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Fourier Series Help

  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
    My new basis of thought is that the a(n) Fourier coefficient can only be positive 6/npi... is this correct... if figure when you add the negative term, the coefficient becomes zero again. At no point can it be negative.

    Correct me if I am wrong.

    Thanks
     
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