Question about damped vibrations in ODEs I am currently studying ODEs and I came across this problem. For the first 2 parts, I just want to know if I am understanding this correctly. My main question is for part (c).

Given the spring-mass system represented by the equation $y'' + 4y' + ky = 0$,
a) for what value of k is the system critically damped?
b) for k greater than the value in (a), is the system over-damped or under-damped?
c) if the solution for $y'' + 4y' + ky = 0$ vanishes at $t = 2$ and $3$ (and not in between), then find the corresponding value of k.

I wanted to ask if anyone could show me how to solve part (c). Here is what I have so far:
a) This is simple I think. The discriminant is $0$ for $k = 4$.
b) for $k > 4$, we will have that $\sqrt{4k} > 4$ and so it will be under-damped.
c) If the system were critically damped or over-damped, then y would vanish at at most one value of t. So the system must be under-damped. Thus $\sqrt{4k} > 4$.
In this case, the characteristic equation $r^2 + 4r + k = 0$ has complex roots, and so the general solution for the ODE will be:
$y = e^{-2t}(c_1\cos{\sqrt{k-4}t} + c_2 \sin{\sqrt{k-4}t})$
Since we know y vanishes at $t = 2,3$ we get the two equations:
$c_1\cos{2\sqrt{k-4}} + c_2 \sin{2\sqrt{k-4}} = 0$
$c_1\cos{3\sqrt{k-4}} + c_2 \sin{3\sqrt{k-4}} = 0$
I understand the problem till here, but I don't see how we can deduce k from this, given that we  have two equations and three unknowns. How do I deduce the value of k from the given information (If my inferences are even correct).
Thank you
 A: The problem becomes a bit easier to manage if you rewrite the solution as $y=e^{-2t}\left(Ae^{it\sqrt{k-4}}+Be^{-it\sqrt{k-4}}\right)$. 
Then the pair of equations become: $$Ae^{2i\sqrt{k-4}}+Be^{-2i\sqrt{k-4}}=0$$$$Ae^{3i\sqrt{k-4}}+Be^{-3i\sqrt{k-4}}=0$$ 
From these, we get: $$e^{4i\sqrt{k-4}}=\frac{-B}{A}=e^{6i\sqrt{k-4}}$$ 
Which means: $$e^{2i\sqrt{k-4}}=1$$ 
Also, if we plug $e^{2i\sqrt {k-4}}=1$ back into the first equation, we get: $$A+B=0\implies A=-B$$ 
So the solution is of the form $y=Ae^{-2t}\left(e^{it\sqrt{k-4}}-e^{-it\sqrt{k-4}}\right)$. 
Since $y$ is not $0$ between $2$ and $3$, $A\neq 0$, but that is all that can be said about $A$ (as question $(c)$ gives no information about the solution's amplitude).
And $e^{2i\sqrt{k-4}}=1$ tells us only that $k$ must be of the form $\pi^2j^2+4$ for some $j\in \mathbb Z$. 
At the same time, if we plug in $k=\pi^2j^2+4$ for any $j\in \mathbb Z$ into the solution expression, we see that $y(2)=0=y(3)$ is satisfied. 
So we cannot solve further for the solution, and the most we can say is $k=\pi^2j^2+4$ for $j\in \mathbb Z$.
