Non zero solutions of second order equation by solving for $a$ Im stuck on this problem:
Let a be a real constant. Consider the equation
$y''+5y'+ay=0$ with boundary conditions $y(0)=0$ and $y(3)=0$
For certain discrete values of $a$, this equation can have non-zero solutions. Find the three smallest values of $a$ for which this is the case.
Enter your answers in increasing order.
I assumed the only time the solutions can be non zero are if we have two complex roots and then solving for those. 
By solving, I know the roots must be equal to $\frac{-2.5\pm \sqrt{6.25-a}}{2}$
Im stuck after that so any help would be appreciated!
 A: The characteristic equation of the homogeneous ODE $y''+5y'+ay=0$ is $r^2+5r+a=0$. The general solution of the equation is $y(x) = \lambda_+e^{r_+ x} + \lambda_-e^{r_- x}$, where
$$
r_\pm = -\frac{5}{2}\pm\sqrt{\frac{25}{4}-a}
$$
and the constants $\lambda_+$, $\lambda_-$ are deduced from the boundary values. Of course, $y=0$ is a solution. So, let us assume that $y\neq 0$. The boundary values yield the linear system
\begin{aligned}
\lambda_+\phantom{e^{3 r_+}} + \lambda_-\phantom{e^{3 r_-}} &= 0 \, ,\\
\lambda_+e^{3 r_+} + \lambda_-e^{3 r_-} &= 0 \, .
\end{aligned}
This system has non-trivial solutions if its determinant is equal to zero, i.e.
$$ \exp\left({3\sqrt{\frac{25}{4}-a}}\right)=\exp\left(-{3\sqrt{\frac{25}{4}-a}}\right) . $$
If $25/4 - a \geq 0$, the only solution for this identity to be true is $a = 25/4$, and $y$ is again trivially equal to zero. However, if $25/4 - a < 0$, then Euler's formula gives
$$
\sin\left({3\sqrt{a-\frac{25}{4}}}\right) = 0 \, ,
$$
i.e.
$$
a \in \left\lbrace \frac{25}{4} + \frac{n^2\pi^2}{9} \, , \quad n \in\Bbb{Z}^* \right\rbrace .
$$
In these cases, a nonzero solution can be found.
The answer is $a_1 = \frac{25}{4} + \frac{\pi^2}{9}$, $a_2 = \frac{25}{4} + \frac{4\pi^2}{9}$, and $a_3 = \frac{25}{4} + \pi^2$.
Note that this post is very similar.
