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## Homework Statement

Find the specified particular solution:

[itex](x^2+2y')y'' + 2xy' = 0, y(0)=1, y'(0)=0[/itex]

## Homework Equations

## The Attempt at a Solution

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

Any clever tricks?