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Trial function for ODE (Exam tomorrow)

  1. Nov 8, 2005 #1
    [itex]y''-3y'+2y=\mathrm{e}^x +1[/itex]

    Find [itex]y[/itex].

    Going through the normal procedure we get the complementary function

    [itex]y = A\mathrm{e}^x + B\mathrm{e}^{2x}[/itex].

    Using the trial function

    [itex]y = Cx\mathrm{e}^x + D[/itex]
    (because of overlap with the complementary function)

    leads to
    [itex]-C\mathrm{e}^x + 2D = \mathrm{e}^x+1[/itex]

    Does this mean

    [itex]C=-1,D=1/2[/itex]??

    If so, the general soln is [itex]y = A\mathrm{e}^x + B\mathrm{e}^{2x} -x\mathrm{e}^x + 1/2[/itex].

    According to Mathematica, however, the solution should be of the form

    [itex]y = A\mathrm{e}^x + B\mathrm{e}^{2x} + 1/2(1-2\mathrm{e}^x -2\mathrm{e}^xx)[/itex].

    Am I using the wrong trial function or what?

    Thanks in advance.

    James
     
  2. jcsd
  3. Nov 8, 2005 #2
    Nevermind, I figured out that Mathematica is using a different set of constants.
     
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