I have the following ODE:

$\displaystyle X''(t) = (1 - \frac{A^}{2w^2})X(t)$

Where A,w are constants strictly greater than 1.

I am asked to find the solution of the ODE and derive the criterion of stability of the point X = 0 in terms of the parameters A and w.
I can do the solving, I just don't understand what the whole stability thing is asking. I thought stability occured at fixed points, ie: where X'=0 not X=0. What am I supposed to do?

Here's what I've done so far:

To find the solution I let $\displaystyle 1 - \frac{A^}{2w^2} = b$ and then solve the equation$\displaystyle X''-bX=0$
This yields an auxiliary equation of $\displaystyle k^2-b=0$ with solutions $\displaystyle k=\pm \sqrt{b}$

Case One:
b>0 then we have solution
$\displaystyle X(t)=\alpha e^{t \sqrt{b}}+\beta e^{t \sqrt{b}}$

Case Two:
b=0 then
$\displaystyle X(t)=\alpha +t \beta$

Case Three:
b<0 then
$\displaystyle X(t)=\alpha cos(t \sqrt{|b|}) + \beta sin(t \sqrt{|b|}) $

(where alpha and beta are constants in all cases)

How do I actually derive this supposed criterion of stability for X=0? Any help would be greatly appreciated, I need to touch up on my differential equations, it's been a reasonably long time since I last studied them.