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Linear second order non-homogeneous ODE question

  1. Jul 31, 2011 #1
    Determine the general solution to the ODE:

    y'' + 2y' = 1 + xe-2x

    I know the solution will be of the form y = yh + yp. The homogeneous solution is y = c1 + c2e-2x.

    For the particular solution, I have been using the method of undetermined coefficients. c3e-2x won't work as it is not linearly independent of the homogeneous solution. So I guess c3xe-2x.

    So y' = c3e-2x(c3 - 2c3x)

    And y'' = 4c3e-2x(x - 1)

    Following on from this I end up with a particular solution:

    yp = ((1 + xe-2x)/(-2e-2x))xe-2x

    However wolfram disagrees with me. My question is, is the method of undetermined coefficients ok to use for this ODE? Or should I be using the method with the Wronskian (it's name escapes me at the moment)

    Cheers
     
  2. jcsd
  3. Jul 31, 2011 #2

    Ray Vickson

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    If you put y(x) = C1(x) + C2(x)*exp(-2x) in the DE, you eventually arrive at a first-order DE in v = dC1/dx + exp(-2x)*dC2/dx.

    RGV
     
  4. Jul 31, 2011 #3
    I'm not sure I understand this. Are you able to explain it a little more?

    Cheers
     
  5. Aug 1, 2011 #4
    I'd have to dig out my book or notes to make sure, but I think your particular solution will also need an e-2x term, so it should be of the form Ae-2x + Bxe-2x. I wouldn't use c's for yh since they're already being used as the constants of integration for yh.
     
  6. Aug 1, 2011 #5

    ehild

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    For the method of variation of the parameters, read http://en.wikipedia.org/wiki/Variation_of_parameters.

    This problem can be solved for y' as the y term is missing. Denote z=y' and solve the first-order equation z'+2z=1+xe-2x, then integrate z=y' to get y.

    ehild
     
  7. Aug 1, 2011 #6
    Aha! Of course. It's all so simple once you know what to do... :rofl:

    Thanks very much!
     
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