Question

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

Answer 1

Given, $\frac{xdx+ydy}{ydx-xdy}=x^2+2y^2+\frac{y^2}{x^2}$

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

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

$\Rightarrow \frac{1}{2}\frac{d(x^2+y^2)}{x^{2}d\left(\frac{y}{x}\right)}=x^2+y^2+y^2+\frac{y^4}{x^2}=x^2+y^2+y^2(1+\frac{y^2}{x^2})$

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

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

Integrating both sides

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

$\Rightarrow \frac{2y}{x}+\frac{1}{x^2+y^2}+C=0$