Question

Solve:

$\left[x\cos \frac{y}{x}+y\sin \frac{y}{x}\right]y=x\left[y\sin \frac{y}{x}-x\cos \frac{y}{x}\right]\times \frac{dy}{dx}$

• A. $xy\sin \frac{y}{x}=c$
• B. $xy\cos \frac{x}{y}=c$
• C. $xy\cos \frac{y}{x}=c$
• D. $xy\sin \frac{x}{y}=c$

Answer 1

The correct option is C $xy\cos \frac{y}{x}=c$

$\left[x\cos \frac{y}{x}+y\sin \frac{y}{x}\right]y=x\left[y\sin \frac{y}{x}-x\cos \frac{y}{x}\right]\times \frac{dy}{dx}$
Given differential eqn can be written as
$\frac{dy}{dx}=\frac{y[x\cos \frac{y}{x}+y\sin \frac{y}{x}]}{x[y\sin \frac{y}{x}-xcos\frac{y}{x}]}$ ....(1)
Substitute $y=vx$
$\frac{dy}{dx}=v+x\frac{dv}{dx}$
So eqn (1) becomes
$v+x\frac{dv}{dx}=\frac{v[\cos v+v\sin v]}{[v\sin v-cosv]}$
$x\frac{dv}{dx}=\frac{2v\cos v}{v\sin v-cosv}$
$\frac{v\sin v-cosv}{v\cos v}dv=\frac{2}{x}dx$
Integrating both sides,
$\int \tan vdv-\int \frac{1}{v}dv=2\int \frac{1}{x}dx$
$-\log \cos v-\log v=2\log x+\log C$
$\Rightarrow \frac{1}{v\cos v}=C{x}^2$
$\Rightarrow xy\cos \left(\frac{y}{x}\right)=\frac{1}{C}$
$\Rightarrow xy\cos \left(\frac{y}{x}\right)=c$