# 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 $6x^2y$

• Why the parentheses around the whole equation? – almagest May 2 '16 at 14:41
• By mistake sorry... – Aditya May 2 '16 at 14:42
• @almagest By mistake edited now.... – Aditya May 2 '16 at 14:51
• You should not be able to get a linear form, those $y^2$ terms will be a problem. Nor is the equation separable. Have you checked whether it is exact? – Ian May 2 '16 at 14:56
• @Ian No i havent checked that.... – Aditya May 2 '16 at 14:57

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.
$$y(x) = - \sqrt {\frac{ 2 c x -1 } { 2 c x + x^2 -1 } }$$
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 $z$.