Solve $2x^{3}ydy+(1-y^{2})((xy)^{2}+y^{2}-1)dx=0$ The question is to solve this
$$2x^{3}ydy+(1-y^{2})(x^2y^{2}+y^{2}-1)dx=0$$
What I tried was to bring this equation in linear differential equation form but failed. I have also tried out rewriting the given in such a way so that each term can be integrated easily but to no avail it isnt exact either since $\frac{dm}{dy}$ is $4y-4y^{3}$ and  $\frac{dn}{dx}$ is $6x^2y$
Please help me to solve this.
Thanks in advance.
 A: I can't seem to get the Mathematica solution. Just chipping away at it,
$$2y\frac{dy}{dx}=\frac{(y^2-1)}{x^3}(x^2y^2+y^2-1)$$
Let $s=y^2$. Then
$$\frac{ds}{dx}=\frac{(s-1)}{x^3}(x^2s+s-1)$$
Let $s-1=t$. Then
$$\frac{dt}{dx}=\frac t{x^3}(x^2(t+1)+t)$$
Let $t=\frac1u$. Then
$$-\frac1{u^2}\frac{du}{dx}=\frac1{ux^3}\left(x^2\left(\frac1u+1\right)+\frac1u\right)$$
$$\frac{du}{dx}+\frac1xu=-\frac1x-\frac1{x^3}$$
The integrating factor is
$$\mu=e^{\int\frac1xdx}=e^{\ln x}=x$$
So
$$\frac d{dx}\left(xu\right)=x\frac{du}{dx}+u=x\left(-\frac1x-\frac1{x^3}\right)=-1-\frac1{x^2}$$
$$xu=-x+\frac1x+C=\frac xt$$
$$t=\frac x{-x+\frac1x+C}=\frac{x^2}{-x^2+1+Cx}=s-1$$
$$s=\frac{1+Cx}{-x^2+1+Cx}=y^2$$
$$y=\pm\sqrt{\frac{1+Cx}{-x^2+1+Cx}}$$
OK, I take it back. That is the Mathematica solution. I found my mistake while typing this up.
A: Mathematica gives the solution as
$$ y(x) = - \sqrt {\frac{ 2 c x -1 } { 2 c x + x^2 -1 } } $$
If you use the sub $z = y^2$,you'll find the equation transform  to
$$ x^3 d z + ( 1-z) ( x^2 z + z -1) dx =0 $$
Then it looks like you should be able to find an integrating factor in $z$.
