Find the specified particular solution:

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

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

Any clever tricks?