Question

Solve the differential equation: $xdy-ydx=\sqrt{x^2+y^2}dx$

Answer 1

Given equation can be written as
$x^2\frac{dy}{dx}-xy=2{\cos }^2\left(\frac{y}{2x}\right),x\ne 0$
$\Rightarrow \frac{x^2\frac{dy}{dx}-xy}{2{\cos }^2\left(\frac{y}{2x}\right)}=1\Rightarrow \frac{se{c}^2\left(\frac{y}{2x}\right)}{2}[x^2\frac{dy}{dx}-xy]=1$
Dividing both sides by $x^3$, we get
$\frac{se{c}^2\left(\frac{y}{2x}\right)}{2}\left[\frac{x\frac{dy}{dx}-y}{x^2}\right]=\frac{1}{x^3}\Rightarrow \frac{d}{dx}\left[\tan \left(\frac{y}{2x}\right)\right]=\frac{1}{x^3}$
Integrating both sides, we get
$\tan \left(\frac{y}{2x}\right)=\frac{-1}{2x^2}+k$
Substituting $x=1,y=\frac{\pi }{2}$, we get
$k=\frac{3}{2},$ therefore, $\tan \left(\frac{y}{2x}\right)=\frac{1}{2x^2}+\frac{3}{2}\tan \left(\frac{y}{2x}\right)=\frac{-1}{2x^2}+\frac{3}{2}$ is the required solution.