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2nd order DE, is there a way to solve this without series?

  1. Jun 17, 2004 #1
    It looks simple enough:

    y'' + x*y = x^2

    However, I tried and I could not find a nontrivial solution to the homogeneous equation:

    y'' + x*y = 0

    Am I right in thinking you need to solve this with series?

    No need to actually do it, I just need to know if it is possible otherwise (like variation of parameters or something else).
     
  2. jcsd
  3. Jun 18, 2004 #2
    can you use integrating factor?
     
  4. Jun 18, 2004 #3
    For a 2nd order equation? I know how to do that for first order but not second order equations.
     
  5. Jun 18, 2004 #4

    Dr Transport

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    The homogeneous equation looks like the Airy equation.....
     
  6. Jun 18, 2004 #5
    Ah, so only power series is it then?

    That's fine. It's just for some reason I thought I wasn't understanding some kind of trick to give a general solution.

    It's almost like I saw an integral of
    [tex]
    \int e^x^2 dx
    [/tex]

    Which of course can only be done with power series, and thought I could integrate it and give a nice general solution.
     
  7. Jun 19, 2004 #6

    arildno

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    You could say that you get a "nice" general solution by dubbing it as Ai(x)..:biggrin:
     
  8. Jun 19, 2004 #7
    If y1 is known, you can use reduction of order to solve this.
     
  9. Jun 19, 2004 #8

    Dr Transport

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    if memory serves me correctly, the Airy equation is proportional to a Bessel function of 1/3 order......Look out there online.
     
  10. Jun 22, 2004 #9
  11. Jun 24, 2004 #10

    selfAdjoint

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    Has anyone tried this one in MAPLE? There might be a Bessel function integrating factor of the homogeneous equation (just interested, that's all, Max's link gives the answer).
     
  12. Jun 24, 2004 #11
    Mathematica gives:

    Edit: Something too long or not properly formatted for PF to handle... But it was pretty much the Airy function.

    cookiemonster
     
    Last edited: Jun 24, 2004
  13. Jun 30, 2004 #12
    The homogenous equation:
    [tex]\frac{d^2y}{dx^2}+xy=0[/tex] is a negative sign off the Airy equation:
    [tex]\frac{d^2y}{dx^2}-xy=0[/tex]

    Therefore the solution of the original DE
    [tex]\frac{d^2y}{dx^2}+xy=x^2[/tex] is given by
    [tex] y = CAiryAi(-x) + DAiryBi(-x)+x[/tex]

    where AiryAi and AiryBi, are independant solutions of the Airy equation.

    Indeed the Airy functions are related to the Bessel functions.

    Finally one can expland the answer as a series with the Gamma function appearing everywhere - nasty.
     
    Last edited: Jun 30, 2004
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