Second-order ODE, reduction of order?

[Combinatus]

Homework Statement

Find the specified particular solution:

$(x^2+2y')y'' + 2xy' = 0, y(0)=1, y'(0)=0$

Homework Equations

The Attempt at a Solution

The equation seems amenable to the substitution $p=y'$, so it can be transformed into $(x^2 + 2p)p' + 2xp=0$, or $(x^2 + 2p)dp + 2xpdp=0$. Since $\frac{\partial (x^2+2p)}{\partial x} = 2x = \frac{\partial (2xp)}{\partial p}$, the latter equation is exact, and is solvable through the method of exact equations. Unfortunately, this yields the rather cumbersome expression $p=\frac{-x^2 \pm \sqrt{k+x^4}}{2}$ for some constant $k$, which I would have a hard time integrating, I think.

Any clever tricks?

 

[tiny-tim]

Hi Combinatus!

erm … p(0) = 0 ! ​

 

[Combinatus]

Hi Combinatus!

erm … p(0) = 0 ! ​

Yes, I just noticed that after looking at my scribbles, and hoped no-one would have had time to reply yet.

Oh well. Thanks.