Question

Solve $\frac{x+y\frac{dy}{dx}}{y-x\frac{dy}{dx}}=x^2+2y^2+\frac{y^4}{x^2}$

Answer 1

Given $\frac{x+y\frac{dy}{dx}}{y-x\frac{dy}{dx}}=\frac{x^4+2x^2{y}^2+y^4}{x^2}$

$\frac{2xdx+2ydy}{2(ydx-xdy)}=\frac{x^4+2x^2{y}^2+y^4}{x^2}$

$\frac{d(x^2+y^2)}{2(x^2+y^2{)}^2}=-\frac{xdy-ydx}{x^2}$

$\frac{d(x^2+y^2)}{2(x^2+y^2{)}^2}=-d\left(\frac{y}{x}\right)$

Integrating on both sides

$\frac{1}{2}\int \frac{d(x^2+y^2)}{(x^2+y^2{)}^2}=-\int d\left(\frac{y}{x}\right)$

$-\frac{1}{2(x^2+y^2)}=-\frac{y}{x}+c$

The original equation is,

$\frac{1}{2(x^2+y^2)}-\frac{y}{x}+c=0$