Sturm-Louville eigenvectors and eigenfunctions I have the following problem
$\frac{d^2y}{dx^2} + \lambda y = 0 , y'(0)=0$ and $y(3)=0$
I'm trying to solve for the eigenvalues $\lambda_n$ for $n=1,2,3...$
and eigenfunctions $y_n$ or $n=1,2,3...$
Im considering all cases for the values of $\lambda$:
$\lambda = 0$: $y= Ax+B$ and $y' = A$ - applying conditions yields $A=0=B$
Now i get stuck on the cases for $\lambda < 0$ and $\lambda > 0$.
I have tried a similar approach to the $\lambda = 0$ case by setting $\lambda = p^2 >0$ for example but I'm unsure where to go from here
Any help or guidance will be greatly appreciated!
edit: my latest attempt

 A: The general solution of this equation is $$y(x) = A \sin \sqrt{\lambda} x+ B \cos \sqrt{\lambda} x ....(1)$$ Then $$y'(x)=A \sqrt{\lambda} \cos {x\sqrt{\lambda}} - B \sin  \sqrt{ \lambda} \sin{x\sqrt{\lambda}}.....(2)$$ Since $y'(0)=0$, $A=0$. So $$ y(x) = B \cos (x \sqrt{\lambda})....(3)$$ The condtion that $y(3)=0$ gives $$ 3 \sqrt{\lambda} = (n+1/2)\pi, n=0,1,2,3...(4)$$ So the eigenvalues are $$\lambda_n=(n+1/2)^2 ~\frac{\pi^2}{9},~~ n=0,1,2,3...$$ Using (4), the corresponding eigenfunctions are given by re-writing (3) as $$ y_n(x)=B \cos\left((n+1/2)\frac{\pi x}{3}\right), n=0,1,2,3,...$$
Note that eigenvalues $\lambda_n$ emerge only when the solution $y(x)$ satisfies the given conditions at $x=0$ and $x=3$. $\lambda_n$ becomes a function of $n$, then only $\lambda_n$ are the allowed (eigen) values of $\lambda$.
One cannot prefix a value for $\lambda$. Then $y(x)$ not sayisfy the given conditions. For instance in you problem $\lambda$ cannot become zero. The least possible value of $\lambda$ is $\lambda_0=\pi^2/36$.
A: Any non-trivial solution can be normalized to satisfy $y'(0)=0,y(0)=1$ because $y'(0)=y(0)=0$ and $y''+\lambda y=0$ would force $y=0$ by uniqueness of solutions of such problems. The unique solution of this modified problem is
$$
        y_{\lambda}(x)=\cos(\sqrt{\lambda}x)
$$
The eigenvalue equation comes from imposing the second endpoint condition:
$$
               \cos(3\sqrt{\lambda})=0.
$$
Negative $\lambda$ turn the $\cos$ to $\cosh$, and there are no solutions. The eigenvalue solutions are $3\sqrt{\lambda}=(n+1/2)\pi$ for $n=0,1,2,3,\cdots$. The eigenvalues are
$$
           \lambda_n = (n+1/2)^2\pi^2/9,\;\;\; n=0,1,2,\cdots
$$
The corresponding eigenfunctions are $y_n(x)=\cos((n+1/2)\pi x/3)$.
