This linear ODE has characteristic equation
[tex]r^2-4r+a=0[/tex]
with roots [tex]r=2\pm\sqrt{4-a}[/tex], which gives solutions of the form
[tex]y_c=C_1e^{(2+\sqrt{4-a})x}+C_2e^{(2-\sqrt{4-a})x}[/tex]
There are three cases to consider:
(1) If [tex]a<4[/tex], then the solution will be exactly what we see above. However, the initial conditions force both [tex]C_1=C_2=0[/tex].
(2) If [tex]a=4[/tex], we're left with
[tex]y_c=C_1e^{2x}+C_2xe^{2x}[/tex]
where [tex]xe^{2x}[/tex] is added to the solution set to account for a second solution that is linearly independent of the first solution. Again, we get [tex]C_1=C_2=0[/tex].
(3) If [tex]a>4[/tex], then the square root introduces a factor of [tex]i[/tex] that admits the solution
[tex]y_c=C_1e^{2x}\cos(\sqrt{4-a}x)+C_2e^{2x}\sin(\sqrt{4-a}x)[/tex]
In this case, we arrive at [tex]C_1=0[/tex], and from the second condition we get
[tex]0=C_2e^{16}\sin(8\sqrt{4-a})[/tex]
In order that [tex]C_2\neq0[/tex], we require that [tex]8\sqrt{4-a}=n\pi[/tex], where [tex]n[/tex] is any integer. Solving for [tex]a[/tex], we get
[tex]8\sqrt{4-a}=n\pi\implies a=\dfrac{256-n^2\pi^2}{64}[/tex]
When [tex]n=0[/tex], we arrive at [tex]a=4[/tex], but remember that we're assuming that [tex]a>4[/tex], so logically the three smallest values of [tex]a[/tex] that are allowed occur for [tex]n=1,2,3[/tex]. ([tex]n^2=(-n)^2[/tex], so we can just look at positive integers [tex]n[/tex].)
Unfortunately, I'm not sure exactly what's going on next. Checking with a computer, the solution is supposed to be
[tex]a=4+4n^2\pi^2[/tex]
(Again, not sure why this is the case, but let's move on.) When [tex]n=1,2,3[/tex], we have the least values, which are, respectively,
[tex]a=4+4\pi^2[/tex]
[tex]a=4+16\pi^2[/tex]
[tex]a=4+36\pi^2[/tex]
