Question

Let be a real constant. Consider the equation $\frac{d^{2}y}{dx^{2}} -4 \frac{dy}{dx} + ay = 0$ with boundary conditions $y(0)=0$ and $y(8)=0$. 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.

Answer 1

This linear ODE has characteristic equation

$r^2-4r+a=0$

with roots $r=2\pm\sqrt{4-a}$, which gives solutions of the form

$y_c=C_1e^{(2+\sqrt{4-a})x}+C_2e^{(2-\sqrt{4-a})x}$

There are three cases to consider:

(1) If $a$, then the solution will be exactly what we see above. However, the initial conditions force both $C_1=C_2=0$.

(2) If $a=4$, we're left with

$y_c=C_1e^{2x}+C_2xe^{2x}$

where $xe^{2x}$ is added to the solution set to account for a second solution that is linearly independent of the first solution. Again, we get $C_1=C_2=0$.

(3) If $a>4$, then the square root introduces a factor of $i$ that admits the solution

$y_c=C_1e^{2x}\cos(\sqrt{4-a}x)+C_2e^{2x}\sin(\sqrt{4-a}x)$

In this case, we arrive at $C_1=0$, and from the second condition we get

$0=C_2e^{16}\sin(8\sqrt{4-a})$

In order that $C_2\neq0$, we require that $8\sqrt{4-a}=n\pi$, where $n$ is any integer. Solving for $a$, we get

$8\sqrt{4-a}=n\pi\implies a=\dfrac{256-n^2\pi^2}{64}$

When $n=0$, we arrive at $a=4$, but remember that we're assuming that $a>4$, so logically the three smallest values of $a$ that are allowed occur for $n=1,2,3$. ($n^2=(-n)^2$, so we can just look at positive integers $n$.)

Unfortunately, I'm not sure exactly what's going on next. Checking with a computer, the solution is supposed to be

$a=4+4n^2\pi^2$

(Again, not sure why this is the case, but let's move on.) When $n=1,2,3$, we have the least values, which are, respectively,

$a=4+4\pi^2$
$a=4+16\pi^2$
$a=4+36\pi^2$