Question

$y(x\cos \left(\frac{y}{x}\right)+y\sin \left(\frac{y}{x}\right))dx=x(y\sin \left(\frac{y}{x}\right)-x\cos \left(\frac{y}{x}\right))dy=0$.

Answer 1

Given,

$y(x\cos \left(\frac{y}{x}\right)+y\sin \left(\frac{y}{x}\right))dx=x(y\sin \left(\frac{y}{x}\right)-x\cos \left(\frac{y}{x}\right))dy=0$

$\frac{dy}{dx}=\frac{y}{x}\frac{(x\cos \left(\frac{y}{x}\right)+y\sin \left(\frac{y}{x}\right))}{x(y\sin \left(\frac{y}{x}\right)-x\cos \left(\frac{y}{x}\right))}$

substitute $y=vx\to \frac{dy}{dx}=x\frac{dv}{dx}+v$

$x\frac{dv}{dx}+v=v\left(\frac{x\cos v+vx\sin v}{vx\sin v-x\cos v}\right)$

$x\frac{dv}{dx}+v=\left(\frac{v\cos v+v^2\sin v}{v\sin v-\cos v}\right)$

$x\frac{dv}{dx}=\left(\frac{v\cos v+v^2\sin v}{v\sin v-\cos v}\right)-v$

$x\frac{dv}{dx}=\frac{v\cos v+v^2\sin v-v^2\sin v-v\cos v}{v\sin v-\cos v}$

$x\frac{dv}{dx}=0$

$\frac{dv}{dx}=0$

$\frac{1}{x}\frac{dy}{dx}-\frac{y}{x}=0$