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Homework Help: Diff EQ

  1. Sep 17, 2006 #1
    I took an ODE course last year, but I seem to have forgotten some stuff. I need to solve this equation:

    [tex]\frac{d^2u}{dt^2} + {\omega}^2u = f_osin({\mu}t)[/tex]

    with the boundry conditions:

    u(0) = 0, du/dt(0) = 0

    When I tried to solve the homogenenous equation first, I got

    [tex]u_g(t)=c_1e^{i{\omega}t}+c_2e^{-i{\omega}t}[/tex]

    I then differentiated and set up the system with the two boundy conditions...but I got c1+c2=0 and c1-c2=0...c1=c2=0.

    This seems wrong. Any help would be appreciated
     
  2. jcsd
  3. Sep 17, 2006 #2

    AKG

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    I think those might be called initial conditions, not boundary conditions, not that it matters. Anyways, you need to find a particular solution, add it to your ug, then find the ci by looking at the initial conditions.
     
  4. Sep 17, 2006 #3
    Here is what I did now:

    let the particular solution be of the form [tex]u_p=Asin({\mu}t})[/tex]

    differentiated twices gives [tex]u_p''=-A{\mu}^2sin({\mu}t})[/tex]

    so now from my origional diff eq I get

    [tex]A({\omega}^2-{\mu}^2)sin({\mu}t)=f_osin({\mu}t)[/tex]

    hence


    [tex]A = \frac{f_o}{{\omega}^2-{\mu}^2}[/tex]

    and therefore

    [tex]u_p=\frac{f_o}{{\omega}^2-{\mu}^2}sin({\mu}t})[/tex]

    so now the general solution is

    [tex]u=c_1e^{i{\omega}t}+c_2e^{-i{\omega}t}+\frac{f_o}{{\omega}^2-{\mu}^2}sin({\mu}t})[/tex]

    plugging in the initial conditions from the first post, I got

    [tex]c_1=\frac{{\mu}f_o}{2({\mu}^2-{\omega}^2)}[/tex]

    [tex]c_2=-\frac{{\mu}f_o}{2({\mu}^2-{\omega}^2)}[/tex]

    and therefore the entire solution to be

    [tex]u=\frac{{\mu}f_o}{2({\mu}^2-{\omega}^2)}e^{i{\omega}t}-\frac{{\mu}f_o}{2({\mu}^2-{\omega}^2)}e^{-i{\omega}t}+\frac{f_o}{{\omega}^2-{\mu}^2}sin({\mu}t})[/tex]
     
    Last edited: Sep 17, 2006
  5. Sep 17, 2006 #4

    AKG

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    I think you're missing a factor of [itex]i\omega[/itex] in the denominators of c1 and c2.
     
  6. Sep 18, 2006 #5
    You are right. Also, is it right to say that the solution exists as long as [tex]|\mu| {\neq} |\omega|[/tex] (assuming that they are both real)?

    Thanks for the help!
     
    Last edited: Sep 18, 2006
  7. Sep 18, 2006 #6

    HallsofIvy

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    If you are going to use [itex]sin(\mu t)[/itex] as you particular solution, why not use [itex]sin(\omega t)[/itex] and [itex]cos(\omega t)[/itex] in the general solution?

    Yes, the solution is what you have as long as [itex]|\mu|\ne|\omega|[/itex]. However, if you are thinking that when they are equal, a solution does not exist, that is not necessarily so. You just need to use a more complicated function as your specific solution.
     
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