Sturm-Louiville eigenvalues I am given this Sturm-Louiville problem: $$\frac{d^2y}{dx^2}+\lambda y=0$$

With boundary conditions: $y'(0) = 0, y(2) = 0$, where $n = 1, 2, 3, ...$

I am now trying to find the eigenvalues. 

I tried using the general solution:

$y = A \cos(px) + B\sin(px)$, where $λ = -p^2 &lt; 0$

I differentiated this and substituted the boundary conditions $y'(0) = 0$ to obtain that $B = 0$.

From this, I used the boundary condition $y(2) = 0$ to obtain:

$A\cos(2p) = 0.$

From this, I arrived at $p = \frac{n  \pi}{4}$.

I was hoping for some kind of confirmation if I am on the right track...I would also like some guidance of where to go from this to obtain the eigenvalues.

I tried to find the eigenvalues by simply finding $-p^2$, but I found that this was wrong. 

Thank you for your time,
I appreciate any help.
$$\frac{d^2y}{dx^2}+\lambda y=0\tag1$$

For $\lambda =0$, the general solution is $y(x)=Ax+B$,  where  $A, B$ are arbitrary constants..

The given conditions are $y'(0) = 0, y(2) = 0$ gives $A=B=0$

Hence we get $y(x)=0$, which is the trivial solution.

For $\lambda\lt 0$, say $\lambda=-p^2$

then the general solution is $y(x)=Ae^{px}+Be^{-px}$, where  $A, B$ are arbitrary constants..

The given conditions are $y'(0) = 0, y(2) = 0$ gives 

$pA-pB=0\implies A-B=0$

$Ae^{2p}+Be^{-2p}=0$ which gives $A=B=0$

Hence we again get $y(x)=0$, which is the trivial solution.

So we discard these two cases and take $\lambda\gt0$.

Let $\lambda=p^2 \gt 0$

The general solution is $y(x)=A \cos(px) + B \sin(px) $, where  $A, B$ are arbitrary constants.

Now the given conditions are $y'(0) = 0, y(2) = 0$

$y'(0) = 0\implies B=0$ and $y(2) = 0\implies A\cos(2p)=0$

If we take $A=0$, then we again get the trivial equation, so we take $A \ne 0$. This gives

$~\cos(2p)=0\implies 2p=(2n-1)\frac{\pi}{2}~~~~~\text{where $~n=1,~2,~3,~\cdots~$ }$
$~\implies ~p=(2n-1)\frac{\pi}{4}~~~~~\text{where $~n=1,~2,~3,~\cdots~$ }$

So the eigen value of the given differential equation $(1)$ is $$~\lambda=p^2=~(2n-1)^2\frac{\pi^2}{16}~~~~~\text{where $~n=1,~2,~3,~\cdots~$ }~$$ 

and the corresponding eigen functions are $$~y(x)~=~A~\cos\left\{(2n-1)\frac{\pi}{4}~x\right\}\qquad \text{where $~n=1,~2,~3,~\cdots~~~$and $~a~$ is a constant}~.$$



You can also take the following 

$~\cos(2p)=0\implies 2p=(2n+1)\frac{\pi}{2}~~~~~\text{where $~n=0,~1,~2,~3,~\cdots~$ }$
$~\implies ~p=(2n+1)\frac{\pi}{4}~~~~~\text{where $~n=0,~1,~2,~3,~\cdots~$ }$

So the eigen value of the given differential equation $(1)$ is $$~\lambda=p^2=~(2n+1)^2\frac{\pi^2}{16}~~~~~\text{where $~n=0,~1,~2,~3,~\cdots~$ }~$$ 

and the corresponding eigen functions are $$~y(x)~=~A~\cos\left\{(2n+1)\frac{\pi}{4}~x\right\}\qquad \text{where $~n=0,~1,~2,~3,~\cdots~~~$and $~a~$ is a constant}~.$$

Note: In both the two cases the values of $~\lambda~$ are same, in the first one $~n~$ does not take the value $~0~$, and in the second case it takes.
Any eigenfunction must be a non-zero constant multiple of the solution of
$$
          y''+\lambda y =0,\;\; y(0)=1,\; y'(0)=0,
$$
which has unique solution
$$
        y(x) = \cos(\sqrt{\lambda}x).
$$
Then, in order to satisfy $y(2)=0$, $\lambda$ must satisfy
$$
           \cos(2\sqrt{\lambda})=0.
$$
 The general solution is
$$
                2\sqrt{\lambda} = (n+1/2)\pi ,\;\; n=0,1,2,3,\cdots, \\
                \lambda = (n+1/2)^2\pi^2/4,\;\;\; n=0,1,2,3,\cdots.
$$
The corresponding eigenfunctions are
$$
           y_n(x) = \cos((n+1/2)\pi x/2),\;\; n=0,1,2,3,\cdots.
$$