General Solution of a differential solution

**Ragingamer** · September 10th 2010, 01:42 PM

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

**mr fantastic** · September 10th 2010, 01:45 PM

Originally Posted by Ragingamer

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

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

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

Now apply the usual technique for such DE's.

**Ragingamer** · September 10th 2010, 02:00 PM

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.

**mr fantastic** · September 10th 2010, 02:15 PM

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/(Sqrt[x^2 + 1] - (x^2 + 1)) - Wolfram|Alpha (click on show steps).