y*y'+x=(x^2+y^2)^.5

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- Sep 10th 2010, 01:42 PMRagingamerGeneral Solution of a differential solution
y*y'+x=(x^2+y^2)^.5

- Sep 10th 2010, 01:45 PMmr fantastic
- Sep 10th 2010, 02:00 PMRagingamer
I did that and got to y'=(x/y)((1+(y/x)^2)^.5-1) and then used a substitution of v=y/x.

Reducing it to y'=(1/v)((1+v^2)^.5-1) and differentiating y=vx to get y'=v+x*dv/dx.

Setting up an integral of v/((1+v^2)^.5-1-v^2)*dv=dx/x But I have attempted many times with this integral and have not gotten the correct solution. - Sep 10th 2010, 02:15 PMmr fantastic
If people posted what they could do and said where they were stuck in the very first post, they would get the help they really needed a lot faster and other people wouldn't waste time doing things that had already been done.

See here: integrate x/(Sqrt[x^2 + 1] - (x^2 + 1)) - Wolfram|Alpha (click on show steps). - Sep 11th 2010, 02:49 AMzzzoak
$\displaystyle u^2=x^2+y^2 $

$\displaystyle uu'=x+yy'$

The initial equation becomes

$\displaystyle u'=1$

$\displaystyle u=x+C$

$\displaystyle y^2=2xC+C^2$