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Problem with a Fourier Series

  1. Apr 15, 2006 #1
    Can anyone help me out with this?

    Find the steady state periodic solution of the following differential equation.

    x''+10x= F(t), where F(t) is the even function of period 4 such that
    F(t)=3 if 0<t<1 , F(t)=-3 if 1<t<2.

    Im basically just having a problem findind the general Fourier series for F(t).
    I know how to do the latter part of the problem.

    My work so far: Knowing this is even, I can eliminate the sin part of the fourier series. So in general I need to solve for the series cofficients of a(0) and a(n)

    for a(o) I get 0. Which makes sense too, even just by inspection of the graph of the function.

    My problem is with a(n). My final result is [6/npi]*[sin(npi/2)]. How do I express that second term in my answer. I noticed that the sign alternates every other odd number. a(n) =0 for every even number.

    Thanks a bunch
  2. jcsd
  3. Apr 15, 2006 #2
    Sorry about that folks. I posted this in the wrong thread.
  4. Apr 15, 2006 #3
    One way is to introduce n = 2p+1 and see that

    [tex]a_n = \frac {6}{(2p+1) \pi}(-1)^p[/tex] where [tex]p = 0,1,2,...[/tex]
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