Question

$x\frac{dy}{dx}\frac{y^2}{x}+y=0$

Answer 1

Given,

$x\frac{dy}{dx}\frac{y^2}{x}+y=0$

$\frac{dy}{dx}{y}^2=-y$

$\frac{dy}{dx}=-\frac{1}{y}$

integrating on both sides, we get,

$y=-\frac{x}{y}$

$y^2+x=0$