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Differential Equation Question

  1. May 19, 2014 #1
    I have a differential equation

    y'' + y' -2y = 3e-2x + 5cosx

    y = yc + yp

    I found

    yc = Ae-2x + Bex

    for A, B arb. Const.

    Then when selecting a trial function to find the particular integral, yp I came up with:

    yp = ae-2x + bcosx + csinx

    However the correct trial function to choose is:

    yp = axe-2x + bcosx + csinx

    Where the ae-2x has been multiplied by x. I was wondering in what instance I would have to multiply the term in the yp equation by x or x2.
     
  2. jcsd
  3. May 19, 2014 #2

    LCKurtz

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    When you plug in your expression for ##y_p##, you know you need to get out a ##3e^{-2x}## term. But if you plug ##ae^{-2x}## into the equation, you know you will get zero because it is in ##y_c##. That's when you want to try a term ##y_p=axe^{-2x}##. There's more to that story but that's the short answer.
     
  4. May 19, 2014 #3

    Zondrina

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    You want the terms in ##y_p## to be linearly independent of the terms in ##y_c##. Note that ##f(x) = 3e^{-2x} +5cos(x)##.

    Your original guess for ##y_p## includes a multiple of one of your homogeneous solutions, namely the ##Ae^{-2x}## term, so it cannot be linearly independent of ##y_c##. Multiplying by ##x## is a quick shortcut to obtaining a linearly independent ##y_p## guess.

    This method was developed by D'alembert, you can find out more by checking out "reduction order".
     
  5. May 20, 2014 #4


    These were both great answers! Thanks a lot!
     
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