I want to show that there are countably infinite number of eigenvalues (that can put in ascending order-- with a minimum value) to the 2nd order ODE x''+\lambda x = 0 subjected to boundary conditions $x(0)=0$ and x'(1)+x(1)=0.
My immediate approach would be to solve it.
1) For $\lambda <0$,
general solution: $a \exp(\mu t)+b\exp(-\mu t)$, where $\mu ^2 = -\lambda$;
boundary conditions equations: $a+b=0$ and $(1+\mu)a \exp(\mu )+(1-\mu) b\exp(-\mu)=0$
So no non-trivial solution.
2) For $\lambda =0$, again, easy to show that there are no non-trivial solutions.
3) For $\lambda >0$,
general solution: $a\sin(\mu t)+b\cos(\mu t)$ where $\mu = \sqrt{\lambda}$
Boundary condition: $b=0$ and (so) $a\mu \cos(\mu) +a\sin(\mu) =0$ Giving $\tan(\mu)=-\mu$ but then considering the intersections of the graphs of $y=\tan(z)$ and $y=-z$, there is no minimum $\mu$??!
Also, are there any more elegant way to do this? (may invoke Sturm-Liouville)
Thank you.