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2nd order non-homgenous DE

  1. Oct 31, 2012 #1

    ElijahRockers

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    Gold Member

    1. The problem statement, all variables and given/known data

    4y''+4y'+y = cos(2t), y(0)=0, y'(0)=0

    2. Relevant equations

    y(t)=yh+yp

    3. The attempt at a solution

    characteristic roots are repeated, m=-1/2, so

    [itex] y_h = A_1 e^{\frac{-t}{2}}+A_2 te^{\frac{-t}{2}}[/itex]

    undetermined coefficients:

    yp = Acos(2t)+Bsin(2t)

    plugging into original equation and solving, I got A= -15/289 and B= 8/289

    using initial conditions, I got A_1 = 15/289 and A_2 = -47/578

    so

    [itex] y(t) = \frac{15}{289}e^{\frac{-t}{2}} - \frac{47}{578}te^{\frac{-t}{2}} - \frac{15}{289}sin(2t) + \frac{8}{289}cos(2t) [/itex]

    should be the answer, but apparently it's wrong. I have checked over my work and can't find the fault.
     
    Last edited: Oct 31, 2012
  2. jcsd
  3. Oct 31, 2012 #2

    Zondrina

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    Homework Helper

    Your process is correct. You solved L[y] = 0 and your assumption for the solution of L[y] = cos(2t) is also good. It's more than likely an arithmetic error.
     
  4. Oct 31, 2012 #3

    ElijahRockers

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    Gold Member

    Checked it a second time. I must've typed something into the calculator wrong at the very end of the question, just my luck.
    [itex]-\frac{47}{578}[/itex] should be [itex]-\frac{1}{34}[/itex] instead.
     
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