I need to solve the ODE $$(2x-y^2)y'=2y$$
This ODE is under the title of ["Variation of parameters"](https://en.wikipedia.org/wiki/Variation_of_parameters#Examples) method, but I can't bring it to the form $y'+p(x)y=q(x)$. I suspect that a change of variables should be applied. The $y^2$ hints that I should try $u=y^2$. That gives:
$$\\(2x-y^2)yy'=2y^2
\\(2x-u)\frac{u}{2}=u
\\ 2x-uu'=2u$$
but I can't see how it helps me.

Suggestions will be welcomed. Thank you!