Solving a DE Using Substitution: y=z^α

[fluidistic]

Homework Statement

I must calculate the general solution to the following (reducible to homogeneous) DE:
$(x^2y^2-1)y'+2xy^3=0$.
Hint: Use the substitution $y=z^\alpha$.

Homework Equations

The one given in the hint.

The Attempt at a Solution

So I've used the hint. This gave me $(x^2z^{2 \alpha }-1)\alpha z^{\alpha -1}+2xz^{3\alpha }=0$.
I factorized by $z^{\alpha -1}$ and I reach that $z^{\alpha -1}=0$ (thus z=0, thus y=0) or $z^{2\alpha } (\alpha x^2 +2xz)=\alpha$.
I ran out of ideas on the second condition. ( I don't think back-substituting $z^\alpha =y$ will help).
Any help is appreciated. Thanks!

 

[lippyka]

Aint:$$z^{2\alpha}(\alpha x^2+2xz)=\alpha\;\;\;\;\;\implies\;\;\;\;\;z=\frac{\alpha}{2x}\pm \sqrt{\frac{\alpha^2}{4x^2}-1}=\frac{\alpha}{2x}\pm\frac{\sqrt{\alpha^2-4x^2}}{2x}$$Now, substitute $y=z^\alpha$ and you'll get two equations from which you can find the constants.