Second-order ODE, reduction of order?

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?

Re: Second-order ODE, reduction of order?

Try letting $\displaystyle \displaystyle u = \frac{dy}{dx} \implies \frac{du}{dx} = \frac{d^2y}{dx^2}$. Then the DE becomes

$\displaystyle \displaystyle \begin{align*} \left(x^2 + 2\frac{dy}{dx}\right)\frac{d^2y}{dx^2} + 2x\,\frac{dy}{dx} &= 0 \\ \left(x^2 + 2u\right)\frac{du}{dx} + 2x\,u &= 0 \\ \frac{du}{dx} + \frac{2x\,u}{x^2 + 2u} &= 0 \\ \frac{du}{dx} &= -\frac{2x\,u}{x^2 + 2u} \\ \frac{dx}{du} &= -\frac{x^2 + 2u}{2x\,u} \\ \frac{dx}{du} &= -\frac{1}{2u}x - \frac{1}{x} \\ \frac{dx}{du} + \frac{1}{2u}x &= -x^{-1} \end{align*} $

This is now a Bernoulli DE which should easily be solvable :)

Re: Second-order ODE, reduction of order?

Nice approach! I'll give it a go in a little while. Also, I think I was just being retarded when I posted this (and possibly still am), as I realized that $\displaystyle p(0)=0$, and so $\displaystyle k=0$, and thus, $\displaystyle y' = \frac{x^2 \pm x^2}{2}$. Ugh.