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Dif eq with particular solution

  1. Feb 18, 2006 #1
    find the particular solution for y''+25y=50sin(5t)

    I am using variation of parameter:

    [tex]y_p=U_1e^{5x}+U_2e^{-5x}[/tex]
    [tex]y_1=e^{5x},y_1'=5e^{5x},y_2=e^{-5x},y_2'=-5e^{-5x}[/tex]
    [tex]U_1=- \int \frac{e^{-5x}50sin(5t)}{-10}=-0.5e^{-5x} (cos5x+sin5x)[/tex]
    [tex]U_2= \int \frac{e^{5x}50sin(5t)}{-10}=0.5e^{5x} (cos5x-sin5x)[/tex]

    [tex]y_p=-0.5e^{-5x} (cos5x+sin5x) e^{5x}+ 0.5e^{5x} (cos5x-sin5x)e^{-5x}[/tex]
    [tex]y_p=-sin5x[/tex]

    I know this is wrong because the check doesnt equal 50sin(5t)

    any ideas?
     
    Last edited: Feb 18, 2006
  2. jcsd
  3. Feb 18, 2006 #2

    arildno

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    Why do you use the e^5x-stuff??
    They aren't solutions to the homogenous problem.
     
  4. Feb 18, 2006 #3

    arildno

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    No. r^2=-25. What is r therefore?
     
  5. Feb 18, 2006 #4
    yep, i finished the problem. had to use Eulers formula, never would have guessed! thanks!
     
  6. Feb 18, 2006 #5

    HallsofIvy

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    Surely your textbook discusses the case when the characteristic roots are complex?
     
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