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:

[tex]f^{\prime\prime}+\lambda f = 0[/tex], [tex]f=f\left(r\right)[/tex]
[tex]f\left(r\right) = f\left(r+\pi\right)[/tex]
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For [tex]\lambda = 0[/tex], the solution is some constant.
For other [tex]\lambda[/tex], I put [tex]\lambda=k^2[/tex], and get

[tex]f^{\prime\prime}+k^2 f = 0[/tex],
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which has the solution

[tex]f\left(r\right)=A\cos\left(kr\right)+B\sin\left(kr\right).[/tex]
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The condition now gives

[tex]A\cos\left(kr\right)+B\sin\left(kr\right)=
A\cos\left(kr+k\pi\right)+B\sin\left(kr+k\pi\right).[/tex]
​

How can I use this to further specify the solution?

 

[LeonhardEuler]

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

 

[Zaare]

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