Hi everyone,
I tried to solve this differential equation, but any method doesnt work me.
$\displaystyle y' = -\frac{x^2y^3 + y}{x^3y^2 - x}.$
I appreciate your help.
everk.
I see some symmetries in your DE of which we might want to take advantage. For example:
$\displaystyle y'=-\dfrac{y}{x}\,\dfrac{x^{2}y^{2}+1}{x^{2}y^{2}-1}.$
I'm thinking a substitution of the form $\displaystyle u=yx$ or $\displaystyle u=y/x$ might be good to try. If we have $\displaystyle u=yx$, then $\displaystyle y=u/x$ and
$\displaystyle y'=\dfrac{xu'-u}{x^{2}}$. It follows, then, that the DE becomes
$\displaystyle \dfrac{xu'-u}{x^{2}}=-\dfrac{u}{x^{2}}\,\dfrac{u^{2}+1}{u^{2}-1},$ or $\displaystyle xu'-u=-u\,\dfrac{u^{2}+1}{u^{2}-1}.$ Therefore,
$\displaystyle xu'=u-u\,\dfrac{u^{2}+1}{u^{2}-1}=\dfrac{u^{3}-u-u^{3}-u}{u^{2}-1}=\dfrac{-2u}{u^{2}-1}.$ The resulting DE $\displaystyle xu'=-\dfrac{2u}{u^{2}-1}$ is separable.
Conversely, if you try $\displaystyle u=y/x,$ then $\displaystyle y=ux$ and thus $\displaystyle y'=u'x+u.$ The DE becomes
$\displaystyle u'x+u=-u\,\dfrac{x^{4}u^{2}+1}{x^{4}u^{2}-1}.$ I don't see how this substitution will help you as much as the other one.