Determine the eigenvalues (and corresponding eigenfunctions) if phi satisfies... I am stuck on how to approach this question: 


Solution:

Any help is appreciated.
Leo
 A: Hints:
The characteristic equation is: $m^2 +\lambda = 0$
At $\lambda = 0$, we have: $m^2 = 0$, so the solution is: $y = x_0 + x_1 t$.
At $\lambda < 0$, we have: $m^2 - \lambda = 0$, thus we have an eigenvalue of $\pm \lambda$, so the solution is: $y = x_0 e^{\sqrt{-\lambda} t} + x_1e^{-\sqrt{-\lambda} t}$, where $-\lambda$ and $\sqrt{-\lambda}$ are positive.
At $\lambda > 0$, we have $m^2 + \lambda = 0$, thus we have a complex eigenvalue of $\pm i\lambda$, so the solution is: $y = x_0 \sin (\sqrt{\lambda}~t) + x_1\cos (\sqrt{\lambda}~t)$.
Can you use the BC's to find $x_0$ and $x_1$, find all three cases together to determine the value of $\lambda$ and finish this off?
A: I suggest you follow the advice in the question and analyze the three cases $\lambda > 0$, $\lambda = 0$ and $\lambda < 0$.
To get you started, assume $\lambda = 0$. Then $\phi'' = 0$ and the solutions are linear, i.e. $\phi = Ax + B$. This can't satisfy the boundary conditions (except for $A=B=0$, but that's not very interesting so we disregard it) so $\lambda = 0$ is not an eigenvalue. 
Now check the other two cases.
