I need to find the eigenvalues and corresponding eigenfunctions of the following differential equation: $$y'' = -sy, \;\;\;\;y(0)=2y'(0), \; y'(1)=0.$$
I've already found the eigenvalues of the function. If $s>0$, then $$y(x) = A\cos(\sqrt sx) + B \sin(\sqrt s x)$$ $$y'(x) = -\sqrt s A \sin(\sqrt s x) + \sqrt sB\cos(\sqrt sx).$$
If we plug in our boundary conditions, we get $$\left.\begin{align}A - 2\sqrt 2 B &= 0 \\ -\sqrt{s}A\sin(\sqrt s) + \sqrt{s}B\cos(\sqrt{s}) & = 0.\end{align}\right\}$$
To solve the eigenvalue problem, I want to avoid the trivial solution of $A=B=0$. In order to accomplish that, I decided to find where there will be more than one solution. So I set the determinant of the coefficient matrix equal to $0$, like so:
$$\begin{vmatrix}1 &- 2\sqrt s \\ -\sin(\sqrt s) & \cos(\sqrt s)\end{vmatrix} = \cos (\sqrt s) - 2\sqrt s \sin(\sqrt s) = 0.$$ Thus, I get eigenvalues $s$ as the solution to the equation $$2\sin(\sqrt s) = \frac{\cos(\sqrt s)}{\sqrt s}.$$
Here's the part I'm stuck at: I need to find the eigenfunction associated with the eigenvalue $s$. I initially tried solving the above equation for $\sqrt s$ and substitute it into $y(x)$, but my answer did not end up correct. I've also tried rearranging the equation to be $\tan(\sqrt s) = \frac{1/2}{\sqrt s}$, and attempted to matched it up with the outline found here, but it still wasn't coming out right. Can anyone provide any pointers on how to find the eigenfunction associated with the eigenvalue $s$?
EDIT: