- #1

- 101

- 8

## Homework Statement

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

## Homework Equations

With help of Fourier series I know that :

C

_{n}(y''(t))= -n

^{2}*C

_{n}(y(t))

C

_{n}(y(t+π)) = exp(nπ)C

_{n}(y(t)) = (-1

^{n}) C

_{n}(y(t))

We get:

-n

^{2}*C

_{n}(y(t)) + a*C

_{n}(y(t)) = (-1

^{n}) C

_{n}(y(t))

C

_{n}(y) is not zero then we have

a= n

^{2}+ (-1

^{n}) , n≠0

## The Attempt at a Solution

*My problem is this expression*

**:**

y''(t)+(n

^{2}+ (-1

^{n}))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)?*Further:

If n≠1 or -1 or 0 then we get :

**y''(t)+(n**My question for

^{2})*y(t)=0**this**term is, shall we think y(t)*(-1

^{n})) = 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)*(-1

^{n})) = y(t+π)