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Trial solution, undetermined coefficients, 2nd order non-homogeneous equation

  1. Aug 5, 2015 #1
    1. The problem statement, all variables and given/known data

    Write a trial solution for the method of undetermined coefficients. Do not determine the coefficients.

    For: [itex]y'' + 2y' + 10y = x^2e^{-x}\cos{3x}[/itex]

    There's a modification performed and I'm not 100% confident as to why.

    2. Relevant equations


    3. The attempt at a solution

    The complementary equation gives a solution of

    [tex]y_c = e^{-x}\left(c_1\cos{3x} + c_2\sin{3x}\right)[/tex]

    Now, before considering a modification to the particular solution [itex]y_p[/itex], let's consider the standard trial solution by the method of undetermined coefficients. I would say it's

    [tex]y_p = e^{-x}\left[(Ax^2+Bx+C)\cos{3x} + (Dx^2+Ex+F)\sin{3x}\right][/tex]

    Now I know that the modification is performed so that "no term in the particular solution can be a solution for the complementary equation"...I do know basically what this means...and I would have thought that for the particular trial solution above, that neither term would suitably solve the complementary equation. Because the particular solution has a polynomial of degree 2 and the complementary solution has a polynomial of degree 0.

    So I've thought, maybe we could solve the complementary equation with the trial solution [itex]y_p[/itex] if we set A,B and/or D,E to = 0,0, so that the polynomials in the trial solutions were reduced from 2nd degree to 0 degree.

    Is this why we need to modify [itex]y_p[/itex] by multiply it by x? Because we can set A,B,D,E to zero in order to solve for the complementary equation? Thus, it won't really matter what degree a polynomial is in the solutions as far as deciding to modify is concerned?
     
  2. jcsd
  3. Aug 6, 2015 #2

    ehild

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    As e-xcos(3x) is solution of the homogeneous (complementary) equation, you have to modify the particular solution by multiplying it with x.
     
    Last edited: Aug 6, 2015
  4. Aug 6, 2015 #3
    Yes I understand that, but in my particular solution I've already multiplied it by a polynomial of 2nd degree...why does it need to be brought up to a polynomial of 3rd degree?
     
  5. Aug 6, 2015 #4

    ehild

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    You had to multiply with second order polynomials as there is the factor x2 in the inhomogeneous term. One more x to multiply with is because e-xcos(3x) is solution of the homogeneous equation. Try to find a solution with your method; you won't be able.
    partsol.jpg
     
    Last edited: Aug 6, 2015
  6. Aug 7, 2015 #5

    vela

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    Denote the derivative operator by D. The operator D+r acting on a term of the form ##x^p e^{-rx}## yields
    $$(D+r) (x^p e^{-rx}) = (p x^{p-1} e^{-rx} - r x^p e^{-rx}) + r x^p e^{-rx} = p x^{p-1} e^{-rx}.$$ The effect of the operator is basically to reduce the power of ##x## by 1 if ##p\ne 0##. If p=0, we have ##(D+r)e^{-rx} = 0##. In other words, ##e^{-rx}## is a solution of the homogeneous differential equation y' + ry = 0.

    Now consider the differential equation ##(D+r)y = x^n e^{-rx}##. It has a homogeneous solution ##y_h = e^{-rx}##. From the result above, you should be able to see that in order to get a term like ##x^n e^{-rx}##, ##y## must contain a term of the form ##x^{n+1} e^{-rx}##. It doesn't matter that ##x^n e^{-rx}##, …, and ##x e^{-rx}## are independent from ##y_h##. When D+r acts on those terms, you end up with ##x^{n-1}e^{-rx}##, …, and ##e^{-rx}##, and there's no linear combination of them that is equal to ##x^n e^{-rx}##.
     
    Last edited: Aug 7, 2015
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