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Differential Equations

Question

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

or

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

or

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

or

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

or

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

or

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

or

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

or

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

or

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

or

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.

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