Find the values of the constant a for which the problem y''(t)+ay(t)=y(t+π), t∈ℝ, has a solution with period 2π
which is not identically zero. Also determine all such solutions
With help of Fourier series I know that :
Cn(y(t+π)) = exp(nπ)Cn(y(t)) = (-1n) Cn(y(t))
-n2*Cn(y(t)) + a*Cn(y(t)) = (-1n) Cn(y(t))
Cn(y) is not zero then we have
a= n2 + (-1n) , n≠0
The Attempt at a Solution
My problem is this expression:
y''(t)+(n2 + (-1n))y(t)=y(t+π)
if a =0 n=1 or -1 we get y''(t)= y(t+π)
In my book this solution is a constant solution but how we say that this solution is constant ?!
Can't we say that y(t+π)=y(t)?
If n≠1 or -1 or 0 then we get :
y''(t)+(n2)*y(t)=0 My question for this term is, shall we think y(t)*(-1n)) = y(t+π) ?!
I know how to solve the rest of this term but I don't get :
1) y''(t)= y(t+π) and why we say this has a constant solution ? How do we solve this actually ?!
2) y(t)*(-1n)) = y(t+π)