# General Solution of a differential solution

• Sep 10th 2010, 01:42 PM
Ragingamer
General Solution of a differential solution
y*y'+x=(x^2+y^2)^.5
• Sep 10th 2010, 01:45 PM
mr fantastic
Quote:

Originally Posted by Ragingamer
y*y'+x=(x^2+y^2)^.5

Your first thought should have been to divide both sides by x:

$\displaystyle \displaystyle \frac{y}{x} \frac{dy}{dx} + 1 = \sqrt{1 + \left( \frac{y}{x} \right)^2}$.

Now apply the usual technique for such DE's.
• Sep 10th 2010, 02:00 PM
Ragingamer
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 PM
mr fantastic
Quote:

Originally Posted by Ragingamer
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.

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&#47;&#40;Sqrt&#91;x&#94;2 &#43; 1&#93; - &#40;x&#94;2 &#43; 1&#41;&#41; - Wolfram|Alpha (click on show steps).
• Sep 11th 2010, 02:49 AM
zzzoak
$\displaystyle u^2=x^2+y^2$
$\displaystyle uu'=x+yy'$
The initial equation becomes
$\displaystyle u'=1$
$\displaystyle u=x+C$
$\displaystyle y^2=2xC+C^2$