Solve the Sturm-Liouville eigenvalue problem Consider the Sturm-Liouville eigenvalue problem on 0≤≤2,
y'' + y = 0 , y(0) = 0 ,y(2)=y'(2).

*

*Assume the infinite set of eigenvalues are ordered so that 0<1<2<…. Let =|0|
and give an implicit equation for 
of the form =()

*Consider the eigenvalue $_$
as →∞ Determine that $_$ ∼ $(2+1)^2$ in this limit, and give an equation for 

My attempt:

*

*Solved the differential equation with the boundary conditions and ended up with  = $\tan^2(2\sqrt{}))$, hence we have that p = $\tan^2(2\sqrt{p}))$ for first case.

*Same solution for this part  = $\tan^2(2\sqrt{}))$, then used that $_$ ∼ $(2+1)^2$ hence ,   $\tan^2(2\sqrt{}))$ ∼ $(2+1)^2$ , hence    ∼ $\tan^2(2\sqrt{}))/(2+1)^2$ , hence ended up with  = $\tan^2(2\sqrt{(2+1)^2}))/(2+1)^2$ , hence  = $\tan^2(2\sqrt{}(2+1))/(2+1)^2$
The solutions for diff equation is correct, but I was told that my the equations I found are not correct , Do you guys have any advice? Where was incorrect and how the correct will look like ?
 A: I think I did this right but got something a little different.
$y''=-\lambda y$.  $r^2=-\lambda\implies r=\pm\sqrt{|\lambda|}$ or $r=\pm i\sqrt{|\lambda|}$
TRIG SOLUTIONS
$y=c_1\sin (rx)+c_2\cos( rx)$
$y(0)=0\implies c_1=0$
$y(2)=c_2\cos(2r). y'(2)=-c_2r\sin(2r)$
So $1=-r\tan{2r}\implies 1=|\lambda|\tan^2(2\sqrt{|\lambda|})$.
Since for increasing $\lambda$, $1/|\lambda|\to 0\implies \tan^2(2\sqrt{|\lambda|})=0$
$\implies \sqrt{|\lambda|}\approx(2p+1)\pi/2\implies |\lambda|\approx(2p+1)^2\pi^2/4$ for integer $p$.
$\alpha=\pi^2/4$
Also note:
$D^2y=-\lambda y\implies D^2 (Dy)=-\lambda(Dy)$. So given a solution to the eigenvalue problem, the derivative is also a solution. Has bearing on the use of Ladder Operators to produce new solutions to an SL problem.
EXP SOLUTIONS
$y=c_3e^{rx}+c_4e^{-rx}$
$y(0)=0\implies c_3=-c_4$
$y(2)=c_3e^{2r}-c_3e^{-2r}$.
$y'(2)=c_3re^{2r}+c_3re^{-2r}$
$y(2)=y'(2)\implies e^{2r}-e^{-2r}=re^{2r}+re^{-2r}\implies e^{4r}-1=re^{4r}+r$
$e^{4r}(1-r)=1+r\implies e^{4r}=\frac{1+r}{1-r}$
Only two solutions one for when $r=0$ and another negative value.
A: Start by solving
$$
     y''+\lambda y = 0,\;\; y(0)=0,\;y'(0)=1.
$$
You can add the normalization $y'(0)=1$ because that just adjusts the scale of the eigenfunction. And it creates an eigenfunction that depends holomorphically on $\lambda$. The solution of this equation is
$$
        y_{\lambda}(x)=\frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}}. 
$$
For $\lambda=0$, this reduces to the limiting case as $\lambda\rightarrow 0$, which is
$$
        y_0 = x.
$$
Then the condition at $x=2$ becomes an equation for the zeros of an entire function of $\lambda$:
$$
        y(2)-y'(2)=0\\
     \frac{\sin(2\sqrt{\lambda})}{\sqrt{\lambda}}-2\cos(2\sqrt{\lambda})=0
$$
This may also be written as
$$
         \tan(2\sqrt{\lambda})=2\sqrt{\lambda}.
$$
$\lambda=0$ is an eigenvalue with solution
$$
           y_0(x)=x.
$$
