## eigenfunctions and eigenvalues

This is probably a straight forward question, but can someone show me how to solve this problem:

$$\frac {d^2} {d \phi^2} f(\phi) = q f(\phi)$$

I need to solve for f, and the solution indicates the answer is:
$$f_{\substack{+\\-}} (\phi) = A e^{\substack{+\\-} \sqrt{q} \phi}$$

I know I've covered this before - just need a refresher on how to solve.
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 One way to solve this problem is to "guess" a solution. Since the form of your equation is very basic, this guess is usually taught as a standard method. As your solution suggests, your guess should be the following: $$f(\phi)=e^{r\phi}$$ Simply replace every function f with that equation above and solve for the unknown, r. Then, once you find r, plug it back into the equation above. Note that the "A" in your answer is an arbitrary constant used to solve with initial conditions.
 thanks - knew it was something simple. I actually remembered the other approach too, where you replace f'' with r^2, f' with r, and f with 1, and then solve for what r is. But either approach gives the same result. Thanks again though.

## eigenfunctions and eigenvalues

$$f (\phi + 2\pi) = f (\phi)$$
$$2 \pi \sqrt{q} = 2 n \pi i$$
 Nevermind, I got it now - didn't realize the relation between 1 and e^(i2n(pi)).... $$A e^{\sqrt{q} \phi} = A e^{\sqrt{q} \left( \phi + 2 \pi \right)}$$ $$e^{\sqrt{q} \phi} = e^{\sqrt{q} \phi} e^{\sqrt{q} 2 \pi}$$ $$1 = e^{\sqrt{q} 2 \pi}$$ $$e^{i 2 n \pi} = e^{\sqrt{q} 2 \pi}$$ $$i 2 n \pi = \sqrt{q} 2 \pi$$ $$i n = \sqrt{q}$$ $$q = -n^2 (n=0,1,2....)$$