Question

Solve:
$x\frac{dy}{dx}=y-x\tan \frac{y}{x}$

Answer 1

Given $x\frac{dy}{dx}=y-x\tan \frac{y}{x}\Rightarrow \tan \frac{y}{x}=\frac{ydx-xdy}{dx}---\left(1\right)$

Now, $\frac{d}{dx}\left(\frac{y}{x}\right)=\frac{1}{x^2}(ydx-xdy)----\left(2\right)$

Suubstituting $\left(2\right)$ in $\left(1\right),x\tan \frac{y}{x}=\frac{x^{2}d}{dx}\left(\frac{y}{x}\right)$

$\Rightarrow \frac{dx}{x}=\frac{d/y/x}{\tan \left(y/x\right)}$

Integrating both side

$\int \frac{dx}{x}=\int \frac{d\left(y/x\right)}{\tan \left(y/x\right)}$

$\Rightarrow \log x=\log \sin \left(\frac{y}{x}\right)+\log c$

$y={\sin }^{-1}\left(\frac{x}{y}\right)$