Differential Equations

Sponsor Area

Question
CBSEENMA12033080

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.

Sponsor Area

Question
CBSEENMA12033081

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
CBSEENMA12033082

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

Solution

The given differential equation is
            y' = ex 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
CBSEENMA12033083

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