Second-order ODE, reduction of order?

  • Thread starter Combinatus
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  • #1
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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?
 

Answers and Replies

  • #2
tiny-tim
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Hi Combinatus! :smile:

erm :redface: … p(0) = 0 ! :wink:
 
  • #3
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Hi Combinatus! :smile:

erm :redface: … p(0) = 0 ! :wink:


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:

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