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.

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- Sep 22nd 2010, 08:31 PMeverkhelp solving differential equation
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. - Sep 23rd 2010, 02:54 AMAckbeet
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. - Sep 23rd 2010, 06:52 AMeverk
Thank you so much for your help. Long time ago I dont resolve any DE.

- Sep 23rd 2010, 06:56 AMAckbeet
You're very welcome. Have a good one!