What is the general Fourier series for an even function with a period of 4?

[Giuseppe]

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

 

[Giuseppe]

Sorry about that folks. I posted this in the wrong thread.

 

[Corneo]

One way is to introduce n = 2p+1 and see that

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