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Question about differential equations?(nonhomogeneous)

  1. Feb 28, 2009 #1
    For this problem I want to find the particular solution;
    but i am not sure if it is right setup.


    y''+9y'=162sin(9t)+1296cos(9t)

    the roots for complementary function are 0,-9
    so it has distinct roots.

    my setup is;
    [Asin(9t)+Bcos(9t)]+[Asin(9t)+Bcos(9t)]
    Axsin(9t)+Bxcos(9t)

    am I going about it the right way?

    Thank You
     
  2. jcsd
  3. Feb 28, 2009 #2

    Mark44

    Staff: Mentor

    No. Here's your particular solution: yp = Asin(9t)+Bcos(9t)
    This would be your particular solution for a right hand side of any of the following:
    c1sin(9t)
    c2cos(9t)
    c1sin(9t) + c2cos(9t)
     
  4. Feb 28, 2009 #3

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    That's true, but it's of no use in finding the particular solution.

    To be honest, I can't figure out what you're thinking. But no, what you have written is not right. All you need is [itex]y_P=A\sin(9t)+B\cos(9t)[/itex]. Plug it into your differential equation and solve for [itex]A[/itex] and [itex]B[/itex].
     
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