c close square brackets

straight y. space straight e to the power of 2 straight x end exponent space equals space fraction numerator 1 over denominator square root of 4 plus 1 end root end fraction straight e to the power of 2 straight x end exponent space sin space open parentheses straight x minus tan to the power of negative 1 end exponent 1 half close parentheses plus straight c

open square brackets because space space integral space space straight e space to the power of ax space sin space bx space dx space equals space fraction numerator 1 over denominator square root of straight a squared plus straight b squared end root end fraction space straight e to the power of ax. space sin space open parentheses bx minus tan to the power of negative 1 end exponent straight b over straight a close parentheses close square brackets

straight y equals space fraction numerator 1 over denominator square root of 5 end fraction sin space open parentheses straight x minus tan to the power of negative 1 end exponent 1 half close parentheses plus ce to the power of negative 2 straight x end exponent

Question 104

Wired Faculty App

Solve the following differential equation:
$sin space straight x space dy over dx plus straight y space cosx space equals space cos space straight x space sin squared straight x$

Solution

The given differential equations is

sinx dy over dx plus straight y space cosx space equals space cosx space sin squared straight x

dy over dx plus straight y space cotx space equals space cos space straight x space sinx

Comparing it with

dy over dx plus Py space equals space straight Q comma

we get.

straight P space equals cot space straight x comma space space space straight Q space equals space space cosx space sinx

therefore space space space space integral space straight P space dx space equals space integral space cot space straight x space dx space equals space log space sinx

therefore space space space straight I. straight F. space equals space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of log space sinx end exponent space equals space sin space straight x

therefore space space space solution space of space differential space equation space is

straight y. space sinx space space equals space integral space cosx. space sinx. space sinx space dx space plus space straight c space space space space open square brackets because space space straight y space straight e to the power of integral straight P space dx end exponent space equals space integral straight Q space straight e to the power of integral straight P space dx end exponent dx plus straight c close square brackets

straight y space sinx space equals space integral sin squared straight x space cosx space dx plus straight c

straight y space sinx space equals space fraction numerator sin cubed straight x over denominator 3 end fraction plus straight c

straight y equals 1 third sin squared straight x plus straight c space cosec space straight x.

Question 105

Wired Faculty App

Find a one parameter family of solutions of each of the following differential equation:
$straight y apostrophe cos squared straight x space equals space tanx minus space straight y$

Solution

The given differential equation is

dy over dx. space cos squared straight x space equals space tanx minus straight y space space space space space space space or space space space dy over dx equals tan space straight x space sec squared straight x space minus space straight y space sec squared straight x

dy over dx plus left parenthesis sec squared straight x right parenthesis. space straight y space equals space tan space straight x space sec squared straight x

Comparing it with

dy over dx plus straight P space straight y space equals space straight Q comma space space we space get comma space straight P space equals space sec squared straight x comma space space straight Q space equals tan space straight x space sec squared straight x

integral straight P space dx space equals space integral sec squared straight x space dx space equals space tan space straight x comma space space space space space therefore space space space space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of tanx

therefore

solution of differential equation is

straight y space straight e to the power of integral Pdx end exponent space equals space integral straight Q space straight e to the power of integral straight P space dx end exponent dx plus straight c

straight y space straight e to the power of tanx space equals space integral tanx space sec squared straight x space. space straight e to the power of tanx dx plus straight c

...(1)
Let

straight I space equals space integral tanx space sec squared straight x. space space straight e to the power of tanx dx

Put tanx = t,

sec squared dx space equals space dt

therefore space space space space space space space space space space space space space space straight I space equals space integral straight t. space straight e to the power of straight t space dt space equals space straight t space straight e to the power of straight t space minus space integral 1. space straight e to the power of straight t space dt space equals space straight t space straight e to the power of straight t space minus space straight e to the power of straight t space equals space left parenthesis straight t minus 1 right parenthesis space straight e to the power of straight t space equals space left parenthesis tanx space minus space 1 right parenthesis straight e to the power of tanx

therefore space space space space from space left parenthesis 1 right parenthesis comma space space space straight y space straight e to the power of tanx space equals space left parenthesis tanx minus 1 right parenthesis space straight e to the power of tanx plus straight c

straight y equals tan space straight x minus 1 space plus space straight c space straight e to the power of negative tanx end exponent

which is required solution.

Question 106

Wired Faculty App

Find a one parameter family of solutions of each of the following differential equation:
$dy over dx equals negative fraction numerator straight x plus straight y space cosx over denominator 1 plus sinx end fraction$

Solution

The given differential equation is

dy over dx equals negative fraction numerator straight x plus straight y space cosx over denominator 1 plus sinx end fraction space space space space space or space space space dy over dx space equals negative fraction numerator straight x over denominator 1 plus sinx end fraction minus straight y fraction numerator cosx over denominator 1 plus sinx end fraction

dy over dx plus open parentheses fraction numerator cos space straight x over denominator 1 plus sinx end fraction close parentheses straight y space equals space minus fraction numerator straight x over denominator 1 plus sinx end fraction

Comparing it with

dy over dx plus Py space equals space straight Q comma space we space get comma

straight P space equals space fraction numerator cosx over denominator 1 plus sinx end fraction comma space space space straight Q space equals negative fraction numerator straight x over denominator 1 plus sinx end fraction

integral straight P space dx space equals space integral fraction numerator cosx over denominator 1 plus sinx end fraction dx space equals space log space left parenthesis 1 plus sinx right parenthesis

therefore space space space space space space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of log space left parenthesis 1 plus space sinx right parenthesis end exponent space equals space log space left parenthesis 1 plus sin space straight x right parenthesis

therefore

solution of differential equation is

straight y space straight e to the power of integral Pdx space end exponent space equals space integral straight Q space straight e to the power of integral Pdx end exponent plus straight c

straight y left parenthesis 1 plus sin space straight x right parenthesis space equals space integral fraction numerator negative straight x over denominator 1 plus sinx end fraction left parenthesis 1 plus sinx right parenthesis space dx plus straight c

straight y space straight y left parenthesis 1 plus sinx right parenthesis space equals space minus integral straight x space dx plus straight c.

straight y left parenthesis 1 plus sinx right parenthesis space equals space minus straight x squared over 2 plus straight c.

Question 107

Wired Faculty App

Solve:
$straight y apostrophe minus 2 straight y space equals space cos space 3 straight x$

Solution

The given differential equation is $dy over dx minus 2 straight y space equals space cos space 3 straight x$
Comparing it with $dy over dx plus straight P space straight y space equals space straight Q comma space space we space get comma space space straight P space equals space minus 2 comma space space straight Q space equals space cos space 3 straight x$
$therefore space space space space space space space integral space straight P space dx space equals space integral left parenthesis negative 2 right parenthesis space dx space equals space minus 2 space straight x comma space space straight I. straight F. space equals space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of negative 2 straight x end exponent$
$therefore$ solution of differential equation is
$straight y. space straight e to the power of negative 2 straight x end exponent space equals space integral cos space 3 straight x. space straight e to the power of negative 2 straight x end exponent space dx plus straight c$ ...(1)
Let I = $integral straight e to the power of negative 2 straight x end exponent space cos space 3 straight x space dx space equals space straight e to the power of negative 2 straight x end exponent. space fraction numerator sin space 3 straight x over denominator 3 end fraction minus integral straight e to the power of negative 2 straight x end exponent left parenthesis negative 2 right parenthesis. space space fraction numerator sin space 3 straight x over denominator 3 end fraction dx$
(Integrating by parts)
$equals space 1 third straight e to the power of negative 2 straight x end exponent space sin space 3 straight x space plus space 2 over 3 integral straight e to the power of negative 2 straight x end exponent space sin space 3 straight x space dx equals space 1 third straight e to the power of negative 2 straight x end exponent space sin space 3 straight x space plus space 2 over 3 space open square brackets straight e to the power of negative 2 straight x end exponent space space open parentheses fraction numerator negative cos space 3 straight x over denominator 3 end fraction close parentheses minus space integral straight e to the power of negative 2 straight x end exponent space left parenthesis negative 2 right parenthesis. space open parentheses fraction numerator negative cos space 3 straight x over denominator 3 end fraction close parentheses close square brackets dx equals space 1 third straight e to the power of negative 2 straight x end exponent space sin space space 3 straight x space minus space 2 over 9 straight e to the power of negative 2 straight x end exponent space cos space 3 straight x space minus 4 over 9 space integral straight e to the power of negative 2 straight x end exponent space cos space 3 straight x space dx$
$therefore space space space straight I space equals space 1 over 9 straight e to the power of negative 2 straight x end exponent left parenthesis space 3 space sin space 3 straight x space space minus 2 space cos space 3 straight x right parenthesis space minus space 4 over 9 straight I space space or space space 13 over 9 straight I space equals space 1 over 9 straight e to the power of negative 2 straight x end exponent left parenthesis 3 space sin space 3 straight x space minus space 2 space cos space 3 straight x right parenthesis$
$rightwards double arrow space space space space space straight I space equals space space 1 over 13 straight e to the power of negative 2 straight x end exponent left parenthesis 3 space sin space 3 straight x minus space 2 space cos space 3 straight x right parenthesis$
$therefore space space space space from space left parenthesis 1 right parenthesis comma space solution space is$
$ye to the power of negative 2 straight x end exponent minus 1 over 13 straight e to the power of negative 2 straight x end exponent left parenthesis 3 space sin space 3 straight x minus space 2 space cos space 3 straight x right parenthesis space plus straight c$
or $straight y equals ee to the power of 2 straight x end exponent space plus space 1 over 13 left parenthesis 3 space sin space 3 straight x space minus space 2 space cos space 3 straight x right parenthesis$

Question 108

Wired Faculty App

Solve:
$straight y apostrophe plus straight y space equals space sin space straight x$

Solution

The given differential equation is $dy over dx plus straight y space equals space sin space straight x$
Comparing it with $dy over dx plus straight P space straight y space space equals straight Q comma space space we space get space comma space straight P space equals space 1 comma space straight Q equals space sin space straight x$
$therefore space space space space space space space integral straight P space dx space equals space integral 1 space dx space equals space straight x comma space space space space straight I. straight F. space equals space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of straight x$
$therefore space$ solution of differential equation is
$straight y. space straight e to the power of straight x space equals integral sinx. space straight e to the power of straight x space dx plus straight c space space space or space space space straight y. space straight e to the power of straight x space equals space fraction numerator 1 over denominator square root of 1 plus 1 end root end fraction straight e to the power of straight x sin space open parentheses straight x minus tan to the power of negative 1 end exponent 1 over 1 close parentheses plus straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space space integral straight e to the power of ax space sin space bx space equals space minus fraction numerator 1 over denominator square root of straight a squared plus straight b squared end root end fraction straight e to the power of ax space sin space open parentheses bx minus tan to the power of negative 1 end exponent straight b over straight a close parentheses close square brackets or space space space space straight y space equals fraction numerator 1 over denominator square root of 2 end fraction sin space open parentheses straight x minus straight pi over 4 close parentheses plus space ce to the power of negative straight x end exponent space space space space space space space$

Question 109

Wired Faculty App

Solve:
$dy over dx plus 2 straight y space equals space sin space 3 straight x$

Solution

The given differential equation is

dy over dx plus 2 straight y space equals sin space 3 straight x

Comparing it with

dy over dx plus Py space equals space straight Q comma space we space get comma

P = 2, Q = sin 3x

integral straight P space dx space equals space integral 2 space dx space equals space 2 straight x

straight I. straight F. space equals space straight e to the power of integral Pdx end exponent space equals space straight e to the power of 2 straight x end exponent

therefore space space space solution space of space differential space equation space is

straight y space straight e to the power of integral straight P space dx end exponent space equals space integral straight Q space straight e to the power of integral straight P space dx end exponent dx plus straight c space space space space space or space space space straight y space straight e to the power of 2 straight x end exponent space equals space integral straight e to the power of 2 straight x end exponent space sin space 3 straight x space dx space plus straight c

therefore space space space space ye to the power of 2 straight x end exponent space equals space fraction numerator 1 over denominator square root of 4 plus 9 end root end fraction space straight e to the power of 2 straight x end exponent space sin space open parentheses 3 straight x minus tan to the power of negative 1 end exponent 3 over 2 close parentheses space plus straight c

open square brackets because space space space integral straight e to the power of ax space sin space bx space dx space equals space fraction numerator 1 over denominator square root of straight a squared plus straight b squared end root end fraction straight e to the power of ax space sin open parentheses bx minus tan to the power of negative 1 end exponent straight b over straight a close parentheses close square brackets

straight y equals space fraction numerator 1 over denominator square root of 13 end fraction sin space open parentheses 3 straight x minus tan to the power of negative 1 end exponent 3 over 2 close parentheses plus space ce to the power of negative 2 straight x end exponent

which is required solution.

Question 110

Wired Faculty App

Solve:
$dy over dx plus 2 straight y space equals space sin space 5 straight x$

Solution

The given differential equation is

dy over dx plus 2 straight y space equals space sin space 5 straight x

Comparing it with

dy over dx plus Py space equals space straight Q comma space we space get comma

P = 2, Q = sin 5x

integral straight P space dx space equals space integral 2 space dx space equals 2 straight x

straight I. straight F. space equals space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of 2 straight x end exponent

therefore

solution of differential equation is

straight y space straight e to the power of integral straight P space dx end exponent space equals space integral straight Q space straight e to the power of integral straight P space dx end exponent dx plus straight c

straight y space straight e to the power of 2 straight x end exponent space equals space integral space sin space 5 straight x. space straight e to the power of 2 straight x end exponent plus straight c

ye to the power of 2 straight x end exponent space equals space fraction numerator 1 over denominator square root of 4 plus 25 end root end fraction straight e to the power of 2 straight x end exponent space sin space open parentheses 5 straight x minus tan to the power of negative 1 end exponent 5 over 2 close parentheses plus straight c space space space space space space space space space space space space space space space space space

open square brackets because space space integral straight e to the power of ax sin space bx space dx space equals space fraction numerator 1 over denominator square root of straight a squared plus straight b squared end root end fraction space straight e to the power of ax space sin space open parentheses bx minus tan to the power of negative 1 end exponent straight b over straight a close parentheses close square brackets

straight y space equals space fraction numerator 1 over denominator square root of 29 end fraction sin space open parentheses 5 straight x minus space tan to the power of negative 1 end exponent 5 over 2 close parentheses plus space straight c space straight e to the power of negative 2 straight x end exponent

Question 111

Wired Faculty App

Solve:
$dy over dx plus 2 straight y space equals space cos space 3 straight x$

Solution

The given differential equation is

dy over dx plus 2 straight y space equals space cos space 3 straight x

Comparing it with

dy over dx plus straight P space straight y space equals space straight Q comma

we get,

straight P space equals 2 comma space space straight Q space equals space cos space 3 straight x

integral straight P space dx space equals space integral space 2 space dx space equals 2 straight x

straight I. straight F. space equals straight e to the power of integral straight P space dx end exponent space equals straight e to the power of 2 straight x end exponent

Solution of given differential equation is

ye to the power of integral straight P space dx end exponent space equals space integral straight Q space straight e to the power of integral straight P space dx end exponent dx plus straight c

ye to the power of 2 straight x end exponent space equals space integral straight e to the power of 2 straight x end exponent space cos space 3 straight x space dx space plus straight c

ye to the power of 2 straight x end exponent space equals space fraction numerator 1 over denominator square root of 4 plus 9 end root end fraction straight e to the power of 2 straight x end exponent space cos space open parentheses 3 straight x minus tan to the power of negative 1 end exponent 3 over 2 close parentheses plus straight c

open square brackets because space space space integral straight e to the power of ax cos space bxdx space equals space fraction numerator 1 over denominator square root of straight a squared plus straight b squared end root end fraction straight e to the power of ax cos space open parentheses bx minus tan to the power of negative 1 end exponent straight b over straight a close parentheses close square brackets

therefore space space space space space space space straight y space equals fraction numerator 1 over denominator square root of 13 end fraction cos space open parentheses 3 straight x minus tan to the power of negative 1 end exponent 3 over 2 close parentheses plus straight c space straight e to the power of negative 2 straight x end exponent

which is required solution.

Question 112

Wired Faculty App

Solve:
$dy over dx plus 3 straight y space equals space sin space 2 straight x$

Solution

The given differential equation is

dy over dx plus 3 straight y space equals space sin space 2 straight x

Comparing it with

dy over dx plus straight P space straight y space equals space straight Q comma space we space get comma space

straight P space equals space 3 comma space space space straight Q space equals sin space 2 straight x

integral straight P space dx space equals space integral 3 space dx space equals space 3 straight x

straight I. straight F. space equals straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of 3 straight x end exponent

Solution of given differential equation is

ye to the power of integral straight P space dx end exponent space equals space integral straight Q space straight e to the power of integral straight P space dx end exponent dx plus straight c

straight y space straight e to the power of 3 straight x end exponent space equals space integral straight e to the power of 3 straight x end exponent space space sin space 2 straight x space dx space plus straight c

ye to the power of 3 straight x end exponent space equals space fraction numerator 1 over denominator square root of 9 plus 4 end root end fraction straight e to the power of 3 straight x end exponent space sin space open parentheses 2 straight x minus tan to the power of negative 1 end exponent 2 over 3 close parentheses plus straight c

open square brackets because space space space integral straight e to the power of ax space sin space bx space dx equals space fraction numerator 1 over denominator square root of straight a squared plus straight b squared end root end fraction straight e to the power of ax space sin space open parentheses bx minus tan to the power of negative 1 end exponent straight b over straight a close parentheses close square brackets

therefore space space space space space straight y space equals space fraction numerator 1 over denominator square root of 13 end fraction sin space open parentheses 2 straight x minus tan to the power of negative 1 end exponent 2 over 3 close parentheses plus space straight c space straight e to the power of negative 3 straight x end exponent

which is required solution.

Question 113

Wired Faculty App

Solve:
$dy over dx plus 3 straight y space equals space cos space 2 straight x$

Solution

The given differential equation is

dy over dx plus 3 straight y space equals space cos space 2 straight x

Comparing it with

dy over dx plus straight P space straight y space equals straight Q comma

we get,
P = 3, Q = cos 2x

integral straight P space dx space equals space integral 3 space dx space equals space 3 straight x

straight I. straight F. space equals space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of 3 straight x end exponent

Solution of given differential equation is

straight y. space straight e to the power of integral straight P space dx end exponent space equals space integral Qe to the power of integral straight P space dx end exponent dx plus straight c

straight y space. straight e to the power of 3 straight x end exponent space equals space integral straight e to the power of 3 straight x end exponent space cos space 2 straight x space dx plus straight c

ye to the power of 3 straight x end exponent space equals space fraction numerator 1 over denominator square root of 9 plus 4 end root end fraction space straight e to the power of 3 straight x end exponent space cos space open parentheses 2 straight x minus tan to the power of negative 1 end exponent 2 over 3 close parentheses plus straight c

ye to the power of 3 straight x end exponent space equals space fraction numerator 1 over denominator square root of 13 end fraction straight e to the power of 3 straight x end exponent space cos space open parentheses 2 straight x minus tan to the power of negative 1 end exponent 2 over 3 close parentheses plus straight c

therefore

straight y space equals fraction numerator 1 over denominator square root of 13 end fraction cos space open parentheses 2 straight x minus tan to the power of negative 1 end exponent 2 over 3 close parentheses plus space straight c space straight e to the power of negative 3 straight x end exponent

is the required solution.

Question 114

Wired Faculty App

Solve:
$dy over dx plus straight y space equals space cos space 2 straight x$

Solution

The given differential equation is $dy over dx plus straight y space equals space cos space 2 straight x$
Comparing it with $dy over dx plus space straight P space straight y space equals space space straight Q comma space space we space get comma space space space straight P space equals space 1 comma space space space straight Q space equals space cos space 2 straight x$
$integral straight P space dx space equals space integral 1. space dx space equals space straight x comma space space space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of straight x$
$therefore$ solution of differential equation is
   $straight y. straight e to the power of integral straight P space dx end exponent space equals space integral straight Q. space straight e to the power of integral straight P space dx end exponent dx plus straight c$
or    $straight y. straight e to the power of straight x space space equals space integral space cos space 2 straight x. space straight e to the power of straight x space dx plus space straight c space space space or space space straight y space straight e to the power of straight x space equals space fraction numerator 1 over denominator 1 plus 4 end fraction straight e to the power of straight x left parenthesis cos space 2 straight x plus 2 space sin space 2 straight x right parenthesis space plus straight c$
or    $straight y space equals space 1 fifth left parenthesis cos space straight x plus 2 space sin space 2 straight x right parenthesis space plus space straight c space straight e to the power of straight x$

Question 115

Wired Faculty App

Solve:
$dy over dx plus straight y space secx space equals space tan space straight x$

Solution

The given differential equation is $dy over dx plus straight y space secx space equals space tan space straight x$
Comparing it with $dy over dx plus Py space equals space straight Q comma space we space get comma space straight P space equals space secx comma space space straight Q space equals tan space straight x$
   $integral straight P space dx space equals space integral secx space dx space equals space log space left parenthesis secx plus tanx right parenthesis$
$therefore space space space space straight I. straight F. space equals space straight e to the power of integral straight P space dx end exponent space equals straight e to the power of log space left parenthesis secx plus tan space straight x right parenthesis space end exponent space equals space secx space plus tanx$
$therefore$ solution of given equation is
   $straight y left parenthesis secx plus tanx right parenthesis space equals space integral tanx space left parenthesis secx plus tanx right parenthesis space dx plus straight c space open square brackets because space space straight y space straight e to the power of integral Pdx end exponent space equals space integral Qe to the power of integral straight P space dx end exponent plus straight c close square brackets$
   $space equals space integral secx space tan space straight x space dx space plus space integral left parenthesis sec squared straight x minus 1 right parenthesis space dx plus straight c$
or    $straight y space left parenthesis secx space plus space tanx right parenthesis space equals space secx plus tanx minus straight x plus straight c space space space space space space space space space space space space open parentheses straight x not equal to space an space odd space multiple space of space straight pi over 2 close parentheses$

Question 116

Wired Faculty App

Solve:
$dy over dx plus straight y space cotx space equals space straight x squared space cotx space plus space 2 straight x$

Solution

The given differential equation is

dy over dx plus straight y space cotx space equals straight x squared cotx space plus space 2 straight x

Comparing it with

dy over dx plus Py space equals space straight Q comma space space we space get comma

P = cot x,

straight Q space equals straight x squared space cot space straight x space plus space 2 straight x

integral straight P space dx space equals space integral cot space straight x space dx space equals space log space sinx. space straight e to the power of integral Pdx end exponent space equals space straight e to the power of log space sinx end exponent space equals space sin space straight x space

therefore

solution of differential equation is

straight y. space sin space straight x space equals space integral left parenthesis straight x squared space cot space straight x space plus space 2 straight x right parenthesis space sinx space dx space plus space straight c space space space space space open square brackets straight y space straight e to the power of integral Pdx end exponent space equals space integral straight Q space straight e to the power of integral straight P space dx end exponent dx plus straight c close square brackets

straight y. space sinx space space equals space integral left parenthesis straight x squared space cosx space plus 2 straight x space sin space straight x right parenthesis space dx space plus space straight c

or space space space straight y space sin space straight x space equals space integral space straight x squared space cosx space dx space space plus integral space 2 space straight x space sinx space dx space plus space straight c

or space space space straight y space sinx space minus straight x squared space sinx space minus space integral 2 straight x space sinx space dx space plus space integral 2 straight x space sinx space dx plus straight c or space space space straight y space sinx space space equals space straight x squared sin space straight x space plus space straight c

Question 117

Wired Faculty App

Solve the differential equation:
$left parenthesis 1 minus straight x squared right parenthesis space dy over dx plus straight x space straight y space equals space straight a space straight x$

Solution

The given differential equation is

left parenthesis 1 minus straight x squared right parenthesis space dy over dx plus straight x space straight y space equals space straight a space straight x space space space or space space space dy over dx plus fraction numerator straight x over denominator 1 minus straight x squared end fraction straight y space equals space straight a fraction numerator straight x over denominator 1 minus straight x squared end fraction

Comparing it with

dy over dx plus straight P space straight y space equals space straight Q comma space space we space get

straight P space equals fraction numerator straight x over denominator 1 minus straight x squared end fraction comma space space space space straight Q space equals space fraction numerator straight a space straight x over denominator 1 minus straight x squared end fraction

integral straight P space dx space equals space integral fraction numerator straight x over denominator 1 minus straight x squared end fraction dx space equals space minus 1 half integral fraction numerator negative 2 straight x over denominator 1 minus straight x squared end fraction dx space equals space minus 1 half log space left parenthesis 1 minus straight x squared right parenthesis

therefore space space space space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of negative 1 half log left parenthesis 1 minus straight x squared right parenthesis end exponent space equals space straight e to the power of log space left parenthesis 1 minus straight x squared right parenthesis to the power of negative 1 half end exponent end exponent space equals space left parenthesis 1 minus straight x squared right parenthesis to the power of negative 1 half end exponent space equals fraction numerator 1 over denominator square root of 1 minus straight x squared end root end fraction

Solution of differential equation is

straight y. space fraction numerator 1 over denominator square root of 1 minus straight x squared end root end fraction space equals integral fraction numerator ax over denominator 1 minus straight x squared end fraction. fraction numerator 1 over denominator square root of 1 minus straight x squared end root end fraction dx plus straight c

fraction numerator straight y over denominator square root of 1 minus straight x squared end root end fraction space equals fraction numerator straight a over denominator negative 2 end fraction integral left parenthesis 1 minus straight x squared right parenthesis to the power of negative 3 over 2 end exponent left parenthesis negative 2 straight x right parenthesis space dx plus straight c

fraction numerator straight y over denominator square root of 1 minus straight x squared end root end fraction space equals negative straight a over 2 fraction numerator left parenthesis 1 minus straight x squared right parenthesis to the power of negative begin display style 1 half end style end exponent over denominator negative begin display style 1 half end style end fraction plus straight c space space space space or space space space fraction numerator straight y over denominator square root of 1 minus straight x squared end root end fraction space equals fraction numerator straight a over denominator square root of 1 minus straight x squared end root end fraction plus straight c

straight y equals straight a plus straight c square root of 1 minus straight x squared end root

Question 118

Wired Faculty App

Find the general solution of the following differential equation:
$straight x dy over dx minus straight a space straight y space equals space straight x plus 1 comma space space space straight x greater than 0$

Solution

The given differential equation is $straight x dy over dx minus ay space equals space straight x plus 1$
Dividing throughout by x, we get,
   $dy over dx minus straight a over straight x straight y space equals space fraction numerator straight x plus 1 over denominator straight x end fraction$ ...(1)
Comparing (1) with $dy over dx plus straight P space straight y space equals space straight Q comma$ we have, $straight P space equals negative straight a over straight x comma space space straight Q space equals fraction numerator straight x plus 1 over denominator straight x end fraction$
$therefore space space space space space integral straight P space dx equals space minus straight a integral 1 over straight x dx space equals negative straight a space logx space equals space log space straight x to the power of negative straight a end exponent therefore space space space straight I. straight F. space equals space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of log space straight x to the power of negative straight a end exponent end exponent space equals space straight x to the power of negative straight a end exponent space equals space 1 over straight x to the power of straight a$
$therefore$ solution of given equation is, $straight y. space 1 over straight x to the power of straight a space equals space integral fraction numerator straight x plus 1 over denominator straight x end fraction 1 over straight x to the power of straight a dx plus straight c$
or    $straight y over straight x to the power of straight a space equals integral open parentheses 1 over straight x to the power of straight a plus 1 over straight x to the power of straight a plus 1 end exponent close parentheses dx plus straight c$
or    $straight y over straight x to the power of straight a space equals fraction numerator straight x to the power of 1 minus straight a end exponent over denominator 1 minus straight a end fraction plus fraction numerator straight x to the power of negative straight a end exponent over denominator negative straight a end fraction plus straight c space space space space or space space space straight y space equals space fraction numerator straight x over denominator 1 minus straight a end fraction minus 1 over straight a plus cx to the power of straight a$

Question 119

Wired Faculty App

Find the general solution of the following differential equation:
$dy over dx plus fraction numerator 4 straight x over denominator straight x squared plus 1 end fraction straight y space equals negative fraction numerator 1 over denominator left parenthesis straight x squared plus 1 right parenthesis cubed end fraction$

Solution

The given differential equation is
$dy over dx plus fraction numerator 4 straight x over denominator straight x squared plus 1 end fraction straight y space equals space minus fraction numerator 1 over denominator left parenthesis straight x squared plus 1 right parenthesis cubed end fraction$
Comparing it with $dy over dx plus straight P space straight y space equals space straight Q comma space space we space get comma space space space straight P space equals space fraction numerator 4 straight x over denominator straight x squared plus 1 end fraction comma space straight Q space equals negative fraction numerator 1 over denominator left parenthesis straight x squared plus 1 right parenthesis cubed end fraction$
$integral straight P space dx space equals space integral fraction numerator 4 straight x over denominator straight x squared plus 1 end fraction dx space equals space 2 integral fraction numerator 2 straight x over denominator straight x squared plus 1 end fraction dx space equals space 2 space log space left parenthesis straight x squared plus 1 right parenthesis space equals space log space left parenthesis straight x squared plus 1 right parenthesis squared integral straight e to the power of Pdx space equals space straight e to the power of log space left parenthesis straight x squared plus 1 right parenthesis squared end exponent equals space left parenthesis straight x squared plus 1 right parenthesis squared$
$therefore$ solution of given differential equation is
$straight y. space straight e to the power of integral straight P space dx end exponent space equals space integral straight Q. space straight e to the power of integral straight P space dx end exponent space dx space plus straight c$
$or space space straight y. space left parenthesis straight x squared plus 1 right parenthesis squared space equals space integral fraction numerator negative 1 over denominator left parenthesis straight x squared plus 1 right parenthesis cubed end fraction. space left parenthesis straight x squared plus 1 right parenthesis squared space dx plus straight c or space space straight y. space open parentheses straight x squared plus 1 close parentheses squared space equals space minus integral fraction numerator 1 over denominator straight x squared plus 1 end fraction dx plus straight c space space or space space space straight y space left parenthesis straight x squared plus 1 right parenthesis squared space equals space minus tan to the power of negative 1 end exponent straight x plus straight c$

Sponsor Area

Question 120

Wired Faculty App

Solve: $open parentheses straight x plus straight y close parentheses space dy over dx space equals space 1.$

Solution

The given differential equation is

open parentheses straight x plus straight y close parentheses dy over dx equals 1 space space space space space space space space space space or space space space straight x plus straight y space equals dy over dx space space space or space space space dx over dy minus straight x space equals space straight y

Comparing it with

dx over dy plus Py space equals straight Q comma space space we space get comma space straight P space equals space minus 1 comma space space straight Q space equals space straight y

integral straight P space dy space equals space minus integral 1 space dy space equals space minus straight y straight e to the power of integral straight P space dy end exponent space equals space straight e to the power of integral space 1 space dy end exponent space equals space straight e to the power of negative straight y end exponent

therefore

solution of given equation is

straight x. space straight e to the power of integral straight P space dy end exponent space equals space integral straight Q. space straight e to the power of integral straight P space dy end exponent space dy space plus straight c

straight x. space straight e to the power of negative straight y end exponent space equals space integral straight y. space straight e to the power of negative straight y end exponent dy plus straight c

or space space space space straight x space straight e to the power of negative straight y end exponent space equals space straight y fraction numerator straight e to the power of negative straight y end exponent over denominator negative 1 end fraction space minus space integral 1. space fraction numerator straight e to the power of negative straight y end exponent over denominator negative 1 end fraction dy plus straight c or space space space straight x space straight e to the power of negative straight y end exponent space equals space minus ye to the power of negative straight y end exponent plus integral straight e to the power of negative straight y end exponent space dy space plus space straight c or space space space xe to the power of negative straight y end exponent space equals space minus ye to the power of negative straight y end exponent plus fraction numerator straight e to the power of negative straight y end exponent over denominator negative 1 end fraction plus straight c or space space space xe to the power of negative straight y end exponent space equals space minus ye to the power of negative straight y end exponent plus fraction numerator straight e to the power of negative straight y end exponent over denominator negative 1 end fraction plus straight c or space space space space xe to the power of negative straight y end exponent space equals space minus ye to the power of negative straight y end exponent minus straight e to the power of negative straight y end exponent plus straight c or space space space space space space space space space straight x space equals space minus straight y space minus space 1 space plus space straight c space straight e to the power of straight y or space space space space straight x plus straight y plus 1 space equals space ce to the power of straight y

Question 121

Wired Faculty App

Solve:
$open parentheses straight x plus 2 straight y cubed close parentheses space straight y apostrophe space equals space straight y comma space space straight y greater than 0$

Solution

The given equation is

left parenthesis straight x plus 2 straight y cubed right parenthesis space dy over dx space equals space straight y

or space space open parentheses straight x plus 2 straight y cubed close parentheses. space space fraction numerator 1 over denominator begin display style dx over dy end style end fraction space equals space straight y space space space space space space or space space space space straight x plus 2 straight y cubed space equals space straight y space dy over dx or space space straight x over straight y plus 2 straight y squared space equals space dx over dy space space space space space space space or space space space space space space space dy over dx minus 1 over straight y. space straight x space equals space 2 space straight y squared

Comparing it with

dx over dy plus Px space equals space straight Q comma space space we space get comma

straight P space equals negative 1 over straight y comma space space straight Q space equals space 2 straight y squared integral straight P space dy space equals space minus integral 1 over straight y dy space equals space minus log space straight y

straight I. straight F. space equals space straight e to the power of integral space straight P space dy end exponent space equals space straight e to the power of negative log space straight y end exponent space equals space straight e to the power of log space straight y to the power of negative 1 end exponent end exponent space equals space straight y to the power of negative 1 end exponent space equals space 1 over straight y

therefore

solution of given equation is

straight x. space 1 over straight y space equals space integral space 2 straight y squared. space 1 over straight y dy plus straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space space straight x. space straight e to the power of integral straight P space dx end exponent space equals space integral straight Q. space straight e to the power of integral straight P space dx end exponent dy plus straight c close square brackets

or space space space space straight x over straight y space equals space 2 integral straight y space dy space plus straight c or space space space straight x over straight y space equals space straight y squared plus straight c comma space space or space space straight x space equals space straight y cubed plus straight c space straight y

which is required solution.

Question 122

Wired Faculty App

Solve:
$straight y space dx space plus space left parenthesis straight x minus straight y right parenthesis squared space dy space equals space 0$

Solution

The given differential equation is

straight y space dx space plus space left parenthesis straight x minus straight y squared right parenthesis space dy space equals space 0

straight y dx over dy plus straight x minus straight y squared space equals space 0 space space space space space or space space space space straight y dx over dy plus straight x space equals space straight y squared space space space or space space dx over dy plus 1 over straight y straight x space equals space straight y

Comparing it with

dx over dy plus Px space equals space straight Q comma space space we space get comma space space straight P space equals space 1 over straight y comma space space straight Q space equals space straight y

therefore space space space space integral straight P space dy space equals space integral 1 over straight y dy space equals log space straight y comma space space space straight e to the power of integral straight P space dy end exponent space equals space straight e to the power of log space straight y end exponent space equals space straight y

therefore space space

solution of differential equation is

straight x space straight e to the power of integral straight P space dy end exponent space equals space integral straight Q. space straight e to the power of integral straight P space dy end exponent space dy space plus space straight c space space or space space space straight x space straight y space equals space integral straight y. space straight y space dy space plus straight c

straight x space straight y space space equals space straight y cubed over 3 plus straight c

Question 123

Wired Faculty App

Find the general solution of the differential equation y dx – (x + 2 y²) dy = 0.

Solution

The given differential equation is
y dx – (x + 2 y²) dy = 0
or

straight y dx over dy minus straight x minus 2 straight y squared space equals space 0 space space space space or space space space straight y dx over dy minus straight x space equals space 2 straight y squared

dx over dy minus 1 over straight y straight x space equals space 2 straight y

Comparing it with

dx over dy plus Px space equals space straight Q comma space space we space get comma space space straight P space equals space minus 1 over straight y comma space space straight Q space equals space 2 straight y

therefore space space space space integral space straight P space dy space equals space minus integral 1 over straight y dy space equals space minus log space straight y space space space space space space space space straight e to the power of integral straight P space dy end exponent space equals space straight e to the power of negative log space straight y end exponent space equals space straight e to the power of log space straight y to the power of negative 1 end exponent end exponent space equals space straight y to the power of negative 1 end exponent space equals space 1 over straight y

Solution of differential equation is

xe to the power of integral straight P space dy end exponent space equals space integral straight Q space straight e to the power of integral straight P space dy end exponent dy space plus space straight c

straight x. space 1 over straight y space equals space integral space 2 straight y. space 1 over straight y dy plus straight c

straight x over straight y space equals space 2 space integral space 1 space dy space plus straight c space space space or space space space straight x over straight y space equals space 2 straight y plus straight c

straight x equals space 2 straight y squared plus straight c space straight y space is space the space required space solution. space

Question 124

Wired Faculty App

Solve the following differential equation:
(1 + y²)dx = (tan^{– 1} y – x) dy

Solution

The given equation is (1 + y²)dx = (tan^{– 1} y – x) dy
or $left parenthesis 1 plus straight y squared right parenthesis space dx over dy space equals space tan to the power of negative 1 end exponent straight y minus straight x space space space or space space left parenthesis 1 plus straight y squared right parenthesis space dy over dx plus straight x space equals space tan to the power of negative 1 end exponent straight y$
or $dx over dy plus fraction numerator 1 over denominator 1 plus straight y squared end fraction straight x space equals space fraction numerator tan to the power of negative 1 end exponent straight y over denominator 1 plus straight y squared end fraction$
$therefore$ Comparing it with $dy over dx plus Px space equals space straight Q$ , we get,
$straight P space equals fraction numerator 1 over denominator 1 plus straight y squared end fraction comma space space straight Q space equals fraction numerator tan to the power of negative 1 end exponent straight y over denominator 1 plus straight y squared end fraction comma space space space integral straight P space dy space equals space integral fraction numerator 1 over denominator 1 plus straight y squared end fraction dy space equals space tan to the power of negative 1 end exponent straight y$
$therefore space space space straight I. straight F. space equals space straight e to the power of integral straight P space dy end exponent space equals space straight e to the power of tan to the power of negative 1 end exponent straight y end exponent therefore space space space solution space of space equation space is$
$straight x. space straight e to the power of tan to the power of negative 1 end exponent straight y end exponent space equals space space integral fraction numerator tan to the power of negative 1 end exponent straight y over denominator 1 plus straight y squared end fraction straight e to the power of tan to the power of negative 1 end exponent straight y end exponent dy plus straight c space space space space space space space space space open square brackets because space space space straight x space straight e to the power of integral straight P space dy end exponent space equals space integral straight Q space straight e to the power of integral straight P space dy end exponent dy space plus straight c close square brackets$
Put $tan to the power of negative 1 end exponent straight y space equals straight t space space space space space space space space space space space space space space space space space space space therefore space space space fraction numerator 1 over denominator 1 plus straight y squared end fraction dy space equals space dt$
$therefore space space straight x space straight e to the power of tan to the power of negative 1 end exponent straight y end exponent space equals space integral straight t space straight e to the power of straight t space dt space plus space straight c space equals space ke to the power of straight t minus space integral straight e to the power of straight t. space dt space plus space straight c space equals space straight t space straight e to the power of straight t space minus space straight e to the power of straight t plus straight c therefore space space space straight x space straight e to the power of tan to the power of negative 1 end exponent straight y end exponent space equals space straight e to the power of tan to the power of negative 1 end exponent straight y end exponent left parenthesis tan to the power of negative 1 end exponent straight y space space minus 1 right parenthesis space plus straight c$

Question 125

Wired Faculty App

Solve the following differential equation:
$square root of 1 minus straight y squared end root space dx equals left parenthesis sin to the power of negative 1 end exponent straight y space minus straight x right parenthesis space dy$

Solution

The given differential equation is
   $square root of 1 minus straight y squared end root space dx space equals space left parenthesis sin to the power of negative 1 end exponent straight y space minus straight x right parenthesis space dy space space space or space space square root of 1 minus straight y squared end root space dy over dx space equals sin to the power of negative 1 end exponent straight y minus straight x$
or    $dx over dy space equals fraction numerator sin to the power of negative 1 end exponent straight y over denominator square root of 1 minus straight y squared end root end fraction minus fraction numerator 1 over denominator square root of 1 minus straight y squared end root end fraction straight x space space space space or space space space dx over dy plus fraction numerator 1 over denominator square root of 1 minus straight y squared end root end fraction straight x space equals space fraction numerator sin to the power of negative 1 end exponent straight y over denominator square root of 1 minus straight y squared end root end fraction$
Comparing it with $dy over dx plus Px space equals space straight Q comma$ we get,
   $straight P space equals space fraction numerator 1 over denominator square root of 1 minus straight y squared end root end fraction comma space space straight Q space equals space fraction numerator sin to the power of negative 1 end exponent straight y over denominator square root of 1 minus straight y squared end root end fraction integral straight P space dy space equals space integral fraction numerator 1 over denominator square root of 1 minus straight y squared end root end fraction dy space equals space sin to the power of negative 1 end exponent straight y comma space space space therefore space space straight e to the power of integral straight P space dy end exponent space equals space straight e to the power of sin to the power of negative 1 end exponent straight y end exponent$
Solution of differential equation is
   $straight x space straight e to the power of integral straight P space dx end exponent space equals space integral straight Q space straight e to the power of integral straight P space dy end exponent dy plus straight c$
or $straight x space straight e to the power of sin to the power of negative 1 end exponent straight y end exponent space equals space integral fraction numerator left parenthesis sin to the power of negative 1 end exponent space straight y right parenthesis over denominator square root of 1 minus straight y squared end root end fraction. space straight e to the power of sin to the power of negative 1 end exponent straight y space end exponent dy plus straight c$ ...(1)
Let $straight I space equals space integral fraction numerator open parentheses sin to the power of negative 1 end exponent space straight y close parentheses over denominator square root of 1 minus straight y squared end root end fraction. space straight e to the power of sin to the power of negative 1 end exponent straight y end exponent dy$
Put $sin to the power of negative 1 end exponent straight y space equals straight t comma$    $therefore space space space space space fraction numerator 1 over denominator square root of 1 minus straight y squared end root end fraction dy space equals space dt$
$therefore space space space space space space space space space space space space space space space space straight I space equals space integral straight t. space straight e to the power of straight t space dt space space space space space space space space space space space space space space space space space space space space space space equals space straight t space straight e to the power of straight t. space integral 1. space straight e to the power of straight t space dt space equals space straight t space straight e to the power of straight t space minus space straight e to the power of straight t space equals space left parenthesis straight t minus 1 right parenthesis space straight e to the power of straight t space space space space space space space space space space space space space space space space space space space space space space equals space left parenthesis sin to the power of negative 1 end exponent straight y space minus space 1 right parenthesis space straight e to the power of sin to the power of negative 1 end exponent end exponent straight y space space space space space space space space space space space space space space space space space space space space space space$
$therefore space space from space left parenthesis 1 right parenthesis comma space space straight x space straight e to the power of sin to the power of negative 1 end exponent straight y end exponent space equals space left parenthesis sin to the power of negative 1 end exponent straight y space minus 1 right parenthesis space straight e to the power of sin to the power of negative 1 end exponent straight y end exponent plus straight c$ ...(2)
Now y(0) = 0 $rightwards double arrow space space space space space straight y space equals space 0 space space when space straight x space equals space 0$
$therefore$    $0 space equals space left parenthesis sin to the power of negative 1 end exponent 0 minus 1 right parenthesis space straight e to the power of sin to the power of negative 1 end exponent 0 end exponent plus straight c$
$rightwards double arrow$    $0 space equals space left parenthesis 0 minus 1 right parenthesis space plus space straight c space space space space space space space space rightwards double arrow space space space straight c space equals space 1$
$therefore$ from (2), solution is
   $straight x space straight e to the power of sin to the power of negative 1 end exponent straight y end exponent space equals space left parenthesis sin to the power of negative 1 end exponent straight y space minus space 1 right parenthesis space space straight e to the power of sin to the power of negative 1 end exponent straight y end exponent plus 1$
or $straight x equals space sin to the power of negative 1 end exponent straight y space minus space 1 plus space straight e to the power of negative sin to the power of negative 1 end exponent straight y end exponent space space space space space space space space or space space space space straight x plus 1 minus space sin to the power of negative 1 end exponent straight y space equals space straight e to the power of negative sin to the power of negative 1 end exponent straight y end exponent$

Question 126

Wired Faculty App

Find a one parameter family of solutions of each of the following differential equation e^–y sec² y dy = dx + x dy.

Solution

The given differential equation is
e^–y sec² y dy = dx + x dy

therefore space space space left parenthesis straight e to the power of negative straight y end exponent space sec squared straight y minus straight x right parenthesis space dy space equals space dx space space space space space space space space space space space or space space space space dx over dy space equals space straight e to the power of negative straight y end exponent space sec squared straight y minus straight x therefore space space space space space space space space dy over dx plus straight x space equals space straight e to the power of negative straight y end exponent space secy

Comparing it with

dx over dy plus Px space equals space straight Q comma space space we space get space straight P space equals space 1 comma space space straight Q space equals space straight e to the power of negative straight y end exponent space sec squared straight y

integral straight P space dy space equals space integral 1 space dy space equals space straight y. space straight e to the power of integral straight P space dy end exponent space equals space straight e to the power of straight y

therefore

solution of differential equation is

xe to the power of integral straight P space dy end exponent space equals space integral Qe to the power of integral straight P space dy end exponent plus straight c space space space space space space or space space space space space straight x space straight e to the power of straight y space equals space integral straight e to the power of negative straight y end exponent space sec squared straight y. space straight e to the power of straight y space dy space plus straight c

straight x space straight e to the power of straight y space equals space integral sec squared straight y space dy space plus space straight c space space space space space space or space space space straight x space straight e to the power of straight y space equals space tan space straight y space plus straight c

Question 127

Wired Faculty App

Find the particular solution of the differential equation:
$dy over dx plus straight y space cotx space equals space 2 straight x plus straight x squared space cotx space space space left parenthesis straight x not equal to 0 right parenthesis$

Solution

The given differential equation is

dy over dx plus straight y space cotx space equals space 2 straight x plus straight x squared cotx thin space left parenthesis straight x not equal to 0 right parenthesis

given that y = 0 when x =

straight pi over 2.

dy over dx plus straight y space cotx space equals space 2 straight x plus straight x squared space cot space straight x

Comparing it with

dy over dx plus Py space equals space straight Q comma

we get,

straight P space equals space cotx comma space space straight Q space equals space 2 straight x plus straight x squared space cotx space space space integral straight P space dx space equals space integral cot space straight x space dx space equals space log space sinx. space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of log space sinx end exponent space equals space sinx

Solution of differential equation is

straight y space straight e to the power of integral straight P space dx end exponent space equals space integral space straight Q space straight e to the power of integral straight P space dx end exponent dx plus straight c

or space space space space space space straight y space sinx space equals space integral left parenthesis 2 straight x plus straight x squared space cot space straight x right parenthesis space sinx space dx plus straight c or space space space space space space straight y space sinx space equals space integral 2 straight x space sinx space dx space space plus space integral straight x squared space cosx space dx or space space space space space straight y space sinx space equals space left parenthesis sinx right parenthesis. space open parentheses fraction numerator 2 straight x squared over denominator 2 end fraction close parentheses space minus space integral space cos space straight x space open parentheses fraction numerator 2 straight x squared over denominator 2 end fraction close parentheses dx plus space integral space straight x squared space cosx space dx space plus space straight c or space space space space straight y space sinx space equals space straight x squared space sinx space minus space integral straight x squared space cosx space dx space plus space integral straight x squared space cos space straight x space dx space plus space straight c or space space space space straight y space sinx space equals space straight x squared space sin space straight x space plus space straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis

Now y = 0 when

straight x space equals space straight pi over 2

therefore space space space space space 0 space equals space straight pi squared over 4 space sin space straight pi over 2 plus straight c space space space space space space rightwards double arrow space space space space 0 space equals space straight pi squared over 4 cross times 1 plus space straight c space space space rightwards double arrow space space space straight c space space equals negative straight pi squared over 4

Question 128

Wired Faculty App

For the given differential equation, find a particular solution satisfying the given condition:
$dy over dx plus 2 straight y space tan space straight x space equals space sin space straight x space colon space straight y space space equals space 0 space space when space straight x space equals space straight pi over 3$

Solution

The given equation is

dy over dx plus 2 straight y space tan space straight x space equals space sin space straight x

Comparing with

dy over dx plus straight P space straight y space equals space straight Q comma space space we space get comma space space straight P space equals space 2 space tanx space comma space space straight Q space equals space sin space straight x

therefore space space space space space integral straight P space dx space equals space integral 2 space tan space straight x space dx space equals space minus 2 space log space cosx space equals space log space open parentheses cosx close parentheses squared therefore space space space space space straight I. straight F. space equals space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of log space left parenthesis cos space straight x right parenthesis to the power of negative 2 end exponent end exponent space equals space left parenthesis cos space straight x right parenthesis to the power of negative 2 end exponent space equals space fraction numerator 1 over denominator cos squared straight x end fraction

therefore

solution of given equation is

straight y. space fraction numerator 1 over denominator cos squared straight x end fraction space equals space integral space sinx. space space fraction numerator 1 over denominator cos squared straight x end fraction dx plus straight c

open square brackets because space space straight y space straight e to the power of integral straight P space dx end exponent space equals space integral straight Q space straight e to the power of integral straight P space dx end exponent dx plus straight c close square brackets

fraction numerator straight y over denominator cos squared straight x end fraction space equals space integral secx space tanx space dx space plus space straight c

or space space space fraction numerator straight y over denominator cos squared straight x end fraction space equals secx space plus space straight c or space space space straight y space equals space cos space straight x space plus space straight c space cos to the power of 2 straight x end exponent space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis because space space space space space space straight y space equals space 0 space space space space when space straight x space equals space straight pi over 3 therefore space space space space space space 0 space equals space 1 half plus straight c space open parentheses 1 half close parentheses squared space space space space space space space space space space space therefore space space space straight c space equals space minus 2 therefore space space space from space left parenthesis 1 right parenthesis comma space space straight y space equals space cosx minus 2 space cos squared straight x.

Question 129

Wired Faculty App

For the given differential equation, find a particular solution satisfying the given condition:
$left parenthesis 1 plus straight x squared right parenthesis space dy over dx space plus space 2 space straight x space straight y space space equals space fraction numerator 1 over denominator 1 plus straight x squared end fraction semicolon space straight y space equals space 0 space space when space straight x space equals space 1$

Solution

The given differential equation is

left parenthesis 1 plus straight x squared right parenthesis space dy over dx plus 2 xy space equals space fraction numerator 1 over denominator 1 plus straight x squared end fraction

dy over dx plus fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction straight y space equals space fraction numerator 1 over denominator left parenthesis 1 plus straight x squared right parenthesis squared end fraction

Comparing it with

dy over dx plus straight P space straight y space equals space straight Q comma space we space get comma space straight P space equals space fraction numerator 2 straight x over denominator straight x squared plus 1 end fraction comma space space straight Q space equals space fraction numerator 1 over denominator left parenthesis straight x squared plus 1 right parenthesis squared end fraction

integral straight P space dx space equals space integral fraction numerator 2 straight x over denominator straight x squared plus 1 end fraction dx space equals space log space left parenthesis straight x squared plus 1 right parenthesis comma space space straight e to the power of integral straight P space dx end exponent space equals straight e to the power of log space left parenthesis straight x squared plus 1 right parenthesis end exponent space equals straight x squared plus 1

therefore

solution of differential equation is

straight y. space straight e to the power of integral straight P space dx end exponent space equals integral straight Q. space straight e to the power of integral straight P space dx end exponent space dx space plus space straight c

straight y. left parenthesis straight x squared plus 1 right parenthesis space equals integral fraction numerator 1 over denominator left parenthesis straight x squared plus 1 right parenthesis squared end fraction. space left parenthesis straight x squared plus 1 right parenthesis space dx space plus space straight c

straight y left parenthesis straight x squared plus 1 right parenthesis space equals space integral fraction numerator 1 over denominator straight x squared plus 1 end fraction dx plus straight c

straight y left parenthesis straight x squared plus 1 right parenthesis space equals space tan to the power of negative 1 end exponent straight x plus straight c

straight y equals fraction numerator tan to the power of negative 1 end exponent straight x over denominator straight x squared plus 1 end fraction plus fraction numerator straight c over denominator straight x squared plus 1 end fraction

...(1)
Now y = 0 when x = 1

therefore space space space space 0 space equals space fraction numerator tan to the power of negative 1 end exponent 1 over denominator 1 plus 1 end fraction plus fraction numerator straight c over denominator 1 plus 1 end fraction space space space rightwards double arrow space space space 0 space equals space straight pi over 4 plus straight c space space space rightwards double arrow space space space straight c space equals space minus straight pi over 4 therefore space space space space from space left parenthesis 1 right parenthesis comma space space straight y space equals space fraction numerator tan to the power of negative 1 end exponent straight x over denominator straight x squared plus 1 end fraction minus fraction numerator straight pi over denominator 4 left parenthesis straight x squared plus 1 right parenthesis end fraction

which is required solution.

Question 130

Wired Faculty App

For the given differential equation, find a particular solution satisfying the given condition:
$dy over dx minus 3 space straight y space cotx space equals space sin space 2 straight x space colon space space straight y space equals space 2 space space when space space straight x space equals space straight pi over 2$

Solution

The given differential equation is

dy over dx minus 3 straight y space cotx space equals space sin space 2 straight x

Comparing it with

dy over dx plus Py space equals straight Q comma space we space get comma space space straight P space equals space minus 3 space cotx space comma space space straight Q space equals space sin space 2 straight x

integral straight P space dx space equals space minus 3 integral cot space straight x space dx space equals space minus 3 space log space sinx

therefore space space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of negative 3 space log space sinx end exponent space equals space straight e to the power of log space left parenthesis sinx right parenthesis cubed end exponent space equals space left parenthesis sin space straight x right parenthesis to the power of negative 3 end exponent space equals fraction numerator 1 over denominator sin cubed straight x end fraction

therefore

solution of given equation is

straight y. space straight e to the power of integral straight P space dx end exponent space equals space integral straight Q. space straight e to the power of integral straight P space dx end exponent dx plus straight c

straight y. space fraction numerator 1 over denominator sin space cubed straight x end fraction space equals space integral sin space 2 straight x. space 1 over sin to the power of 3 straight x end exponent dx plus space straight c

fraction numerator straight y over denominator sin cubed straight x end fraction space equals space 2 space integral fraction numerator cosx over denominator sin squared straight x end fraction dx plus straight c

fraction numerator straight y over denominator sin cubed straight x end fraction space equals space 2 space integral left parenthesis sinx right parenthesis to the power of negative 2 end exponent space cosx space dx space plus space straight c

fraction numerator straight y over denominator sin cubed straight x end fraction space equals 2 fraction numerator left parenthesis sin space straight x right parenthesis to the power of negative 1 end exponent over denominator negative 1 end fraction plus straight c

fraction numerator straight y over denominator sin cubed straight x end fraction space equals space minus fraction numerator 2 over denominator sin space straight x end fraction plus straight c

straight y space equals space minus 2 space sin squared straight x plus space straight c space sin cubed straight x

...(1)
Initially, y = 2 when

straight x equals space straight pi over 2

therefore

2 space equals space minus 2 sin squared straight pi over 2 plus straight c space sin cubed. space straight pi over 2

rightwards double arrow

2 space equals space minus 2 plus straight c

open square brackets because space space sin space straight pi over 2 space equals space 1 close square brackets

rightwards double arrow

c = 4
Putting c = 4 in (1), we get,

straight y space equals negative 2 space sin squared straight x plus 4 space sin cubed straight x comma space which space is space required space solution. space

Question 131

Wired Faculty App

Find a particular solution of the differential equation:
$dy over dx plus straight y space cotx space equals space 4 straight x$
$cosec space straight x space left parenthesis straight x not equal to 0 right parenthesis comma space space given space that space straight y space equals space 0 space space space when space straight x space equals space straight pi over 2.$

Solution

The given differential equation is

dy over dx plus straight y space cotx space equals space 4 space straight x space cosec space straight x

dy over dx plus left parenthesis cotx right parenthesis. space straight y space equals space 4 space straight x space cosec space straight x

Comparing it with

dy over dx plus Py space equals space straight Q comma space we space get comma

straight P space space equals space cot space straight x comma space space space space straight Q space equals space 4 straight x space cosec space straight x

integral straight P space dx space equals space integral cotx space dx space equals space log space sinx space

therefore space space space space space space space space space space space space space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of log space sinx end exponent space equals space sin space straight x

therefore space

solution of differential equation is

straight y space straight e to the power of integral straight P space dx end exponent space equals space integral straight Q. space straight e to the power of integral straight P space dx end exponent dx plus straight c

or space space space space space space space space straight y space sinx space equals space integral 4 straight x space cosecx. space sinx space dx space plus straight c or space space space space space space space space straight y space sinx space space equals space 4 space integral space straight x space dx space plus space straight c or space space space space space space space space straight y space sinx space space equals space 2 straight x squared plus straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space.... left parenthesis 1 right parenthesis Now space straight y space equals space 0 space space space when space straight x space equals space straight pi over 2

therefore space space space space space space space space space space space space space space space space space 0 space equals space 2 cross times straight pi squared over 4 plus straight c space space space space space space space space space space space space space space rightwards double arrow space space space space space space space space straight c space equals space minus straight pi squared over 2

therefore

from (1),

straight y space sinx space equals space 2 straight x squared minus straight pi squared over 2

which is required solution.

Question 132

Wired Faculty App

Find a particular solution of the differential equation $left parenthesis straight x plus 1 right parenthesis space dy over dx space equals space 2 straight e to the power of negative straight y end exponent minus 1 comma space space$ given that y = 0, when x = 0.

Solution

The given differential equation is

left parenthesis straight x plus 1 right parenthesis space straight y apostrophe space space equals space 2 space straight e to the power of negative straight y end exponent minus 1 space space space space or space space left parenthesis straight x plus 1 right parenthesis space dy over dx space equals space 2 straight e to the power of negative straight y end exponent minus 1

dy over dx space equals space fraction numerator 2 over denominator straight x plus 1 end fraction space straight e to the power of negative straight y end exponent minus fraction numerator 1 over denominator straight x plus 1 end fraction

straight e to the power of straight y dy over dx plus fraction numerator 1 over denominator straight x plus 1 end fraction straight e to the power of straight y space equals space fraction numerator 2 over denominator straight x plus 1 end fraction

Put

straight e to the power of straight y space equals space straight z comma space space space space space therefore space space straight e to the power of straight y dy over dx space equals space dz over dx

therefore

dz over dx plus fraction numerator 1 over denominator straight x plus 1 end fraction straight z space equals fraction numerator 2 over denominator straight x plus 1 end fraction

Comparing it with

dz over dx plus Pz space equals space straight Q comma space space we space get comma

straight P space equals fraction numerator 1 over denominator straight x plus 1 end fraction. space space space straight Q space space equals space fraction numerator 2 over denominator straight x plus 1 end fraction

straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of integral fraction numerator 1 over denominator straight x plus 1 end fraction end exponent space equals space straight e to the power of log left parenthesis straight x plus 1 right parenthesis end exponent space equals space straight x plus 1

therefore

solution of differential equation is

straight z space. space straight e to the power of integral straight P space dx end exponent space equals space integral space straight Q. space straight e to the power of integral straight P space dx end exponent space plus straight c space

straight z left parenthesis straight x plus 1 right parenthesis space equals space integral open parentheses fraction numerator 2 over denominator straight x plus 1 end fraction close parentheses space left parenthesis straight x plus 1 right parenthesis space dx plus straight c

or space space space space space space straight z left parenthesis straight x plus 1 right parenthesis space equals space integral 2 dx plus straight c

or space space space space space straight z left parenthesis straight x plus 1 right parenthesis space equals space 2 straight x plus straight c or space space space straight e to the power of straight y left parenthesis straight x plus 1 right parenthesis space equals space 2 straight x plus straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis

Now y = 0 when x = 0

therefore space space space space space space straight e to the power of 0 left parenthesis 0 plus 1 right parenthesis space equals space 0 plus straight c space space space space space space space space space space space rightwards double arrow space space space straight c space equals space 1 therefore space space space from space space left parenthesis 1 right parenthesis comma space space space straight e to the power of straight y space left parenthesis straight x plus 1 right parenthesis space equals space 2 straight x plus 1

which is required solution.

Question 133

Wired Faculty App

Solve the differential equation, given that y = 1 where x = 2
$straight x dy over dx plus straight y space equals space straight x cubed.$

Solution

The given differential equation is
$straight x dy over dx plus straight y space equals space straight x cubed$
or $dy over dx plus 1 over straight x straight y space equals space straight x squared$
Comparing it with $dy over dx plus Py space equals space straight Q comma space space space we space get comma space space space straight P space equals space 1 over straight x comma space space straight Q space equals space straight x squared$
$integral straight P space dx space equals space integral 1 over straight x dx space equals space log space straight x$
$therefore space space space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of log space straight x end exponent space equals space straight x$
$therefore$ solution of given differential equation is
$ye space to the power of integral straight P space dx end exponent space equals space integral space straight Q. space straight e to the power of integral straight P space dx end exponent space dx space plus straight c space space space space or space space space straight y. space straight x space equals space integral straight x squared. space space straight x space dx space plus space straight c$
or $xy equals space integral straight x cubed dx plus straight c space space or space space straight x space straight y space equals space straight x to the power of 4 over 4 plus straight c space space space or space space space straight y space equals space straight x cubed over 4 plus straight c over straight x$
Now $straight y space equals space 1 space space space space when space straight x space equals space 2$
$therefore space space space space space space space space space space space 1 space equals space 8 over 4 plus straight c over 2 space space space or space space space 1 space equals space 2 plus straight c over 2 space space space space space space rightwards double arrow space space space space straight c space equals space minus 2$
$therefore space space$ solution is $straight y space equals space straight x cubed over 4 minus straight x over 2.$

Question 134

Wired Faculty App

Find the equation of a curve passing through the point (0, 1). If the slope of the tangent to the curve at any point (x, y) is equal to the sum of x coordinate (abscissa) and the product of the x coordinate and y coordinate (ordinate) of that point.

Solution

Let y = f (x) be equation of curve.
Now

dy over dx

is the slope of the tangent to the curve at point (x, y)
From the given condition,

dx over dy space equals space straight x plus xy

dy over dx minus straight x space straight y space equals space straight x

Comparing

dy over dx plus Py space equals space straight Q

with

dy over dx minus straight x space straight y space equals space straight x comma space

we get, P = -x, Q = x

therefore space space space space space space space integral straight P space dx space equals negative integral straight x space dx space equals space minus straight x squared over 2 comma space space space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of negative straight x squared over 2 end exponent

Solution of differential equation is

straight y space straight e to the power of integral straight P space dx end exponent space equals space integral straight Q space straight e to the power of integral straight P space dx end exponent dx plus straight c

straight y space straight e to the power of negative straight x squared over 2 end exponent space equals space integral space straight x space straight e to the power of negative straight x squared over 2 end exponent dx plus straight c

...(1)
Let

straight I space equals space integral space straight x space space straight e to the power of negative straight x squared over 2 end exponent dx plus straight c

Put

negative straight x squared over 2 space equals space straight t comma space space space space therefore space space space space space space minus straight x space dx space space equals space dt space space space space rightwards double arrow space space space space straight x space dx space equals space minus dt

therefore

straight I space equals space minus integral straight e to the power of straight t dt space equals space minus straight e to the power of straight t space equals space minus straight e to the power of negative straight x squared over 2 end exponent

therefore

straight y space straight e to the power of negative straight x squared over 2 end exponent space equals space minus straight e to the power of negative straight x squared over 2 end exponent plus straight c

straight y space equals negative 1 plus space straight c space straight e to the power of straight x squared over 2 end exponent

...(2)
Since the curve passes through (0, 1)

therefore

straight I space equals space minus 1 plus space straight c space straight e to the power of 0 space space space space rightwards double arrow space space space 2 space equals space straight c

therefore

from (2),

straight y equals negative 1 plus 2 space straight e to the power of straight x squared over 2 end exponent

straight y plus 1 space equals space 2 space straight e to the power of straight x squared over 2 end exponent

, which is required equation of curve.

Question 135

Wired Faculty App

Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.

Solution

Let y = f (x) be the equation of curve.
We know that

dy over dx

represents the slope of the tangent to the curve at the point (x, y).
From the given condition.

dy over dx space equals space straight x plus straight y

dy over dx minus straight y space equals straight x

Comparing

dy over dx plus straight P space straight y space equals space straight Q space space space with space dy over dx minus straight y space equals space straight x comma space we space get comma space space straight P space equals space minus 1 comma space space straight Q space equals space straight x

therefore space space space space space space space space integral straight P space dx space equals space minus integral 1 space dx space equals space minus straight x comma space space space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of negative straight x end exponent

Solution of differential equation is

straight y space straight e to the power of integral straight P space dx end exponent space equals integral straight Q space straight e to the power of integral straight P space dx end exponent dx space plus space straight c

straight y space straight e to the power of negative straight x end exponent space equals space integral space straight x space straight e to the power of negative straight x end exponent space dx plus straight c space space space or space space space space straight y space straight e to the power of negative straight x end exponent space equals space straight x fraction numerator straight e to the power of negative straight x end exponent over denominator negative 1 end fraction minus integral 1. space fraction numerator straight e to the power of negative straight x end exponent over denominator negative 1 end fraction dx plus straight c

straight y space straight e to the power of negative straight x end exponent space equals space minus straight x space straight e to the power of negative straight x end exponent plus integral space straight e to the power of negative straight x end exponent dx plus straight c

straight y space straight e to the power of negative straight x end exponent space equals space minus xe to the power of negative straight x end exponent plus straight e to the power of negative straight x end exponent plus straight c

straight y space equals space minus straight x minus 1 plus straight c over straight e to the power of negative straight x end exponent

straight x plus straight y plus 1 space equals space straight c space straight e to the power of straight x

...(1)
Now curve passes through origin (0, 0).

therefore space space space 0 plus 0 plus 1 space equals space straight c space straight e to the power of 0 space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space straight c space equals space 1

therefore

from (1),

straight x plus straight y plus 1 space equals space straight e to the power of straight x

is the required equation of curve.

Question 136

Wired Faculty App

The integrating factor of the differential equation $straight x dy over dx minus straight y space equals space 2 straight x squared$ is

Solution

The given differential equation is

straight x dy over dx minus straight y space equals 2 straight x squared space space space space space or space space space dy over dx minus 1 over straight x straight y space equals 2 straight x

Comparing it with

dy over dx plus Py space equals space straight Q comma space space we space get comma space straight P space equals space minus 1 over straight x comma space space straight Q space equals space 2 straight x

I.F. =

straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of negative integral space 1 over straight x dx end exponent space equals space straight e to the power of negative space log space straight x end exponent space equals space straight e to the power of log space straight x to the power of negative 1 end exponent end exponent space equals space straight x to the power of negative 1 end exponent space equals space 1 over straight x

therefore space space space left parenthesis straight C right parenthesis space is space correct space answer.

Question 137

Wired Faculty App

The integrating factor of the differential equation:
$left parenthesis 1 minus straight y squared right parenthesis dx over dy plus straight y space straight x space equals space straight a space straight y space space left parenthesis negative 1 less than straight y less than 1 right parenthesis$

$fraction numerator 1 over denominator straight y squared minus 1 end fraction$
$fraction numerator 1 over denominator square root of straight y squared minus 1 end root end fraction$
$fraction numerator 1 over denominator 1 minus straight y squared end fraction$
$fraction numerator 1 over denominator square root of 1 minus straight y squared end root end fraction$

Solution

fraction numerator 1 over denominator square root of 1 minus straight y squared end root end fraction

The given differential equation is $left parenthesis 1 minus straight y squared right parenthesis space dx over dy plus space straight y space straight x space equals space straight a space straight y$
or $dx over dy plus fraction numerator straight y over denominator 1 minus straight y squared end fraction straight x space equals space fraction numerator straight a space straight y over denominator 1 minus straight y squared end fraction$
Comparing it with $dx over dy plus Px space equals space straight Q comma space space we space get comma space space straight P space equals space fraction numerator straight y over denominator 1 minus straight y squared end fraction comma space space space straight Q space equals space fraction numerator straight a space straight y over denominator 1 minus straight y squared end fraction$
I.F. = $straight e to the power of integral straight P space dy end exponent space equals space straight e to the power of integral fraction numerator straight y over denominator 1 minus straight x squared end fraction dy end exponent space equals space straight e to the power of 1 half integral subscript 1 minus straight y squared end subscript superscript negative 2 straight y end superscript dy end exponent space equals space straight e to the power of negative 1 half log space left parenthesis 1 minus straight x squared right parenthesis end exponent$
$equals space straight e to the power of log space left parenthesis 1 minus straight x squared right parenthesis to the power of negative 1 half end exponent end exponent space equals space left parenthesis 1 minus straight y squared right parenthesis to the power of negative 1 half end exponent space equals space fraction numerator 1 over denominator square root of 1 minus straight y squared end root end fraction$
$therefore space space space space left parenthesis straight D right parenthesis space is space correct space answer.$

Question 138

Wired Faculty App

The general solution of a differential equation of the type $dx over dy plus straight P subscript 1 straight x space equals space straight Q subscript 1$ is
$straight y space straight e to the power of integral straight P subscript 1 dy end exponent space equals space integral space open parentheses straight Q subscript 1 space straight e to the power of integral straight P subscript 1 dy end exponent close parentheses space dy plus straight C$
$straight y. space straight e to the power of integral straight P subscript 1 dx end exponent space equals space integral space open parentheses straight Q subscript 1 space straight e to the power of integral straight P subscript 1 dx end exponent close parentheses dx plus straight C$
$straight x. straight e to the power of integral straight P subscript 1 dy end exponent space equals space integral open parentheses straight Q subscript 1 straight e to the power of integral straight P subscript 1 dy end exponent close parentheses dy plus straight C$
$straight x space straight e to the power of integral straight P subscript 1 dx end exponent space equals space integral open parentheses straight Q subscript 1 space straight e to the power of integral straight P subscript 1 dx end exponent close parentheses dx plus straight C$

Solution

straight x. straight e to the power of integral straight P subscript 1 dy end exponent space equals space integral open parentheses straight Q subscript 1 straight e to the power of integral straight P subscript 1 dy end exponent close parentheses dy plus straight C

The given differential equation is $dx over dy plus straight P subscript 1 straight x space equals space straight Q subscript 1$
Multiplying through by $straight e to the power of integral straight P subscript 1 dy end exponent comma space$ we get,
$dx over dy straight e to the power of integral straight P subscript 1 dy end exponent space plus space straight P subscript 1 straight x space straight e to the power of integral straight P subscript 1 dy end exponent space equals straight Q subscript 1 space straight e to the power of integral straight P subscript 1 end exponent space dy$
or $straight d over dy left parenthesis space straight x space straight e to the power of integral straight P subscript 1 dy end exponent right parenthesis space equals space straight Q subscript 1 space straight e to the power of integral straight P subscript 1 end exponent dy$
$open square brackets because space space straight d over dy left parenthesis straight x space straight e to the power of integral straight P subscript 1 dy end exponent right parenthesis space equals space xe to the power of integral straight P subscript 1 dy. end exponent straight d over dy left parenthesis integral straight P subscript 1 space dy right parenthesis plus straight e to the power of integral straight P subscript 1 dy end exponent dx over dy space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space xe to the power of integral straight P subscript 1 dy end exponent space space straight P subscript 1 space plus space straight e to the power of integral straight P subscript 1 dy end exponent space dx over dy space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space dx over dy straight e to the power of integral straight P subscript 1 dy end exponent space plus space straight P subscript 1 space xe to the power of integral straight P subscript 1 dy end exponent close square brackets space space space$
Integrating both sides w.r.t y, we get,
$xe to the power of integral straight P subscript 1 dy end exponent space equals space integral straight Q subscript 1 space straight e to the power of integral straight P subscript 1 dy end exponent dy plus straight C space which space is space required space solution. space$
$therefore space space space left parenthesis straight C right parenthesis space is space correct space answer. space$

Question 139

Wired Faculty App

The general solution of the differential equation e^x dy + (ye^x + 2x) dx = 0 is
x e^y + x² = C
x e^y + y² = C

y e^x + x² = C

y e^y + x² = C

Solution

y e^x + x² = C

The given differential equation is

straight e to the power of straight x dy space plus space left parenthesis straight y space straight e to the power of straight x space plus space 2 straight x right parenthesis space dx space equals space 0

straight e to the power of straight x dy over dx plus space straight y space straight e to the power of straight x space plus space 2 space straight x space equals 0 space space space space or space space space dy over dx plus straight y plus fraction numerator 2 straight x over denominator straight e to the power of straight x end fraction space equals space 0 space space or space space dy over dx plus straight y space equals space minus 2 xe to the power of negative straight x end exponent

Comparing it with

dy over dx plus Py space equals space straight Q comma

we get, P = 1,

straight Q space equals negative 2 xe to the power of negative straight x end exponent

therefore space space space space space space space space space space integral straight P space dx space equals space integral space 1 space dx space equals space apostrophe straight x comma space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of straight x

Solution of given differential equation is

ye space to the power of integral straight P space dx end exponent space equals space integral straight Q space straight e to the power of integral straight P space dx end exponent dx plus straight C space space space or space space space straight y space straight e to the power of straight x space equals space integral left parenthesis negative 2 straight x space straight e to the power of straight x right parenthesis. space straight e to the power of straight x dx plus straight C

straight y space straight e to the power of straight x space equals space minus 2 space integral space straight x space dx space plus space straight C space space space space or space space space straight y space straight e to the power of straight x space equals space minus straight x squared plus straight C

space ye to the power of straight x plus straight x squared space equals space straight C

therefore space space left parenthesis straight C right parenthesis space is space correct space answer.

Question 140

Wired Faculty App

Solve the differential equation $left parenthesis 1 plus straight x squared right parenthesis dy over dx plus straight y equals straight e to the power of tan to the power of negative 1 end exponent straight x end exponent$

Solution

Given differential equation is:
$left parenthesis 1 plus straight x squared right parenthesis dy over dx plus straight y equals straight e to the power of tan to the power of negative 1 end exponent straight x end exponent rightwards double arrow dy over dx plus fraction numerator straight y over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction space equals space fraction numerator straight e to the power of tan to the power of negative 1 end exponent straight x end exponent over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction$
This is a linear differential equation of the form
$dy over dx plus Py space equals space straight Q$
$where space straight P space equals space fraction numerator 1 over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction space and space straight Q space equals fraction numerator straight e to the power of tan to the power of negative 1 end exponent straight x end exponent over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction Therefore comma space straight I. straight F. space equals space straight e to the power of integral Pdx end exponent space equals space straight e to the power of tan to the power of negative 1 end exponent straight x end exponent Thus space the space solution space is comma straight y left parenthesis straight I. straight F right parenthesis space equals space integral straight Q thin space left parenthesis straight I. straight F right parenthesis space dx rightwards double arrow straight y open parentheses straight e to the power of tan to the power of negative 1 end exponent straight x end exponent close parentheses space equals integral fraction numerator straight e to the power of tan to the power of negative 1 end exponent straight x end exponent over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction cross times straight e to the power of tan to the power of negative 1 end exponent straight x end exponent dx Substitute space straight e to the power of tan to the power of negative 1 end exponent straight x end exponent space equals space straight t semicolon$
$straight e to the power of tan to the power of negative 1 end exponent straight x end exponent space cross times fraction numerator 1 over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction dx space equals space dt Thus comma space straight y open parentheses straight e to the power of tan to the power of negative 1 end exponent straight x end exponent close parentheses space equals space integral tdt rightwards double arrow space space straight y open parentheses straight e to the power of tan to the power of negative 1 end exponent straight x end exponent close parentheses space equals straight t squared over 2 plus straight C rightwards double arrow space straight y open parentheses straight e to the power of tan to the power of negative 1 end exponent straight x end exponent close parentheses space equals space open parentheses straight e to the power of tan to the power of negative 1 end exponent straight x end exponent close parentheses squared over 2 plus straight C$

Question 141

Wired Faculty App

$If space straight y space equals space Pe to the power of ax plus Qe to the power of bx$ show that
$straight d to the power of 2 straight y end exponent over dx squared minus left parenthesis straight a plus straight b right parenthesis dy over dx plus aby space equals 0$

Solution

Given that
$straight y equals Pe to the power of ax plus Qe to the power of bx Differentiating space the space above space function space with space respect space to space straight x comma dy over dx equals Pae to the power of ax plus Qbe to the power of bx Differentiating space once space again comma space we space have comma fraction numerator straight d squared straight y over denominator dx squared end fraction equals space Pa squared straight e to the power of ax plus Qb squared straight e to the power of bx$
Let us now find $left parenthesis straight a plus straight b right parenthesis dy over dx colon$
$left parenthesis straight a plus straight b right parenthesis dy over dx equals left parenthesis straight a plus straight b right parenthesis open parentheses Pae to the power of ax plus Qbe to the power of bx close parentheses space space rightwards double arrow space left parenthesis straight a plus straight b right parenthesis dy over dx equals Pa squared straight e to the power of ax plus Qabe to the power of bx plus Pabe to the power of bx plus Qb squared straight e to the power of bx rightwards double arrow space left parenthesis straight a plus straight b right parenthesis dy over dx equals Pa squared straight e to the power of ax plus left parenthesis straight P plus straight Q right parenthesis abe to the power of bx plus Qb squared straight e to the power of bx space space space$
Also we have,
$aby space equals ab open parentheses Pe to the power of ax plus Qe to the power of bx close parentheses space equals space ab Thus comma space fraction numerator straight d squared straight y over denominator dx squared end fraction minus left parenthesis straight a plus straight b right parenthesis dy over dx plus aby equals Pa squared straight e to the power of ax plus Qb squared straight e to the power of bx minus Pa squared straight e to the power of ax minus left parenthesis straight P plus straight Q right parenthesis abe to the power of bx minus Qb squared straight e to the power of bx plus abPe to the power of ax plus abQe to the power of bx equals 0$

Question 142

Wired Faculty App

$If space straight x equals cost left parenthesis 3 minus 2 cos squared straight t right parenthesis space and space straight y equals sint left parenthesis 3 minus 2 sin squared straight t right parenthesis comma space find space the space value space of space dy over dx at space straight t equals straight pi over 4.$

Solution

straight x equals cost space left parenthesis 3 minus 2 cos squared straight t right parenthesis and straight y equals sint left parenthesis 3 minus sin squared straight t right parenthesis We space need space to space find space dy over dx colon dy over dx equals fraction numerator begin display style dy over dx end style over denominator begin display style dx over dt end style end fraction Let space us space find space dx over dt colon straight x equals cost left parenthesis 3 minus 2 cos squared straight t right parenthesis dx over dt equals cost left parenthesis 4 cost space sint right parenthesis plus left parenthesis 3 minus 2 cos squared straight t right parenthesis left parenthesis negative sint right parenthesis rightwards double arrow dx over dt equals negative 3 sint plus 4 cos squared tsint plus 2 cos squared tsint Let space us space find space dy over dt colon straight y equals sint left parenthesis 3 minus 2 sin squared straight t right parenthesis dy over dt equals sint left parenthesis negative 4 sint space cost right parenthesis plus left parenthesis 3 minus 2 sin squared straight t right parenthesis space left parenthesis cost right parenthesis rightwards double arrow dy over dt equals 3 cost minus 4 sin squared tcost minus 2 sin squared tcost Thus comma space dy over dx equals fraction numerator 3 cost minus 4 sin squared tcost minus 2 sin squared tcost over denominator negative 3 sint plus 4 cos squared tsint plus 2 cos squared tsint end fraction rightwards double arrow space dy over dx equals fraction numerator 3 cost minus 6 sin squared tcost over denominator negative 3 sint plus 6 cos squared tsint end fraction rightwards double arrow dy over dx equals fraction numerator 3 cost left parenthesis 1 minus 2 sin squared straight t right parenthesis over denominator negative 3 sint left parenthesis 1 minus 2 cos squared straight t right parenthesis end fraction rightwards double arrow dy over dx equals fraction numerator 3 cost left parenthesis 1 minus 2 sin squared straight t right parenthesis over denominator 3 sint left parenthesis 2 cos squared straight t minus 1 right parenthesis end fraction rightwards double arrow dy over dx equals fraction numerator cost space over denominator sint end fraction open square brackets because 2 cos squared straight t minus 1 equals 1 minus 2 sin squared straight t close square brackets rightwards double arrow dy over dx equals cot space straight t rightwards double arrow open parentheses dy over dx close parentheses subscript straight t equals straight pi over 4 end subscript equals cot straight pi over 4 equals 1

Question 143

Wired Faculty App

Find the particular solution of the differential equation $log open parentheses dy over dx close parentheses equals 3 straight x plus 4 straight y comma space given space that space straight y space equals space 0 comma space when space straight x equals space 0.$

Solution

Consider the differential equation,
$log open parentheses dy over dx close parentheses equals 3 straight x plus 4 straight y$
Taking exponent on both the sides, we have
$straight e to the power of log open parentheses dy over dx close parentheses end exponent space equals space straight e to the power of 3 straight x plus 4 straight y end exponent space rightwards double arrow dy over dx equals straight e to the power of 3 straight x plus 4 straight y end exponent space rightwards double arrow dy over dx equals straight e to the power of 3 straight x end exponent. straight e to the power of 4 straight y end exponent rightwards double arrow dy over straight e to the power of 4 straight y end exponent equals straight e to the power of 3 straight x end exponent. dx$
Integration in both the sides, we have
$integral dy over straight e to the power of 4 straight y end exponent space equals space integral straight e to the power of 3 straight x end exponent dx fraction numerator straight e to the power of negative 4 straight y end exponent over denominator negative 4 end fraction equals straight e to the power of 3 straight x end exponent over 3 plus straight C We space need space to space find space the space particular space solution We space have comma space straight y space equals space 0 comma space when space straight x space equals space 0 fraction numerator 1 over denominator negative 4 end fraction space equals 1 third plus straight C rightwards double arrow straight C space equals negative 1 fourth minus 1 third rightwards double arrow straight C space equals space fraction numerator negative 3 minus 4 over denominator 12 end fraction equals negative 7 over 12 Thus comma space the space solution space is space straight e to the power of 3 straight x end exponent over 3 plus straight e to the power of negative 4 straight y end exponent over 4 equals 7 over 12$

Question 144

Wired Faculty App

Write the degree of the differential equation $straight x cubed open parentheses fraction numerator straight d squared straight y over denominator dx squared end fraction close parentheses squared space plus space open parentheses dy over dx close parentheses to the power of 4 space equals space 0$

Solution

straight x cubed open parentheses fraction numerator straight d squared straight y over denominator dx squared end fraction close parentheses squared plus open parentheses dy over dx close parentheses to the power of 4 space equals space 0

We know that the degree of a differential equation is the highest power (exponent) of the highest order derivative in it.
The highest order derivative present in the given differential equation is

fraction numerator straight d squared straight y over denominator dx squared end fraction.

Its power is 2. So, the degree of the given differential equation is 2.

Question 145

Wired Faculty App

The amount of pollution content added in air in city due to x-diesel vehicles is given by $straight P left parenthesis straight x right parenthesis space equals space 0.005 straight x cubed plus 0.02 straight x squared plus 30 straight x.$ Find the marginal increase in pollution content when 3 diesel vehicles are added and write which value is indicated in the above questions.

Solution

straight P left parenthesis straight x right parenthesis space equals space 0.005 straight x cubed plus 0.02 straight x squared plus 30 straight x

Differentiating w.r.t.x,
Marginal increase in pollution content

equals fraction numerator dP left parenthesis straight x right parenthesis over denominator dx end fraction equals 0.015 straight x squared plus 0.04 straight x plus 30

Putting x = 3 in (1),

open parentheses fraction numerator dp left parenthesis straight x right parenthesis over denominator dx end fraction close parentheses subscript straight x equals 3 end subscript space equals space 0.015 cross times 9 plus 0.04 cross times 3 plus 30 equals 30.255

Therefore, the value of marginal increase pollution content is 30.255.
Increase in number of vehicles increases the pollution. We should aim at saving the environment by reducing the pollution by decreasing the vehicle density on road.

Question 146

Wired Faculty App

Show that the function f in $straight A space equals space straight R to the power of minus open curly brackets 2 over 3 close curly brackets space defined space as space straight f left parenthesis straight x right parenthesis space equals space fraction numerator 4 straight x plus 3 over denominator 6 straight x minus 4 end fraction space is space one minus one space and space onto. space Hence space find space straight f to the power of negative 1 end exponent.$

Solution

straight f left parenthesis straight x right parenthesis space equals space fraction numerator 4 straight x plus 3 over denominator 6 straight x minus 4 end fraction Let space straight f left parenthesis straight x subscript 1 right parenthesis space equals space straight f left parenthesis straight x subscript 2 right parenthesis rightwards double arrow fraction numerator 4 straight x subscript 1 plus 3 over denominator 6 straight x subscript 1 minus 4 end fraction equals fraction numerator 4 straight x subscript 2 plus 3 over denominator 6 straight x subscript 2 minus 4 end fraction rightwards double arrow space space 24 straight x subscript 1 straight x subscript 2 minus 16 straight x subscript 1 plus 18 straight x subscript 2 minus 12 space equals space 24 straight x subscript 1 straight x subscript 2 plus 18 straight x subscript 1 minus 16 straight x subscript 2 minus 12 rightwards double arrow space space 18 straight x subscript 2 plus 16 straight x subscript 2 space equals space 18 straight x subscript 1 plus 16 straight x subscript 1 rightwards double arrow 34 straight x subscript 2 space equals 34 straight x subscript 1 rightwards double arrow straight x subscript 1 space equals space straight x subscript 2

Since,

fraction numerator 4 straight x plus 3 over denominator 6 straight x minus 4 end fraction

is a real number, therefore, for every y in the co-domain of f, there exists a number x in

straight R to the power of minus open curly brackets 2 over 3 close curly brackets space such space that space straight f left parenthesis straight x right parenthesis space equals space straight y space equals space fraction numerator 4 straight x plus 3 over denominator 6 straight x minus 4 end fraction

Therefore, f(x) is onto.
Hence,

straight f to the power of negative 1 end exponent

exists.
Now,

let space straight y space equals space fraction numerator 4 straight x plus 3 over denominator 6 straight x minus 4 end fraction rightwards double arrow 6 xy minus 4 straight y space equals space 4 straight x plus 3 rightwards double arrow space 6 xy minus 4 straight x space equals space 4 straight y plus 3 rightwards double arrow straight x left parenthesis 6 straight y minus 4 right parenthesis space equals space 4 straight y plus 3 rightwards double arrow space space space straight x space equals space fraction numerator 4 straight y plus 3 over denominator 6 straight y minus 4 end fraction rightwards double arrow space space straight y equals space fraction numerator 4 straight x plus 3 over denominator 6 straight x minus 4 end fraction space space left square bracket interchanging space the space variables space straight x space and space straight y right square bracket rightwards double arrow straight f to the power of negative 1 end exponent left parenthesis straight x right parenthesis space equals space fraction numerator 4 straight x plus 3 over denominator 6 straight x minus 4 end fraction left square bracket put space straight y space equals space straight f to the power of negative 1 end exponent left parenthesis straight x right parenthesis right square bracket

Question 147

Wired Faculty App

Differentiate the following function with respect to x:
$left parenthesis log space straight x right parenthesis to the power of straight x plus straight x to the power of log straight x$

Solution

$Let space straight y space equals left parenthesis logx right parenthesis to the power of straight x plus straight x to the power of logx space space space space space space space space... left parenthesis 1 right parenthesis Now space let space straight y subscript 1 equals left parenthesis log space straight x right parenthesis to the power of straight x space and space straight y subscript 2 equals straight x to the power of logx rightwards double arrow space space straight y space equals space straight y subscript 1 plus straight y subscript 2 space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis$
Differentiating (2), w.r.t.x,
$dy over dx equals dy subscript 1 over dx plus dy subscript 2 over dx space space space space space space space space space space... left parenthesis 3 right parenthesis$

Now consider

straight y subscript 1 space equals space left parenthesis log space straight x right parenthesis to the power of straight x

Taking log on both sides,

logy subscript 1 space equals space straight x space log left parenthesis log space straight x right parenthesis

Differentiating, w.r.t.x, we get

1 over straight y subscript 1 dy subscript 1 over dx equals straight x cross times 1 over logx cross times 1 over straight x plus 1 cross times log left parenthesis log space straight x right parenthesis rightwards double arrow space space dy subscript 1 over dx space equals space straight y subscript 1 open parentheses 1 over logx plus log left parenthesis log space straight x right parenthesis close parentheses rightwards double arrow dy subscript 1 over dx equals left parenthesis logx right parenthesis to the power of straight x open parentheses fraction numerator 1 over denominator log space straight x end fraction plus log space left parenthesis log space straight x right parenthesis close parentheses space space space space space space... left parenthesis 4 right parenthesis

Now comma space consider space straight y subscript 2 space equals straight x to the power of logx logy subscript 2 space equals space left parenthesis log space straight x right parenthesis left parenthesis log space straight x right parenthesis space equals left parenthesis log space straight x right parenthesis squared Differentiating space straight w. straight r. straight t. straight x comma space we space get 1 over straight y subscript 2 dy subscript 2 over dx equals 2 left parenthesis log space straight x right parenthesis cross times 1 over straight x rightwards double arrow space dy subscript 2 over dx space equals straight y subscript 2 open parentheses fraction numerator 2 logx over denominator straight x end fraction close parentheses equals straight x to the power of logx open parentheses fraction numerator 2 logx over denominator straight x end fraction close parentheses... left parenthesis 5 right parenthesis

Using equations (3), (4) and (5), we get:

dy over dx equals left parenthesis log space straight x right parenthesis to the power of straight x open square brackets 1 over logx plus log left parenthesis log space straight x right parenthesis close square brackets plus straight x to the power of logx open parentheses fraction numerator 2 logx over denominator straight x end fraction close parentheses

Question 148

Wired Faculty App

$If space straight y space equals space log space open square brackets straight x space plus space square root of straight x squared plus space straight a squared end root close square brackets comma space show space that space left parenthesis straight x squared space plus space straight a squared right parenthesis space fraction numerator straight d squared straight y over denominator dx squared end fraction space plus straight x dy over dx space equals space 0$

Solution

Given,

$straight y space equals space log space open square brackets straight x space plus space square root of straight x squared space plus space straight a squared end root close square brackets space straight y space equals space log space open square brackets straight x space plus space square root of straight x squared space plus straight a squared end root close square brackets...... space left parenthesis straight i right parenthesis Differentiating space left parenthesis 1 right parenthesis space straight w. straight r. straight t space space straight x comma space we space get dy over dx space equals space fraction numerator 1 plus fraction numerator straight x over denominator square root of straight x squared plus straight a squared end root end fraction over denominator straight x space plus space square root of straight x squared space plus space straight a squared end root end fraction rightwards double arrow space dy over dx space equals space fraction numerator 1 over denominator square root of straight x squared space plus space straight a squared end root end fraction space space space space space..... space left parenthesis ii right parenthesis rightwards double arrow space straight x dy over dx space equals space fraction numerator straight x over denominator square root of straight x squared space plus space straight a squared end root end fraction space space space space space space.... space left parenthesis iii right parenthesis Again comma space differentiating space left parenthesis ii right parenthesis space straight w. straight r. straight t space straight x comma space we space get fraction numerator straight d squared straight y over denominator dx squared end fraction space equals space fraction numerator negative begin display style fraction numerator 2 straight x over denominator 2 left parenthesis straight x squared space plus space straight a squared right parenthesis to the power of begin display style 1 half end style end exponent end fraction end style over denominator left parenthesis straight x squared space plus space straight a squared right parenthesis end fraction rightwards double arrow space fraction numerator straight d squared straight y over denominator dx squared end fraction space equals space minus fraction numerator straight x over denominator left parenthesis straight x squared space plus space straight a squared right parenthesis to the power of begin display style 3 over 2 end style end exponent end fraction rightwards double arrow space left parenthesis straight x squared space plus space straight a squared right parenthesis fraction numerator straight d squared straight y over denominator dx squared end fraction space equals space minus fraction numerator straight x over denominator square root of straight x squared space plus space straight a squared end root end fraction space space... space left parenthesis iv right parenthesis adding space equation space left parenthesis iii right parenthesis space and space left parenthesis iv right parenthesis comma space we space get left parenthesis straight x squared space plus space straight a squared right parenthesis space fraction numerator straight d squared straight y over denominator dx squared end fraction space plus space straight x dy over dx space equals space minus fraction numerator straight x over denominator square root of straight x squared plus straight a squared end root end fraction space space plus fraction numerator straight x over denominator square root of straight x squared space plus space straight a squared end root end fraction space equals space 0 rightwards double arrow space left parenthesis space straight x squared space plus space straight a squared right parenthesis space fraction numerator straight d squared straight y over denominator dx squared end fraction space plus space straight x dy over dx space equals space 0$

Question 149

Wired Faculty App

If $straight x equals straight a space sint space and space straight y space equals space straight a open parentheses cost plus log space tan straight t over 2 close parentheses space find space fraction numerator straight d squared straight y over denominator dx squared end fraction$

Solution

straight y space equals space straight a open parentheses cost plus log space tan straight t over 2 close parentheses rightwards double arrow dy over dt space equals space straight a open square brackets straight d over dt left parenthesis cost right parenthesis plus straight d over dt left parenthesis log space tan straight t over 2 right parenthesis close square brackets space equals space straight a open square brackets negative sint plus cot straight t over 2 cross times sec squared straight t over 2 cross times 1 half close square brackets equals space straight a open square brackets negative sint plus fraction numerator 1 over denominator 2 sin begin display style straight t over 2 end style cos begin display style straight t over 2 end style end fraction close square brackets equals straight a open parentheses negative sint plus 1 over sint close parentheses equals straight a open parentheses fraction numerator negative sin squared straight t plus 1 over denominator sint end fraction close parentheses equals straight a fraction numerator cos squared straight t over denominator sint end fraction

straight x equals straight a space sint dx over dt equals straight a straight d over dt left parenthesis sin space straight t right parenthesis space equals space straight a space cost therefore space dy over dx equals fraction numerator open parentheses begin display style dy over dt end style close parentheses over denominator open parentheses begin display style dx over dt end style close parentheses end fraction equals space fraction numerator open parentheses straight a begin display style fraction numerator cos squared straight t over denominator sint end fraction end style close parentheses over denominator acost end fraction space equals cost over sint equals cot space straight t fraction numerator straight d squared straight y over denominator dx squared end fraction equals negative cosec squared straight t dt over dx equals negative cosec squared straight t cross times fraction numerator 1 over denominator straight a space cost end fraction equals negative fraction numerator 1 over denominator straight a space sin squared straight t space cost end fraction

Question 150

Wired Faculty App

Show that the differential equation $2 ye to the power of straight x divided by straight y end exponent dx plus left parenthesis straight y minus 2 straight x space straight e to the power of straight x divided by straight y end exponent right parenthesis space dy space equals space 0$ is homogeneous. Find the particular solution of this differential equation, given that x = 0 when y = 1.

Solution

2 ye to the power of straight x divided by straight y end exponent dx plus left parenthesis straight y minus 2 straight x space straight e to the power of straight x divided by straight y end exponent right parenthesis dy space equals space 0 dy over dx space equals space fraction numerator 2 xe to the power of begin display style straight x over straight y end style end exponent minus straight y over denominator 2 ye to the power of begin display style straight x over straight y end style end exponent end fraction space space space space... left parenthesis 1 right parenthesis

Let space straight F left parenthesis straight x comma space straight y right parenthesis space equals space fraction numerator 2 xe to the power of begin display style straight x over straight y end style end exponent minus straight y over denominator 2 ye to the power of begin display style straight x over straight y end style end exponent end fraction

Then comma space straight F left parenthesis λx comma space λy right parenthesis space equals space fraction numerator straight lambda open parentheses 2 xe to the power of begin display style straight x over straight y end style end exponent minus straight y close parentheses over denominator straight lambda open parentheses 2 ye to the power of begin display style straight x over straight y end style end exponent close parentheses end fraction equals straight lambda degree open square brackets straight F left parenthesis straight x comma space straight y right parenthesis close square brackets

Thus, F(x, y) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation.
Let x = vy
Differentiating w.r.t. y, we get

dx over dy equals straight v plus straight y dv over dy

Substituting the value of x and

dx over dy

in equation (1), we get

straight v plus straight y dv over dy equals fraction numerator 2 vye to the power of straight v minus straight y over denominator 2 ye to the power of straight v end fraction space equals fraction numerator 2 ve to the power of straight v minus 1 over denominator 2 straight e to the power of straight v end fraction

or space space straight y dv over dy space equals fraction numerator 2 ve to the power of straight v minus 1 over denominator 2 straight e to the power of straight v end fraction minus straight v or space space straight y dv over dy equals negative fraction numerator 1 over denominator 2 straight e to the power of straight v end fraction or space space 2 straight e to the power of straight v dv space equals fraction numerator negative dy over denominator straight y end fraction or space space integral 2 straight e to the power of straight v. dv space equals space minus integral dy over straight y or space space 2 straight e to the power of straight v space equals space minus log space open vertical bar straight y close vertical bar plus straight C

Substituting the value of v, we get

2 straight e to the power of straight x over straight y end exponent plus log space open vertical bar straight y close vertical bar space equals space straight C space space... left parenthesis 2 right parenthesis

Substituting x = 0 and y = 1 in equation (2), we get

2 straight e degree plus log space open vertical bar 1 close vertical bar space equals space straight C space rightwards double arrow space straight C space equals space 2

Substituting the value of C in equation (2), we get

2 straight e to the power of straight x over straight y end exponent plus log space open vertical bar straight y close vertical bar space equals space 2 comma

which is the particular solution of the given differential equation.

Question 151

Wired Faculty App

Find the differential equation representing the family of curves y = ae^bx ^{+ 5} , where a and b are arbitrary constants.

Solution

$y = {ae}^{bx} x e^{5} y = {ae}^{bx} x e^{5} y = {αe}^{bx} where e^{5} a = α Differentiate w . r . t .' x' \frac{dy}{dx} = α {be}^{bx} ∴ \frac{dy}{dx} = by \frac{\frac{dy}{dx}}{y} = b again differentiate w . r . t' x' \frac{y \frac{d^{2} y}{{dx}^{2}} - \frac{dy}{dx} x \frac{dy}{dx}}{y^{2}} = 0 y \frac{d^{2} y}{{dx}^{2}} - {(\frac{dy}{dx})}^{2} = 0$

Question 152

Wired Faculty App

If y = sin (sin x), prove that $\frac{d^{2} y}{{dx}^{2}} + \tan x \frac{dy}{dx} + y \cos^{2} x = 0$

Solution

As y = sin (sin x)

$\Rightarrow \frac{dy}{dx} = \cos (\sin x) \cos x . . . . (i) and \frac{d^{2} y}{{dx}^{2}} = \cos (\sin x) (- \sin x) - \cos^{2} x (\sin (sinx)) Now \frac{d^{2} y}{{dx}^{2}} + \tan x \frac{dy}{dx} + y \cos^{2} x = - sinx \cos (\sin x) - \cos^{2} x \sin (\sin x) + \frac{\sin x}{\cos x} x \cos x \cos (\sin x) + \sin (\sin x) \cos^{2} x = - \sin x \cos (\sin x) - \cos^{2} x sion (\sin x) + \sin x \cos (\sin x) + \cos^{2} x \sin (\sin x) = 0 = R . H . S Hence we have proved that \frac{d^{2} y}{{dx}^{2}} + \tan x \frac{dy}{dxx} + y \cos^{2} x = 0 for y = \sin (\sin x)$

Question 153

Wired Faculty App

Find the particular solution of the differential equation e^x tan ydx + (2 -e^x) sec² ydy = 0, given that $y = \frac{π}{4}$ when x = 0

Solution

$e^{x} \tan y dx + (2 - e^{x}) \sec^{2} y dy = 0 e^{x} \tan y dx = (e^{x} - 2) \sec^{2} y dy = 0 e^{x} \tan y dx = (e^{x} - 2) \sec^{2} y dy \int \frac{e^{x} dx}{e^{x} - 2} = \int \frac{\sec^{2} y dy}{\tan y} In | e^{x} - 2 | = In | \tan y | + In C In | e^{x} - 2 | = In (C \tan y) e^{x} - 2 = C \tan y Given; x = 0, y = \frac{π}{4} e^{o} - 2 = C \tan (\frac{π}{4}) e^{o} - 2 = - C \tan (\frac{π}{4}) 1 - 2 = C x 1 \Rightarrow C = - 1 ∴ e^{x} - 2 = - \tan e^{x} - 2 + \tan y = 0$

Question 154

Wired Faculty App

Find the particular solution of the differential equation $\frac{dy}{dx} + 2 y \tan x = \sin x$ , given that y = 0 when $x = \frac{π}{3}$

Solution

$\frac{dy}{dx} + (2 \tan x) y = \sin x \frac{dy}{dx} + py = Q P = 2 \tan x and Q = \sin x I . F = e^{\int Pdx} = e^{2 \int \tan x dx} = e^{2 In \sec x} = e^{In \sec^{2} x} = \sec^{2} x soln . y (I . F) = \int Q (I . F) dx y . \sec^{2} x = \int \sin x . \sec^{2} x dx y \sec^{2} x = \int \tan x \sec x dx y \sec^{2} x = \sec x + C Given y = 0 x = \frac{π}{3} \sec \frac{π}{3} + C = 0 C = - \sec \frac{π}{3} = - 2 ∴ y \sec^{2} x = \sec x - 2 y \sec^{2} x - \sec x + 2 = 0$

Question 155

Wired Faculty App

If $(x^{2} + y^{2})^{2} = xy, find \frac{dy}{dx}$

Solution

Given:

$(x^{2} + y^{2})^{2} = xy x^{4} + y^{4} + 2 x^{2} y^{2} = xy diff w . r . t x 4 x^{3} + 4 y^{3} \frac{dy}{dx} + 2 (2 x^{2} y \frac{dy}{dx} + 2 {xy}^{2}) = (x \frac{dy}{dx} + y) 4 y^{3} \frac{dy}{dx} + 4 x^{2} y \frac{dy}{dx} - x \frac{dy}{dx} = y - 4 x^{3} - 4 {xy}^{2} \frac{dy}{dx} (4 y^{3} + 4 x^{2} y - x) = y - 4 x^{3} - 4 {xy}^{2} \frac{dy}{dx} = \frac{y - 4 x^{3} - 4 {xy}^{2}}{4 y^{3} + 4 x^{2} y - x}$

Question 156

Wired Faculty App

If $x = a (2 θ - \sin 2 θ) and y = a (1 - \cos 2 θ), find \frac{dy}{dx} when θ = \frac{π}{3}$

Solution

$y = a [1 - \cos 2 θ], \frac{dy}{dθ} = a (0 - 2 \sin 2 θ) ∴ \frac{dy}{dθ} = - 2 a \sin 2 θ x = a (2 θ - \sin 2 θ), \frac{dy}{dθ} = a (2 - 2 \cos 2 θ) \Rightarrow \frac{dy}{dx} = \frac{\frac{dy}{dθ}}{\frac{dx}{dθ}} = \frac{- 2 a \sin 2 θ}{2 a [1 - \cos 2 θ]} (∵ \frac{dx}{dθ} \neq 0) \Rightarrow \frac{- 2 \sin θ cosθ}{2 \sin^{2} θ} = - \cot θ ∴ \frac{dy}{dx} = - \cot (\frac{π}{3}) = - \frac{1}{\sqrt{3}}$

Question 157

Wired Faculty App

Differentiate the following with respect of x:
$y = \tan^{- 1} (\frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}})$

Solution

$Let x = \cos 2 θ \Rightarrow θ = \frac{1}{2} \cos^{- 1} x . . . . . . . . . . . . . . . . . (1) ∴ \sqrt{1 + x} = \sqrt{1 + \cos 2 θ} = \sqrt{1 + 2 \cos^{2} θ - 1} = \sqrt{2} cosθ \sqrt{1 - x} = \sqrt{1 - \cos 2 θ} = \sqrt{1 - 1 - 2 \sin^{2} θ} = \sqrt{2} sinθ Let y = \tan^{- 1} [\frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}}] = \tan^{- 1} [\frac{\sqrt{2} cosθ - \sqrt{2} sinθ}{\sqrt{2} cosθ + \sqrt{2} sinθ}] = \tan^{- 1} [\frac{1 - tanθ}{1 + tanθ}] = \tan^{- 1} \{\tan (\frac{π}{4} - θ)\} = \frac{π}{4} - θ = \frac{π}{4} - \frac{1}{2} \cos^{- 1} x . . . . . . . . . From (1) ∴ \frac{dy}{dx} = - \frac{1}{2} (\frac{- 1}{\sqrt{1 - x^{2}}}) = \frac{1}{2 \sqrt{1 - x^{2}}}$

Question 158

Wired Faculty App

Solve the following differential equation:
(x² − y²⁾ dx + 2xy dy = 0 given that y = 1 when x = 1

Solution

( x² - y² ) dx + 2xy dy = 0

$\Rightarrow \frac{dy}{dx} = \frac{y^{2} - x^{2}}{2 xy} . . . . . . . . . (1)$

It is a homogeneous differential equation.

Let y = vx ..........(2)

$∴ \frac{dy}{dx} = v + x \frac{dv}{dx} . . . . . . . . . . . (3)$

Substituting (2) and (3) in (1), we get:

$v + x \frac{dv}{dx} = \frac{v^{2} x^{2} - x^{2}}{2 x . vx} v + x \frac{dv}{dx} = \frac{x^{2} (v^{2} - 1)}{2 {vx}^{2}} = \frac{v^{2} - 1}{2 v} 2 v^{2} + 2 vx \frac{dv}{dx} = v^{2} - 1 2 vx \frac{dv}{dx} = - v^{2} - 1 (\frac{2 v}{v^{2} + 1}) dv = - \frac{dx}{x}$

Integrating both sides, we get:

$\int \frac{2 v}{v^{2} + 1} dv = - \int (\frac{1}{x}) dx \log |v^{2} + 1| = - \log |x| + \log C \log |v^{2} + 1| = \log |\frac{C}{x}| v^{2} + 1 = \frac{C}{x} x (v^{2} + 1) = C$

$x [{(\frac{y}{x})}^{2} + 1] = C y^{2} + x^{2} = Cx . . . . . . . . . (4)$

It is given that when x = 1, y = 1

(1)² + (2)² = C(1)

$\Rightarrow$ C = 2

Thus, the required solution is y² + x² = 2x.

Question 159

Wired Faculty App

Solve the following differential equation:

$\frac{dy}{dx} = \frac{x (2 y - x)}{x (2 y + x)}$ , if y = 1 when x = 1

Solution

We need to solve the following differential equation

$\frac{dy}{dx} = \frac{x (2 y - x)}{x (2 y + x)} \frac{dy}{dx} = \frac{2 y - x}{2 y + x} . . . . . . . . (1)$

It is a homogeneous differential equation.

Let y = vx ..........(2)

$∴ \frac{dy}{dx} = v + x \frac{dv}{dx} . . . . . . . . . . (3)$

Substituting (2) and (3) in (1), we get:

$v + x \frac{dv}{dx} = \frac{x (2 v - 1)}{x (2 v + 1)} x \frac{dv}{dx} = \frac{2 v - 1}{2 v + 1} - v x \frac{dv}{dx} = \frac{- 2 v^{2} = v - 1}{2 v + 1} (\frac{2 v + 1}{- 2 v^{2} + v - 1}) dv = (\frac{1}{x}) dx (\frac{2 v + 1}{2 v^{2} - v + 1}) dv = (- \frac{1}{x}) dx$

Integrating both sides,

$\int \frac{1}{2} (\frac{4 v - 1 + 3}{2 v^{2} - v + 1}) dv = \int (\frac{1}{x}) dx \int \frac{1}{2} (\frac{4 v - 1}{2 v^{2} - v + 1}) dv + \int \frac{3}{2} (\frac{1}{2 v^{2} - v + 1}) dv = \int (- \frac{1}{x}) dx$

$\int \frac{1}{2} (\frac{4 v - 1}{2 v^{2} - v + 1}) dv + \int \frac{3}{4} [\frac{1}{v^{2} - \frac{v}{2} + \frac{1}{2}}] = \int (- \frac{1}{x}) dx$

$\frac{1}{2} \log |2 v^{2} - v + 1| + \frac{3}{4} \int [\frac{1}{v^{2} - \frac{v}{2} + \frac{1}{16} + \frac{7}{16}}] dv = - \log |x| + C \frac{1}{2} \log |2 v^{2} - v + 1| + \frac{3}{4} \int [\frac{dv}{{(v - \frac{1}{4})}^{2} + {(\frac{\sqrt{7}}{4})}^{2}}] = - \log |x| + C \frac{1}{2} \log |2 v^{2} - v + 1| + \frac{3}{4} x \frac{4}{\sqrt{7}} \tan^{- 1} (\frac{v - \frac{1}{4}}{\frac{\sqrt{7}}{4}}) = - \log |x| + C \frac{1}{2} \log |2 v^{2} - v + 1| + \frac{3}{\sqrt{7}} \tan^{- 1} (\frac{4 v - 1}{\sqrt{7}}) = C - \log |x|$

$Put v = \frac{y}{x} \frac{1}{2} \log |2 {(\frac{y}{x})}^{2} - (\frac{y}{x}) + 1| + \frac{3}{\sqrt{7}} \tan^{- 1} [\frac{\frac{4 y}{x} - 1}{\sqrt{7}}] = C - \log |x| \frac{1}{2} \log |\frac{2 y^{2} - xy + x^{2}}{x^{2}}| + \frac{3}{\sqrt{7}} \tan^{- 1} (\frac{4 y - x}{\sqrt{7} x}) = C - \log |x| . . . . . . . . . . (4) Now y = 1 when x = 1 \frac{1}{2} \log |\frac{2 (1)^{2} - 1 x 1 + 1^{2}}{1^{2}}| + \frac{3}{\sqrt{7}} \tan^{- 1} (\frac{4 x 1 - 1}{\sqrt{7} x 1}) = C - \log |1| \frac{1}{2} \log 2 + \frac{3}{\sqrt{7}} \tan^{- 1} (\frac{3}{\sqrt{7}}) = C . . . . . . . . . . . . . . . . . (5)$

Therefore, from (4) and (5) we get:

$\frac{1}{2} \log |\frac{2 y^{2} - xy + x^{2}}{x^{2}}| + \frac{3}{\sqrt{7}} \tan^{- 1} (\frac{4 y - x}{\sqrt{7} x}) = \frac{1}{2} \log 2 + \frac{3}{\sqrt{7}} \tan^{- 1} (\frac{3}{\sqrt{7}}) - \log |x| \frac{1}{2} \log |\frac{2 y^{2} - xy + x^{2}}{x^{2}}| - \frac{1}{2} \log 2 + \log |x| = \frac{3}{\sqrt{7}} [\tan^{- 1} \frac{3}{\sqrt{7}} - \tan^{- 1} (\frac{4 y - x}{\sqrt{7} x})] \frac{1}{2} \log |\frac{2 y^{2} - xy + x^{2}}{x^{2}} x^{2}| = \frac{3}{\sqrt{7}} \tan^{- 1} (\frac{\frac{3 x - 4 y + x}{\sqrt{7} x}}{1 + \frac{3 x (4 y - x)}{7 x}}) \frac{1}{2} \log |\frac{2 y^{2} - xy + x^{2}}{2}| = \frac{3}{\sqrt{7}} \tan^{- 1} (\frac{\frac{4 (x - y)}{\sqrt{7} x}}{\frac{7 x + 12 y - 3 x}{7 x}}) \frac{1}{2} \log |\frac{2 y^{2} - xy + x^{2}}{2}| = \frac{3}{\sqrt{7}} \tan^{- 1} (\frac{\sqrt{7} (x - y)}{(x + 3 y)})$

Question 160

Wired Faculty App

Solve the following differential equation:
$\cos^{2} x \frac{dy}{dx} + y = tanx$

Solution

$\cos^{2} x \frac{dy}{dx} + y = \tan x \frac{dy}{dx} + \sec^{2} x . y = \tan x . \sec^{2} x$

It is a linear differential equation of the first order.

comparing it with $\frac{dy}{dx} + py = Q,$ we get:

P = sec²x and Q = tanx.sec²x

$Integration factor = e^{\int P dx} = e^{\int \sec^{2} x dx} = e^{\tan x} The solutoin of the given linear differential equation is given as : y e^{\tan x} = \int \tan x . \sec^{2} x . e^{\tan x} dx + C Let \tan x = t \Rightarrow \sec^{2} x dx = dt y e^{t} = \int t . e^{t} . dt + C y e^{t} = t . e^{t} - \int 1 . e^{t} . dt + C y e^{t} = t . e^{t} - e^{t} + C y e^{t} = e^{t} (t - 1) + C y e^{\tan x} = e^{\tan x} \tan x - 1 + C y = \tan x - 1 + C e^{- \tan x}$

Question 161

Wired Faculty App

Solve the following differential equation:
$(1 + x^{2}) \frac{dy}{dx} + y = \tan^{- 1} x$

Solution

$(1 + x^{2}) \frac{dy}{dx} + y = \tan^{- 1} x$

The equation can be expressed as

$\frac{dy}{dx} + \frac{y}{1 + x^{2}} = \frac{\tan^{- 1} x}{1 + x^{2}} . . . . . . . . . (i)$

This is a linear differential equation of the type $\frac{dy}{dx} + Py = Q$

$So, I . F = e^{\int \frac{1}{1 + x^{2}} dx} = e^{\tan^{- 1} x} Solution of (i) y e^{\tan^{- 1} x} = \int e^{\tan^{- 1} x} (\frac{\tan^{- 1} x}{1 + x^{2}}) dx . . . . . . . . (ii) For R . H . S ., let \tan^{- 1} x = t \Rightarrow \frac{1}{1 + x^{2}} dx = dt$

Substituting in equation (ii)

${ye}^{\tan^{- 1} x} = \int e^{t} . tdt \Rightarrow y . e^{\tan^{- 1} x} = [{te}^{t} - e^{t}] + C \Rightarrow y . e^{\tan^{- 1} x} = e^{\tan^{- 1} x} (\tan^{- 1} x - 1) + C \Rightarrow y = (\tan^{- 1} x - 1) + {Ce}^{- \tan^{- 1} x}$

Question 162

Wired Faculty App

Find the particular solution, satisfying the given condition, for the following differential equation:
$\frac{dy}{dx} - \frac{y}{x} + cosec (\frac{y}{x}) = 0; y = 0 when x = 1$

Solution

$\frac{dy}{dx} - \frac{y}{x} + cosec (\frac{y}{x}) = 0 . . . . . . . (i) y = 0 When x = 1 Let \frac{y}{x} = t \Rightarrow y = xt \Rightarrow \frac{dy}{dx} = x \frac{dt}{dx} + t By substituting \frac{dy}{dx} in equation (i) (x \frac{dt}{dx} + t) - t + cosect = 0 \Rightarrow x \frac{dt}{dx} = - cosect \Rightarrow \int \frac{dt}{\cos ect} + \int \frac{dx}{x} = 0 \Rightarrow - cost + logx = C \Rightarrow - \cos (\frac{y}{x}) + logx = C$

Using y = 0 When x= 1

$- 1 + 0 = C \Rightarrow C = - 1 So the solution is : \cos (\frac{y}{x}) = logx + 1$

Question 163

Wired Faculty App

What is the degree of the following differential equation?
5x $5 x {(\frac{dy}{dx})}^{2} - \frac{d^{2} y}{{dx}^{2}} - 6 y = \log x$

Solution

$Given differential equation is 5 x {(\frac{dy}{dx})}^{2} - \frac{d^{2} y}{{dx}^{2}} - 6 y = \log x In the above equation the highest order derivative is \frac{d^{2} y}{{dx}^{2}} and its power is 1 . Thus, the degree of differential equation 5 x {(\frac{dy}{dx})}^{2} - \frac{d^{2} y}{{dx}^{2}} - 6 y = \log x is 1 .$

Question 164

Wired Faculty App

Find the particular solution of the differential equation satisfying the given conditions: x² dy + (xy + y² )dx = 0; y = 1 when x = 1.

Solution

$x^{2} dy + (xy + y^{2}) dx = 0 x^{2} dy = - (xy + y^{2}) dx \Rightarrow \frac{dy}{dx} = - \frac{- (xy + y^{2})}{x^{2}} . . . . . . . . . . (i)$

This is a homogeneous differential equation.

Such type of equations can be reduced to variable separable form by the substitution y = vx.

Differentiating w.r.t.x we get,

$\frac{d}{dx} (y) = \frac{d}{dx} (vx) \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx} substituting the value of y and \frac{dy}{dx} in equation (i), we get : v + x \frac{dv}{dx} = \frac{- [x \times vx + {(vx)}^{2}]}{x^{2}} \Rightarrow x \frac{dv}{dx} = - v^{2} - 2 v = - v (v + 2) \Rightarrow \frac{dv}{v (v + 2)} = - \frac{dx}{x} \Rightarrow \frac{1}{2} [\frac{1}{v} - \frac{1}{v + 2}] dv = - \frac{dx}{x}$

Integrating both sides, we get:

$\frac{1}{2} [\log v - \log (v + 2)] = - \log x + \log C \Rightarrow \frac{1}{2} \log (\frac{v}{v + 2}) = \log \frac{C}{x} \Rightarrow \frac{v}{v + 2} = {(\frac{C}{x})}^{2} Substituting v = \frac{y}{x}$

$\Rightarrow \frac{\frac{y}{x}}{\frac{y}{x} + 2} = {(\frac{C}{x})}^{2} \Rightarrow \frac{y}{y + 2 x} = \frac{C^{2}}{x^{2}} \Rightarrow \frac{x^{2} y}{y + 2 x} = D . . . . . . . . . . . . (ii)$

Now, it is given that y = 1 at x = 1.

$\Rightarrow \frac{1}{1 + 2} = D \Rightarrow D = \frac{1}{3} Substituting D = \frac{1}{3} in equation (ii), we get \frac{x^{2} y}{y + 2 x} = \frac{1}{3} \Rightarrow y + 2 x = 3 x^{2} y So, the required solution is y + 2 x = 3 x^{2} y$

Question 165

Wired Faculty App

Find the general solution of the differential equation,
$x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x$

Solution

$x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x$

Dividing all the terms of the equation by x log x, we get

$\Rightarrow \frac{dy}{dx} + \frac{y}{x \log x} = \frac{2}{x^{2}}$

This equation is in the form of a linear differential equation

$\frac{dy}{dx} + py = Q, where P = \frac{1}{x \log x} and Q = \frac{2}{x^{2}} Now, I . F = e^{\int pdx} = e^{\int \frac{1}{x \log x} dx} = e^{\log (\log x)} = \log x$

The general solution of the given differential equation is given by

y x I.F. = $\int$ ( Q x I.F. ) dx + C

$\Rightarrow y \log x = \int (\frac{2}{x^{2}} \log x) dx \Rightarrow y \log x = 2 \int (\log x \times \frac{1}{x^{2}}) dx . = 2 [\log x \times \int \frac{1}{x^{2}} dx - \int \{\frac{d}{dx} (\log x) \times \int \frac{1}{x^{2}} dx\} dx] = 2 [\log x (- \frac{1}{x}) - \int (\frac{1}{x} \times (- \frac{1}{x})) dx] = 2 [- \frac{\log x}{x} + \int \frac{1}{x^{2}} dx] = 2 [- \frac{\log x}{x} - \frac{1}{x}] + C So the required general solution is y \log x = - \frac{2}{x} (1 + \log x) + C$

Question 166

Wired Faculty App

Find the particular solution of the differential equation satisfying the given conditions:
$\frac{dy}{dx} = y \tan x$ , given that y = 1 when x= 0.

Solution

$\frac{dy}{dx} = y \tan x \Rightarrow \frac{dy}{y} = \tan x dx$

On integration, we get

$\int \frac{dy}{y} = \int \tan x dx \Rightarrow \log y = \log (\sec x) + \log C . . . . . . . . (i) \Rightarrow \log y = \log (C \sec x) \Rightarrow y = C \sec x$

Now, it is given that y = 1 when x = 0

$\Rightarrow 1 = C x \sec 0 \Rightarrow 1 = C x 1 ∴ C = 1$

substituting C = 1 in equation (i), we get

y = sec x as the required particular solution.

Question 167

Wired Faculty App

Solve the following differential equation:
e^x tan y dx + ( 1 - e^x) sec² y dy = 0

Solution

The given differential equation is:

e^x tan y dx + ( 1 - e^x) sec² y dy = 0

$\Rightarrow$ e^x tan y dx = - ( 1 - e^x) sec² y dy

$\Rightarrow$ e^x tan y dx = ( e^x - 1 ) sec² y dy

$\Rightarrow \frac{e^{x}}{e^{x} - 1} dx = \frac{\sec^{2} y}{\tan y} dy$

On integrating on both sides, we get

$\int \frac{e^{x}}{e^{x} - 1} dx = \int \frac{\sec^{2} y}{\tan y} dy . . . . . . . . . . (i) Let I_{1} = \int \frac{\sec^{2} y}{\tan y} dy Put \tan y = t \Rightarrow \sec^{2} y dy = dt ∴ \int \frac{\sec^{2} y}{\tan y} dy = \int \frac{dt}{t} = \log |t| = \log \tan y . . . . . . . . . (ii) Let I_{2 =} \int \frac{e^{x}}{e^{x} - 1} dx$

Put e^x - 1 = u

$∴$ e^x dx = du

$\int \frac{e^{x}}{e^{x} - 1} dx = \int \frac{du}{u}$

= log u

= log ( e^x - 1 ) ............(iii)

From (i), (ii), and (iii), we get

log tan y = log ( e^x - 1 ) + log C

$\Rightarrow$ log tan y = log C ( e^x - 1 )

$\Rightarrow$ tan y = C ( e^x - 1 )

The solution of the given differential equation is tan y = C ( e^x - 1 ).

Question 168

Wired Faculty App

Solve the following differential equation:
$\cos^{2} x \frac{dy}{dx} + y = \tan x$

Solution

$\cos^{2} x \frac{dy}{dx} + y = \tan x \Rightarrow \frac{dy}{dx} + \sec^{2} x . y = \sec^{2} x \tan x This equation is in the form of \frac{dy}{dx} + py = Q Here P = \sec^{2} x and Q = \sec^{2} x \tan x Integrating factor, I . F = e^{\int p dx} = e^{\int \sec^{2} x dx} = e^{\tan x}$

The general solution can be given by

y ( I. F ) = $\int$ ( Q x I. F ) dx + C ..........(i)

Let tan x = t

$\frac{d}{dx} (\tan x) = \frac{dt}{dx} \Rightarrow \sec^{2} x = \frac{dt}{dx} \Rightarrow \sec^{2} x dx = dt$

Therefore, equation (i) becomes:

$y e^{\tan x} = \int (e^{t} . t) dt \Rightarrow y e^{\tan x} = \int (e^{t} . t) dt + C \Rightarrow y e^{\tan x} = t . \int e^{t} dt - \int (\frac{d}{dt} (t) . \int e^{t} dt) dt + C \Rightarrow y e^{\tan x} = t . e^{t} - \int e^{t} dt + C \Rightarrow y e^{\tan x} = t . e^{t} - e^{t} + C \Rightarrow y e^{\tan x} = (t - 1) e^{t} + C \Rightarrow y e^{\tan x} = (\tan x - 1) e^{\tan x} + C \Rightarrow y = (\tan x - 1) + C e^{- \tan x}, where C is an arbitary constant .$

Question 169

Wired Faculty App

How many real solutions does the equation x⁷+ 14x⁵+ 16x³+ 30x – 560 = 0 have?

7

1

3

5

Solution

x⁷+ 14x⁵+ 16x³+ 30x – 560 = 0
Let f(x) = x⁷+ 14x⁵+ 16x³+ 30x
⇒ f′(x) = 7x⁶+ 70x⁴+ 48x²+ 30 > 0 ∀ x.
∴ f (x) is an increasing function ∀ x.

Question 170

Wired Faculty App

Let the orthocentre and centroid of a triangle be A(–3, 5) and B(3, 3) respectively. If C is the circumcentre of this triangle, then the radius of the circle having line segment AC as diameter, is

$\frac{3 \sqrt{5}}{2}$

$\sqrt{10}$

$2 \sqrt{10}$

$3 \sqrt{\frac{5}{2}}$

Solution

$3 \sqrt{\frac{5}{2}}$

A(-3,5)

B(3,3)

so,

$A B = \sqrt[2]{10} N o w, a s, A C = \frac{3}{2} A B S o, R a d i u s = \frac{3}{4} A B = \frac{3}{2} \sqrt{10} = 3 \sqrt{\frac{5}{2}}$

Mathematics Part Ii Chapter 9 Differential Equations

NCERT Solution For Class 12 Mathematics Mathematics Part Ii

At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, – 3). Find the equation of the curve given that it passes through ( 2, 1).

Find the equation of the curve passing through the point whose differential equation is sin x cos y dx + cos x sin y dy = 0.

Find the equation of a curve passing through the point (0, 0) and whose differential equation is y' = ex sin x.

The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

In a bank, principal increases continuously at the rate of 5% per year. In how many years Rs. 1000 double itself ?

In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs. 100 double itself in 10 years.

In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs. 100 double itself in 10 years. (e0·5 = 1·648).

In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20.000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009 ?

The general solution of the differential equation is ex + e– y = C ex + ey = C e– x + ey = C e– x + e–y = C

Solve

Solve

Solve the following differential equation:

Solve the following differential equation:

Solve the following differential equation:

Solve the differential equation

Solve: sin (x+y) .

Solve:

Solve:

Solve:

Solve the following initial value problem:(x + y + 1)2 dy = dx, y ( –1) = 0

Solve the following initial value problemcos (x + y) dy = dx, y (0) = 0

For the following differential equation, given below, find particular solution satisfying the given condition:

For the following differential equation, given below, find particular solution satisfying the given condition:

For the following differential equation, given below, find particular solution satisfying the given condition:

For the following differential equation, given below, find particular solution satisfying the given condition:

Solve:

Solve the following differential equation:

Solve the differential equation:

Solve the differential equation:

Solve the following differential equation:(y2 – x2) dy = 3 x y dx

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:(x - y) y' = x + 2 y

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:(x2 + y2) y' = 8 x2 - 3 x y + 2 y2

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:(3 x y + y2) dx = (x2 + x y) dy

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:2 x y dx + (x2 + 2 y2) dy = 0

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:

Show that the given differential equation is homogeneous and solve it:

Show that the given differential equation is homogeneous and solve it:

Show that the given differential equation is homogeneous and solve it:(x-y) dy - (x+y) dx = 0

Show that the given differential equation is homogeneous and solve it:

Show that the given differential equation is homogeneous and solve it:

Show that the given differential equation is homogeneous and solve it:

Solve

Find a one parameter family of solutions of each of the following differential equation:y2 + x2 y' = x y y'

Find a one parameter family of solutions of each of the following differential equation:(y2 – 2xy) dx = (x2 – 2xy) dy

Find a one parameter family of solutions of each of the following differential equation:y2 dx + (x2 – x y + y2) dy = 0

Find a one parameter family of solutions of each of the following differential equation: (x2 + x y) dy = (x2 + y2) dx

Solve the following differential equation:.

Solve

Show that the differential equation is homogeneous and solve it.

Show that the differential equation:

Solve:

Solve:

Solve:

Solve:

For the given differential equations, find the particular solution satisfying the given condition:

For the given differential equation, find the particular solution satisfying the given condition:

For the given differential equation, find the particular solution satisfying the given condition:

For the given differential equation, find the particular solution satisfying the given condition:

For the given differential equation, find the particular solution satisfying the given condition:

For the given differential equation, find the particular solution satisfying the given condition:

For the given differential equation, find the particular solution satisfying the given condition:

Find a particular solution of the differential equation(x – y) (dx + dy) = dx – dy. given that y = – 1, when x = 0.

Solve the following differential equation:

Solve the following initial value problem:

Show that the family of curves for which the slope of the tangent at any point (x, y) on it is is given by

The general solution of the differential equation is xy = C x = Cy2 y = Cx y = Cx2

A homogeneous differential equation of the from can be solved by making the substitution. y = vx v = yx x = vy x = v

Which of the following is a homogeneous differential equation? (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0 (xy) dx – (x3 + y3) dy = 0 (x3 + 2 y2) dx + 2xy dy = 0 y2 dx + (x2 – x y – y2) dy =0

Solve :

Solve the differential equation:

Solve the differential equation:

Solve

Find the equation of the curve passing through the point $open parentheses 0 comma space straight pi over 4 close parentheses$ whose differential equation is sin x cos y dx + cos x sin y dy = 0.

Find the equation of a curve passing through the point (0, 0) and whose differential equation is y' = e^x sin x.

In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs. 100 double itself in 10 years. (e^0·5 = 1·648).

The general solution of the differential equation $dy over dx space equals space straight e to the power of straight x plus straight y end exponent$ is
e^x + e^{– y} = C
e^x+ e^y = C
e^{– x} + e^y = C
e^{– x}+ e^–y = C

Solve $dy over dx space equals space left parenthesis 4 straight x plus straight y plus 1 right parenthesis squared.$

Solve $left parenthesis straight x minus straight y right parenthesis squared space dy over dx space equals space straight a squared.$

Solve the following differential equation:
$open parentheses straight x plus straight y close parentheses squared dy over dx space equals straight a squared$

Solve the following differential equation:
$open parentheses straight x plus straight y plus 2 close parentheses space dy over dx space equals 2$

Solve the following differential equation:
$dy over dx plus 1 space equals space straight e to the power of straight x plus straight y end exponent$

Solve the differential equation $dy over dx space equals sin left parenthesis straight x plus straight y right parenthesis space space or space sin to the power of negative 1 end exponent open parentheses dy over dx close parentheses space equals space straight x plus straight y$

Solve: sin (x+y) $dy over dx space equals 1$ .

Solve:
$dy over dx space equals space cos space left parenthesis straight x plus straight y right parenthesis space space or space space cos to the power of negative 1 end exponent open parentheses dy over dx close parentheses space equals space straight x plus straight y$

Solve:
$cos space left parenthesis straight x plus straight y right parenthesis space dy over dx space equals space 1.$

Solve:
$dy over dx space equals space tan space left parenthesis straight x plus straight y right parenthesis$

Solve the following initial value problem:
(x + y + 1)² dy = dx, y ( –1) = 0

Solve the following initial value problem
cos (x + y) dy = dx, y (0) = 0

For the following differential equation, given below, find particular solution satisfying the given condition:
$dy over dx equals straight y space tanx space semicolon space space straight y space equals space 1 space space when space straight x space equals space 0$

Solve: $straight x squared dy over dx space equals straight x squared plus 5 xy plus 4 straight y squared.$

Solve the following differential equation:
$straight x squared dy over dx space equals space 2 xy plus straight y squared$

Solve the differential equation:
$straight x squared dy over dx space equals straight y left parenthesis straight x plus straight y right parenthesis.$

Solve the differential equation:
$left parenthesis straight y plus straight x right parenthesis space dy over dx space equals space straight y minus straight x.$

Solve the following differential equation:
(y² – x²) dy = 3 x y dx

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
(x - y) y' = x + 2 y

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
(x² + y²) y' = 8 x² - 3 x y + 2 y²

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
(3 x y + y²) dx = (x² + x y) dy

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
2 x y dx + (x² + 2 y²) dy = 0

Show that the given differential equation is homogeneous and solve it:
$left parenthesis straight x squared plus xy right parenthesis space dy space equals space left parenthesis straight x squared plus straight y squared right parenthesis space dx$

Show that the given differential equation is homogeneous and solve it:
$straight y apostrophe space equals space fraction numerator straight x plus straight y over denominator straight x end fraction$

Show that the given differential equation is homogeneous and solve it:
(x-y) dy - (x+y) dx = 0

Show that the given differential equation is homogeneous and solve it:
$open parentheses straight x squared minus straight y squared close parentheses space dx space plus space 2 xy space dy space equals space 0$

Show that the given differential equation is homogeneous and solve it:
$straight x squared dy over dx space equals space straight x squared minus 2 straight y squared plus straight x space straight y$

Solve $straight x space dy space minus space straight y space dx space equals space square root of straight x squared plus straight y squared end root space dx.$

Find a one parameter family of solutions of each of the following differential equation:
y² + x² y' = x y y'

Find a one parameter family of solutions of each of the following differential equation:
(y² – 2xy) dx = (x² – 2xy) dy

Find a one parameter family of solutions of each of the following differential equation:
y² dx + (x² – x y + y²) dy = 0

Find a one parameter family of solutions of each of the following differential equation:
(x² + x y) dy = (x² + y²) dx

Solve $open parentheses straight x space sin space straight y over straight x close parentheses dy space equals open parentheses straight y space sin space straight y over straight x minus straight x close parentheses dx$

Show that the differential equation $straight x space cos space open parentheses straight y over straight x close parentheses dy over dx space equals straight y space cos space open parentheses straight y over straight x close parentheses plus straight x$ is homogeneous and solve it.

Solve:
$straight y space straight e to the power of straight x over straight y end exponent dx space equals open parentheses xe to the power of straight x over straight y end exponent plus straight y squared close parentheses space space dy comma space space space straight y not equal to 0$

Solve:
$straight x dy over dx minus straight y plus straight x space tan straight y over straight x equals 0$

Solve:
$straight y apostrophe space equals space straight y over straight x plus sin straight y over straight x$

Solve:
$dy over dx space equals straight y over straight x plus tan straight y over straight x open parentheses 0 less than straight y over straight x less than straight pi over 2 close parentheses$

For the given differential equation, find the particular solution satisfying the given condition:
$2 xy plus straight y squared minus 2 straight x squared dy over dx equals 0 semicolon space space space straight y space equals 2 space space when space straight x space equals space 1.$

Find a particular solution of the differential equation
(x – y) (dx + dy) = dx – dy. given that y = – 1, when x = 0.

The general solution of the differential equation $fraction numerator straight y space dx space minus space straight x space dy over denominator straight y end fraction space equals space 0$ is

xy = C

x = Cy²
y = Cx
y = Cx²

A homogeneous differential equation of the from $dx over dy space equals space straight h open parentheses straight x over straight y close parentheses$ can be solved by making the substitution.

y = vx
v = yx

x = vy

x = v

Which of the following is a homogeneous differential equation?

(4x + 6y + 5) dy – (3y + 2x + 4) dx = 0

(xy) dx – (x³ + y³) dy = 0

(x³ + 2 y²) dx + 2xy dy = 0

y² dx + (x² – x y – y²) dy =0

Solve : $dy over dx plus straight y space equals space sin space straight x comma space space left parenthesis straight x space element of space straight R right parenthesis$

Solve the differential equation:
$straight x dy over dx minus straight y minus 2 straight x cubed space equals space 0.$

Solve the differential equation:
$dy over dx minus 2 straight y space equals space 3 straight x.$

Solve $dy over dx plus straight y over straight x space equals space straight e to the power of straight x comma space space left parenthesis straight x greater than 0 right parenthesis.$

Solve the differential equation:
$straight x dy over dx plus 2 straight y space equals straight x squared.$

Solve:
$straight v plus 2 straight y space equals 4 straight x$

Solve:
$dy over dx plus straight y over straight x equals straight x squared.$

Solve:
$dy over dx plus 4 straight y space equals space 5 straight x$

Solve:
$dy over dx plus straight y space equals space 1$

Solve:
$fraction numerator dy apostrophe over denominator dx end fraction minus straight y over straight x equals space 2 straight x squared$

Solve:
$2 straight x dy over dx plus straight y space equals space 6 straight x cubed space space or space space space dy over dx plus fraction numerator straight y over denominator 2 straight x end fraction equals 3 straight x squared.$

Solve $dy over dx plus fraction numerator 1 over denominator straight x space logx end fraction straight y space equals space 1 over straight x.$

Solve:
$straight x dy over dx plus straight y space equals space straight x space log space straight x$

Solve:
$straight x dy over dx plus 2 straight y space equals space straight x squared log space straight x$

Solve:
$straight x space log space straight x space dy over dx plus straight y space equals space 2 over straight x log space straight x$

Find one-parameter families of solution curves of the following differential equations:
$straight y apostrophe plus 3 straight y space equals straight e to the power of negative 2 straight x end exponent$

Find one-parameter families of solution curves of the following differential equation:
$straight y apostrophe space minus space straight y space equals space cos space straight x$

Find one-parameter families of solution curves of the following differential equation:
y'+3 y = e^mx(m: a given real number)

Find one-parameter families of solution curves of the following differential equation:
y' - y = cos 2x

Find one-parameter families of solution curves of the following differential equation:
x y' - y = (x+1) e^-x

Find one-parameter families of solution curves of the following differential equation:
x y' + y = x⁴

Find one-parameter families of solution curves of the following differential equation:
x y' log x + y = log x

Find one-parameter families of solution curves of the following differential equation:
$dy over dx minus fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction comma space space straight y space equals space straight x squared plus 2$

Find one-parameter families of solution curves of the following differential equation:
$straight y apostrophe plus straight y space cosx space equals space straight e to the power of sin space straight x end exponent space cos space straight x$

Solve the following differential equation:
$dy over dx space plus 2 space straight y space tan space straight x space equals space sin space straight x$

For the differential equation, find the general solution:
$dy over dx plus 2 straight y space equals sin space straight x$

Solve the following differential equation:
$sin space straight x space dy over dx plus straight y space cosx space equals space cos space straight x space sin squared straight x$

Find a one parameter family of solutions of each of the following differential equation:
$straight y apostrophe cos squared straight x space equals space tanx minus space straight y$

Find a one parameter family of solutions of each of the following differential equation:
$dy over dx equals negative fraction numerator straight x plus straight y space cosx over denominator 1 plus sinx end fraction$

Solve:
$straight y apostrophe minus 2 straight y space equals space cos space 3 straight x$

Solve:
$straight y apostrophe plus straight y space equals space sin space straight x$

Solve:
$dy over dx plus 2 straight y space equals space sin space 3 straight x$

Solve:
$dy over dx plus 2 straight y space equals space sin space 5 straight x$