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Math Help - Second-order ODE, reduction of order?

  1. #1
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    Second-order ODE, reduction of order?

    Find the specified particular solution:

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

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    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 + 2xpdx=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?
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  2. #2
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    Re: Second-order ODE, reduction of order?

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

    \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
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  3. #3
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    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 p(0)=0, and so k=0, and thus, y' = \frac{x^2 \pm x^2}{2}. Ugh.
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