Question

Solve the differential equation:-
$(xdy-ydx)y\sin \left(\frac{y}{x}\right)=(ydx+xdy)x\cos \left(\frac{y}{x}\right)$.

Answer 1

The given differential equation can be written as

$\frac{dy}{dx}=\frac{y}{x}\frac{\left[xcos\right(\frac{y}{x})+ysin(\frac{y}{x}\left)\right]}{\left[ysin\right(\frac{y}{x})-xcos(\frac{y}{x}\left)\right]}$

This being a homogeneous differential equation, we can solve the equation by putting

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

Now putting this, then the equation changes to

$v+x\frac{dv}{dx}=v\frac{\left[cos\right(v)+vsin(v\left)\right]}{\left[vsin\right(v)-cos(v\left)\right]}$

$x\frac{dv}{dx}=v\frac{\left[cos\right(v)+vsin(v\left)\right]}{\left[vsin\right(v)-cos(v\left)\right]}-v$

$x\frac{dv}{dx}=\frac{v\left[cos\right(v)+vsin(v)-vsin(v)+cos(v\left)\right]}{vsin\left(v\right)-cos\left(v\right)}$

$x\frac{dv}{dx}=\frac{2vcos\left(v\right)}{vsin\left(v\right)-cos\left(v\right)}$

$\Rightarrow \frac{1}{2}\int \left[tan\right(v)-\frac{1}{v}]dv-\int \frac{dx}{x}=C$

$\Rightarrow \frac{1}{2}\left[ln|sec\right(v)|-ln|v|]-ln|x|=C$

$\Rightarrow -ln|vcos\left(v\right)|-2ln|x|=C$

$\Rightarrow ln|x^2\frac{y}{x}cos\left(\frac{y}{x}\right)|=-C=ln\left(C{}_1\right)$

$\Rightarrow xycos\left(\frac{y}{x}\right)=C{}_1$

Here both $C$ and $C{}_1$ are constants.