# difficult(for me) DE

• Apr 3rd 2010, 01:38 AM
noteiler
difficult(for me) DE
can someone here help me out with an idea how to solve this DE :
$\displaystyle 2xyy' - x^2y'^2 - y^2 = 2S$
S is real constant. i know that this is DE that cannot be solved according to derivative, or it is but if i solved it, i mean if i write it down this way :
$\displaystyle y'=F(x,y)$
i think it would be impossible to solve it, but this DE can't be written down in form $\displaystyle y=F(x,y') or x=F(y,y')$ easily either. of course i can try to solve it some of thease ways, but maybe there is some trick hiding here?
• Apr 3rd 2010, 04:29 AM
shawsend
How about solving for $\displaystyle y'$. Don't worry about the $\displaystyle \pm$ part, just use $\displaystyle +$ for now. Tell you what, I went through it quick so you can check for errors if you want to. Doing that I get:

$\displaystyle \left(y+\sqrt{y^2+k}\right)dx-xdy=0$

and I think an integrating factor in terms of y can be found to make it exact.
• Apr 3rd 2010, 05:35 AM
Jester
Quote:

Originally Posted by noteiler
can someone here help me out with an idea how to solve this DE :
$\displaystyle 2xyy' - x^2y'^2 - y^2 = 2S$
S is real constant. i know that this is DE that cannot be solved according to derivative, or it is but if i solved it, i mean if i write it down this way :
$\displaystyle y'=F(x,y)$
i think it would be impossible to solve it, but this DE can't be written down in form $\displaystyle y=F(x,y') or x=F(y,y')$ easily either. of course i can try to solve it some of thease ways, but maybe there is some trick hiding here?

Note that your ODE can be written as

$\displaystyle \left(xy'-y\right)^2 = -2 S$

or

$\displaystyle x y' - y = \pm \sqrt{-2S}.$

It's now linear!
• Apr 3rd 2010, 05:49 AM
noteiler
thanks alot, i knew that i didn't see something