Periodic Solutions to DE with Extra Condition: Solving for Unknown Constants

[Zaare]

I don't know how to use the extra condition given in the this problem:

f^{\prime\prime}+\lambda f = 0, f=f\left(r\right)
f\left(r\right) = f\left(r+\pi\right)
​

For \lambda = 0, the solution is some constant.
For other \lambda, I put \lambda=k^2, and get

f^{\prime\prime}+k^2 f = 0,
​

which has the solution

f\left(r\right)=A\cos\left(kr\right)+B\sin\left(kr\right).
​

The condition now gives

A\cos\left(kr\right)+B\sin\left(kr\right)=<br /> A\cos\left(kr+k\pi\right)+B\sin\left(kr+k\pi\right).
​

How can I use this to further specify the solution?

 

[LeonhardEuler]

Well, you know that \cos{x} + \sin{x} = \cos{(x+2n\pi)} + \sin{(x+2n\pi)} for all x only when n is an integer. What does that tell you about k, and therefore \lambda?

 

[Zaare]

Oh, right, so k must be an even integer.
Thank you.