Solve the Sturm-Liouville problem $y'' + λy = 0$ given $y(0) = 0$, $y'(6) = 0$? Given 
$$y''(x) + \lambda \, y(x) = 0, \quad 0 < x < 6, \quad \lambda \in \mathbb{R},$$ 
with the boundary conditions $y(0) = 0$, $y'(6) = 0$, how do you find the eigenvalues and eigenfunctions? I tried doing it for $λ < 0$ and I got 
$$\lambda = -k^2 = -\frac{1}{144}\cdot π^2\cdot(4n-1)^2.$$
But I'm not sure if that's right. Any help would be appreciated, thanks 
 A: Your problem can be made quasi-independent of the lenght of the interval, $L = 6$, by introducing $X = x/L,$ hence obtaining (prove it):
$$ y'' + \xi \, y = 0, \quad y = y(X), \quad 0 < X< 1, \quad \xi := L^2 \lambda,$$
and, of course, $y(0) =0, \, y'(1) = 0$.
Then we have to analyze the cases: $\xi <0, \ \xi   = 0, \ \xi > 0$ since the solution of the equation strongly depends on the character of $\xi$. If we enumerate this cases as a), b) and c), then the solution (and its derivative) turns to be, respectively:

*

*a) For $\xi < 0 $, the roots of the charactristic polynomial indicates:
$$y(X) = A \cosh{ \xi' X} + B \sinh{\xi'X}, $$
$$y'(X) = A \xi' \sinh{ \xi' X} + B \xi' \cosh{\xi'  X}, $$ where I made $\xi' = \sqrt{|\xi|}$. Then, applying the initial conditions we come up with: $A = B = 0$, which is the (undesired) trivial solution.


*b) For $\xi = 0 $, it is easy to see that:
$$y(X) = A X + B, \quad y'(X) = A, $$ which yields again the trivial solution after applying the correspondent initial conditions.


*c) Interesting things occur when $\xi >0 $ and the roots of the characteristic equation are given by $\pm i \sqrt{\xi}$ and, thus:
$$y(X) = A \cos{\sqrt{\xi} X} +B \sin{\sqrt{\xi} X}, $$
$$y'(X) = -A  \sqrt{\xi} \sin{\sqrt{\xi} X} +B   \sqrt{\xi}  \cos{\sqrt{\xi} X}. $$ Setting $y(0) = 0$ yields $A = 0$ and, ultimately, $y'(0) = 0$, to:
$$ 0 = B \sqrt{\xi} \cos{\sqrt{\xi}}, $$ which is true for:

*

*$B = 0$, which give us again the undesired trivial solution.

*$\xi = 0$, which is a contradiction (remember that $\xi> 0 $).

*$\cos{\sqrt{\xi}} = 0$, which is true iff $\sqrt{\xi}$ is an odd multiple of $\pi/2$, i.e.:
$$\xi_n = (2 n + 1)^2 \frac{\pi^2}{4}, \quad n \in \{0,1,2,\ldots\} \equiv \mathbb{N} \cup \{0\},$$ with associated solution:
$$y_n(X) = B_n \sin\left({(2 n + 1)\frac{\pi}{2} \, X}\right).$$ In terms of your original variables, we would have:
$$y_n(x) = B_n \sin\left({(2 n + 1)\frac{\pi}{2L} \, x}\right).$$
Hope this helps.
Cheers!
A: If we try a solution of the form $y(x) = e^{rx}$ we obtain the characteristic equation $r^2 + \lambda = 0$. Regardless of whether $\lambda < 0$, $\lambda = 0$, or $\lambda > 0$, we get a solution to the differential equation $y''(x) + \lambda y(x) = 0$. However, we also require that the initial conditions $y(0) = 0$ and $y'(6) = 0$ are met; this is where the dependence on $\lambda$ is used.
If $\underline{\lambda < 0}:$ Let $\lambda = -k^2$ where $k > 0$, then the differential equation has characteristic equation $r^2 - k^2 = 0$, so $r = \pm k$ and therefore the general solution is $$y(x) = Ae^{kx} + Be^{-kx}.$$
Let's see if the initial conditions are satisfied. Note that $$y(0) = Ae^0+Be^0 = A+B,$$ so $A+B = 0$ and therefore $B = -A$. As 
$$y'(x) = Ake^{kx}-Bke^{-kx} = Ake^{kx}+Ake^{-kx},$$
we have 
$$y'(6) = Ake^{6k}+Ake^{-6k}.$$ 
As $k > 0$, $e^{6k} >0$, and $e^{-6k} > 0$, the only way we can have $Ake^{6k}+Ake^{-6k} = 0$ is if $A = 0$, but then $y(x) = 0$ is the trivial solution.
Therefore, there are no non-trivial solutions for $\lambda < 0$.
If $\underline{\lambda = 0}$: The characteristic equation is $r^2 = 0$, so the general solution is
$$y(x) = (A + Bx)e^{0x} = A+Bx.$$
In order for $y(0) = 0$, we need $A = 0$ so $y(x) = Bx$ and therefore $y'(x) = B$. As $y'(6) = 0$, we also require $B = 0$ so $y(x) = 0$ is the trivial solution.
Therefore, there are no non-trivial solutions for $\lambda = 0$.
If $\underline{\lambda > 0}$: Let $\lambda = k^2$, $k > 0$, then . . . if you follow the procedure I have used in the above cases, you should obtain some non-trivial solutions, but I'll leave that to you (I don't want to do all of it for you).
