Where are the other solutions to this ODE disappearing in this analysis? I am asked to solve the following ODE involving constants $\alpha, L, V_0 > 0$ and $E < 0$:
$$-\psi'' - \alpha V_0\psi\cdot [\delta(x)+\delta(x-L)] = \alpha E\psi.$$
In particular, we want solutions $\psi:\mathbb{R}\to \mathbb{C}$ that are:


*

*Continuous

*$L^2$


We are given that the solutions look something like this (shifted): 

Now pretend we didn't know this. Here is my attempt at arriving at these functions: 

Away from $0$ and $L$, the ODE is just the wave equation, whose solutions are always of the form $a_1e^{-i\sqrt{\alpha E}x}+a_2e^{i\sqrt{\alpha E}x}.$ So we know $\psi$ is also of this form piece-wise:
  $$\psi(x)=\left\lbrace \begin{array}{ll}
    A_1e^{\sqrt{|\alpha E|}x}+A_2e^{-\sqrt{|\alpha E|}x} & x<0 \\
    B_1e^{\sqrt{|\alpha E|}x}+B_2e^{-\sqrt{|\alpha E|}x} & 0\leq x< L \\
    C_1e^{\sqrt{|\alpha E|}x}+C_2e^{-\sqrt{|\alpha E|}x} & x\geq  L \\
\end{array} \right.$$
  (I expanded a little here, since $E<0\Rightarrow \sqrt{\alpha E}=i\sqrt{|\alpha E|}$)
  
  
*
  
*For $\psi$ to be integrable, we also need the exponentials with infinite > support be integrable. Then $A_2=C_1=0.$
  
  
  Then we can simplify it:
  $$\psi(x)=\left\lbrace \begin{array}{ll}
    A_1e^{\sqrt{|\alpha E|}x} & x<0 \\
    B_1e^{\sqrt{|\alpha E|}x}+B_2e^{-\sqrt{|\alpha E|}x} & 0\leq x< L \\
    C_2e^{-\sqrt{|\alpha E|}x}& x\geq L \\
\end{array} \right.$$
  All that's left to do is to find the space of coefficients $A_i,B_i,C_i$ so that it is continuous and satisfies the DE.
  
  
*
  
*For $\psi$ to be continuous, $\lim_{x\to 0^+}\psi(x)=\lim_{x\to 0^-}\psi(x)$ and likewise at $L$. So $A_1=B_1+B_2,$ and a similar equation at $L.$ 
  
*To satisfty the ODE at $0$ and $L$, because of the deltas, we know that the derivative must have a discintinuity of size $-\alpha V_0 \psi$ there. In particular: $(\lim_{x\to 0^+}\psi'(x))-(\lim_{x\to 0^-}\psi'(x))=-\alpha V_0 \psi(0),$ so computing these terms, $(\sqrt{|\alpha E|}B_1 - \sqrt{|\alpha E|}B_2)- (\sqrt{|\alpha E|}A_1) = -\alpha V_0A_1$. We have a similar equation at $L$.
  

Now we have 4 equations relating all the coefficients to each other. I solved this system and plotted the solutions. The problem is, varying values of $\alpha, L, V_0, E$, I can only get solutions that look like the 'even' graph. 
Where did the odd solutions disappear in this analysis and how do you get them back?
 A: As stated, the solution to the ODE is
$$
\psi(x) = 
\begin{cases}
A_1 e^{\sqrt{-\alpha E} x} & x < 0,\\
B_1 e^{\sqrt{-\alpha E} x} + B_2 e^{-\sqrt{-\alpha E} x} & 0 < x < L, \\
C_2 e^{-\sqrt{-\alpha E} x} & L < x.\\
\end{cases}
$$
Writing the jump and continuity conditions in matrix form, we have
$$
\begin{pmatrix}
1 & -1 & -1 & 0\\
0 & e^{2 L \sqrt{-\alpha E}} & 1 & -1\\
1-\sqrt{-\frac{\alpha V_0^2}{E}} & -1 & 1 & 0\\
0 & e^{2 L \sqrt{-\alpha E}} & -1 & 1-\sqrt{-\frac{\alpha V_0^2}{E}}
\end{pmatrix}
\begin{pmatrix}A_1 \\ B_1 \\ B_2 \\ C_2\end{pmatrix} = 0,
$$
so either you have the trivial solution or the determinant of the matrix is zero. Hence,
$$
f(E) = e^{2 L\sqrt{-\alpha E}}\left(1 - \frac{4 E}{\alpha V_0^2} - 4\sqrt{-\frac{E}{\alpha V_0^2}}\right) = 1.
$$
This is an interesting equation. It's clear that is trivially satisfied if $E = 0$, but this leads to a nonphysical solution. 
The function $f$ is nonnegative, has a unique double root at $E = -\frac{\alpha V_0^2}{4}$ and grows as $c_1|E|e^{c_2\sqrt{|E|}}$. This means that it exists at least one solution  of the transcendental equation such that $f(E_*) = 1$. Moreover,
$$-\alpha V_0^2 < E_* < -\frac{\alpha V_0^2}{4}.$$
Also, if $\alpha V_0 L > 2$, then $$\lim_{E\to 0^-}f'(E) = -\infty,$$ wich implies that $f$ has a local maximum $E_M$  such that $f(E_M) > 1$, i.e., 
$$E_M = -\tfrac{(\alpha V_0 L - 2)^2}{4 \alpha L^2}.$$
As consequence, there is a second solution such that $f(E^*) = 1$, where 
$$-\frac{\alpha V_0^2}{4} < E^* < -\tfrac{(\alpha V_0 L - 2)^2}{4 \alpha L^2}.$$
Combining this results, we have:

Lemma. If $0 < \alpha V_0 L \le 2$, there exist one eigenvalue, $$-\alpha V_0^2 < E_* < -\frac{\alpha V_0^2}{4}.$$ If $2 < \alpha V_0 L$, there exist two eigenvalues, $$-\alpha V_0^2 < E_* < -\frac{\alpha V_0^2}{4} < E^* < -\tfrac{(\alpha V_0 L - 2)^2}{4 \alpha L^2}.$$

For $L=1$, $\alpha = 1$, $V_0 = \frac{3}{2}$, this is how the eigenfunction looks like:

For $L=1$, $\alpha = 1$, $V_0 = \frac{5}{2}$, this are the two eigenfunctions:

Nice!
