This linear ODE has characteristic equation
![r^2-4r+a=0](https://tex.z-dn.net/?f=r%5E2-4r%2Ba%3D0)
with roots
![r=2\pm\sqrt{4-a}](https://tex.z-dn.net/?f=r%3D2%5Cpm%5Csqrt%7B4-a%7D)
, which gives solutions of the form
![y_c=C_1e^{(2+\sqrt{4-a})x}+C_2e^{(2-\sqrt{4-a})x}](https://tex.z-dn.net/?f=y_c%3DC_1e%5E%7B%282%2B%5Csqrt%7B4-a%7D%29x%7D%2BC_2e%5E%7B%282-%5Csqrt%7B4-a%7D%29x%7D)
There are three cases to consider:
(1) If
![a](https://tex.z-dn.net/?f=a%3C4)
, then the solution will be exactly what we see above. However, the initial conditions force both
![C_1=C_2=0](https://tex.z-dn.net/?f=C_1%3DC_2%3D0)
.
(2) If
![a=4](https://tex.z-dn.net/?f=a%3D4)
, we're left with
![y_c=C_1e^{2x}+C_2xe^{2x}](https://tex.z-dn.net/?f=y_c%3DC_1e%5E%7B2x%7D%2BC_2xe%5E%7B2x%7D)
where
![xe^{2x}](https://tex.z-dn.net/?f=xe%5E%7B2x%7D)
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](https://tex.z-dn.net/?f=C_1%3DC_2%3D0)
.
(3) If
![a>4](https://tex.z-dn.net/?f=a%3E4)
, then the square root introduces a factor of
![i](https://tex.z-dn.net/?f=i)
that admits the solution
![y_c=C_1e^{2x}\cos(\sqrt{4-a}x)+C_2e^{2x}\sin(\sqrt{4-a}x)](https://tex.z-dn.net/?f=y_c%3DC_1e%5E%7B2x%7D%5Ccos%28%5Csqrt%7B4-a%7Dx%29%2BC_2e%5E%7B2x%7D%5Csin%28%5Csqrt%7B4-a%7Dx%29)
In this case, we arrive at
![C_1=0](https://tex.z-dn.net/?f=C_1%3D0)
, and from the second condition we get
![0=C_2e^{16}\sin(8\sqrt{4-a})](https://tex.z-dn.net/?f=0%3DC_2e%5E%7B16%7D%5Csin%288%5Csqrt%7B4-a%7D%29)
In order that
![C_2\neq0](https://tex.z-dn.net/?f=C_2%5Cneq0)
, we require that
![8\sqrt{4-a}=n\pi](https://tex.z-dn.net/?f=8%5Csqrt%7B4-a%7D%3Dn%5Cpi)
, where
![n](https://tex.z-dn.net/?f=n)
is any integer. Solving for
![a](https://tex.z-dn.net/?f=a)
, we get
![8\sqrt{4-a}=n\pi\implies a=\dfrac{256-n^2\pi^2}{64}](https://tex.z-dn.net/?f=8%5Csqrt%7B4-a%7D%3Dn%5Cpi%5Cimplies%20a%3D%5Cdfrac%7B256-n%5E2%5Cpi%5E2%7D%7B64%7D)
When
![n=0](https://tex.z-dn.net/?f=n%3D0)
, we arrive at
![a=4](https://tex.z-dn.net/?f=a%3D4)
, but remember that we're assuming that
![a>4](https://tex.z-dn.net/?f=a%3E4)
, so logically the three smallest values of
![a](https://tex.z-dn.net/?f=a)
that are allowed occur for
![n=1,2,3](https://tex.z-dn.net/?f=n%3D1%2C2%2C3)
. (
![n^2=(-n)^2](https://tex.z-dn.net/?f=n%5E2%3D%28-n%29%5E2)
, so we can just look at positive integers
![n](https://tex.z-dn.net/?f=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](https://tex.z-dn.net/?f=a%3D4%2B4n%5E2%5Cpi%5E2)
(Again, not sure why this is the case, but let's move on.) When
![n=1,2,3](https://tex.z-dn.net/?f=n%3D1%2C2%2C3)
, we have the least values, which are, respectively,
![a=4+4\pi^2](https://tex.z-dn.net/?f=a%3D4%2B4%5Cpi%5E2)
![a=4+16\pi^2](https://tex.z-dn.net/?f=a%3D4%2B16%5Cpi%5E2)