# Solve $2xy+y^2-2x^2\frac{dy}{dx}=0$; $y=2$ when $x=1$

$2xy+y^2-2x^2\dfrac{dy}{dx}=0$; $y=2$ when $x=1$.

My reference gives the solution $y=\dfrac{2x}{1-\log x}$, but is it really the solution ?