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Second-order ODE, reduction of order?

  1. Mar 21, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the specified particular solution:

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



    2. Relevant equations



    3. 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?
     
  2. jcsd
  3. Mar 21, 2012 #2

    tiny-tim

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    Hi Combinatus! :smile:

    erm :redface: … p(0) = 0 ! :wink:
     
  4. Mar 21, 2012 #3

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

    Oh well. Thanks. :smile:
     
    Last edited: Mar 21, 2012
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