help solving differential equation

**everk** · Sep 22nd 2010, 08:31 PM

Hi everyone,
I tried to solve this differential equation, but any method doesnt work me.
$y' = -\frac{x^2y^3 + y}{x^3y^2 - x}.$
I appreciate your help.
everk.

**Ackbeet** · Sep 23rd 2010, 02:54 AM

I see some symmetries in your DE of which we might want to take advantage. For example:

$y'=-\dfrac{y}{x}\,\dfrac{x^{2}y^{2}+1}{x^{2}y^{2}-1}.$

I'm thinking a substitution of the form $u=yx$ or $u=y/x$ might be good to try. If we have $u=yx$ , then $y=u/x$ and

$y'=\dfrac{xu'-u}{x^{2}}$ . It follows, then, that the DE becomes

$\dfrac{xu'-u}{x^{2}}=-\dfrac{u}{x^{2}}\,\dfrac{u^{2}+1}{u^{2}-1},$ or $xu'-u=-u\,\dfrac{u^{2}+1}{u^{2}-1}.$ Therefore,

$xu'=u-u\,\dfrac{u^{2}+1}{u^{2}-1}=\dfrac{u^{3}-u-u^{3}-u}{u^{2}-1}=\dfrac{-2u}{u^{2}-1}.$ The resulting DE $xu'=-\dfrac{2u}{u^{2}-1}$ is separable.

Conversely, if you try $u=y/x,$ then $y=ux$ and thus $y'=u'x+u.$ The DE becomes

$u'x+u=-u\,\dfrac{x^{4}u^{2}+1}{x^{4}u^{2}-1}.$ I don't see how this substitution will help you as much as the other one.

**everk** · Sep 23rd 2010, 06:52 AM

Thank you so much for your help. Long time ago I dont resolve any DE.

**Ackbeet** · Sep 23rd 2010, 06:56 AM

You're very welcome. Have a good one!

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