Sturm-Liouville Problem eigenvalues and eigenfunctions $$y''+4y'+\lambda y=0 \ $$ with boundary conditions:
$$ y(0)= 0 \ $$ 
$$ y'(1)+2y(1)=0 \ $$
So far I have put it into the form $$ (e^{4x} y')' + \lambda e^{4x}y = 0 \ $$
How would I go about finding the eigenvalues and eigenfunctions for this problem?
 A: The general solution to the equation is given by
\begin{equation}
y(x) = c_+ e^{\mu_+ x} + c_- e^{\mu_- x}, \qquad (1)
\end{equation}
with $\mu_\pm$ solutions to the characteristic equation $\mu^2 + 4 \mu + \lambda = 0$, i.e.
\begin{equation}
 \mu_\pm = -2 \pm \sqrt{4-\lambda}. \qquad (2)
\end{equation}
Now, the boundary condition at $x=0$ implies that $c_+ + c_- = 0$. The boundary condition at $x=1$ gives, incorporating the fact that $c_- = - c_+$,
\begin{equation}
 c_+ \Big((2 + \mu_+)e^{\mu_+} - (2 + \mu_-)e^{\mu_-} \Big) = 0. \qquad (3)
\end{equation}
The first obvious choice to solve this equation would be $c_+ = 0$, but that would mean that $y(x) = 0$ -- which is of course a solution to the ODE (for every value of $\lambda$), albeit not a very interesting one.
Assuming therefore that $c_+ \neq 0$, and using the expression for $\mu_\pm$ $(2)$, we see that $(3)$ becomes
\begin{equation}
 \frac{\sqrt{4-\lambda}}{ e^{2}}\left(e^{\sqrt{4-\lambda}} + e^{- \sqrt{4-\lambda}} \right) = 0. \qquad (4)
\end{equation}
Another easy solution presents itself: we can choose $\lambda = 4$. However, a tricky thing happens in this case: the two roots of the characteristic equation $(2)$ coincide, and therefore $e^{\mu_+ x}$ and $e^{\mu_- x}$ are not two linearly independent solutions to $(1)$ anymore. It's a nice exercise to find the two independent solutions to $(1)$ in that case, and work through the above to see what the boundary conditions bring. To give you the answer: it turns out that the only solution obeying the boundary conditions is again the trivial solution $y(x) = 0$.
So, what values for $\lambda$, except for $\lambda = 4$, can we choose to solve $(4)$? Here, Euler's formula comes to the rescue. Remember that we can write $\cos(x)$ in terms of exponentials as
\begin{equation}
 \cos(x) = \frac{1}{2} e^{i\,x} + \frac{1}{2} e^{-i\,x}.
\end{equation}
We have a similar structure in $(4)$. Rewriting the exponents as $\sqrt{4-\lambda} = i \sqrt{\lambda - 4}$, we can rewrite $(4)$ as
\begin{equation}
 \frac{2\sqrt{4-\lambda}}{ e^{2}} \cos\left(\sqrt{\lambda-4}\right) = 0. \qquad (5)
\end{equation}
Now it's clear what we have to choose for $\lambda$: any value for which $\sqrt{\lambda - 4}$ makes the cosine vanish will do. That means that
\begin{equation}
 \sqrt{\lambda - 4} = \pi\left(n+\frac{1}{2}\right),\quad n \in \mathbb{N}
\end{equation}
solves $(5)$. Thus, the eigenvalues $\lambda$ are given by
\begin{equation}
 \lambda = 4 + \pi^2 \left(n+\frac{1}{2}\right)^2, \quad n\in \mathbb{N}.
\end{equation}
Finding the corresponding eigenfunctions shouldn't be a problem from this point on.
A: If you follow the usual process for solving linear ODEs with constant coefficients, you arrive at $y=C_1e^{r_1x}+C_2e^{r_2x}$ for the roots $r_1$ and $r_2$ of $r^2+4r+\lambda=0$. These are:
$$r_{1,2}=-2\pm\sqrt{4-\lambda}$$
If $\lambda=4$ this is a double root and the solution should actually be $y=C_1e^{-2x}+C_2xe^{-2x}$, which only satisfies the given boundary conditions for $C_1=C_2=0$. So $\lambda=4$ is not a eigenvalue.
If $\lambda\ne 4$ you can use the 1st boundary condition to show $C_1=-C_2$ and obtain a constraint equation on $\lambda$ from the 2nd.
$$y=C(e^{r_1x}-e^{r_2x})$$
$$y'=C(r_1e^{r_1x}-r_2e^{r_2x})$$
$$y'(1)+2y(1)=C((r_1+2)e^{r_1}-(r_2+2)e^{r_2})=0$$
In terms of $\lambda$ this becomes:
$$e^{-2}\sqrt{4-\lambda}\left(e^{\sqrt{4-\lambda}}+e^{-\sqrt{4-\lambda}}\right)=0$$
If $\lambda<4$ this equation cannot be satisfied, so $\lambda>4$ (or it is complex, but I think that is ruled out by Sturm-Louiville theorems/assumptions) and so we can rewrite it as (dividing out the nonzero factors):
$$e^{i\sqrt{\lambda-4}}+e^{-i\sqrt{\lambda-4}}=2\cos\sqrt{\lambda-4}=0$$
So the eigenvalues are the solutions $\lambda>4$ to $\cos\sqrt{\lambda-4}=0$, which are:
$$\lambda_n=4+\pi^2\left({1\over 2}+n\right)^2$$
And the corresponding eigenfunctions are:
$$y_n=e^{-2x}\sin\left(\left({1\over 2}+n\right)\pi x\right)$$
For non-negative integers $n$.
