Solve differential equation $(xy'+y)^2=x^2y'$

Question

Solve the differential equation

$$(xy'+y)^2=x^2y'$$

Please any help. I tried to express one of $x,y,y'$ using other two variables and then solve this. But I get something complicated.

Answer 1

$$(xy'+y)^2=x^2y'$$ $$((xy)')^2=x(xy'+y)-xy$$ Substitute $xy=z$ $$(z')^2=xz'-z $$ Solve this equation $$(z')^2-xz'=-z $$ $$(z')^2-xz'+\frac {x^2}{4}=-z +\frac {x^2}{4}$$ $$(z'-\frac {x}{2})^2=-z +\frac {x^2}{4}$$ $$(z'-\frac {x}{2})^2=-(z -\frac {x^2}{4})$$ Substitute $u=(-z +\frac {x^2}{4})$ $$(u')^2=u$$ then integrate this equation $u'= \pm u^{1/2}$

For $u'=u^{1/2}$ you get $$\boxed{y=K-\frac {K^2}x}$$