Help using an energy function to determine stability of an ODE I'm struggling to understand applications of "energy functions" for investigating ODEs. I'm working on the follow problem:

Investigate stability of the equilibrium points of the system $$x^{\prime\prime} = (x-a)(x^2-a)$$ for all real parameters $a$.
Hint: Rewrite the equation as $x^{\prime\prime} = -V^{\prime}(x)$ for a suitable potential energy function $V$ and then use your intuition about particles moving in potentials.

So putting aside the fact that I don't have much intuition about particles moving in potentials... Here's what I've done so far:
First, we can split it up into a system of two variables on the plane:
$$x^\prime = y$$
$$y^\prime = (x-a)(x^2-a)$$
which obviously has its only fixed points at $(a,0), (\sqrt{a},0), (-\sqrt{a},0)$
I've created the energy function $$E = \frac{y^2}{2}-\frac{x^4}{4}+\frac{ax^3}{3}+\frac{ax^2}{2}-ax$$
I know that since I've found an energy function, this is a conservative system, and so it can only have saddles and centres for fixed points. I think that I can also say something about level curves of the energy function, but I'm not sure.
Generally in this sort of situation I would linearize the system and analyze stability that way, but I imagine the fact that the system is conservative has some special bearing on the question. Can someone explain what to do next with the energy function?
 A: Following the hint, $V$ must be an antiderivative of $-(x-a)(x^2-a)$, thus (taking the arbitrary constant to be $0$)
$$ V =  - \dfrac{x^4}{4} -\dfrac{a x^3}{3} + \dfrac{ x^2}{2} - a^2 x $$
The total energy (kinetic + potential) is conserved:
$$ E = \frac{y^2}{2} + V = \frac{y^2}{2} - \dfrac{x^4}{4} -\dfrac{a x^3}{3} + \dfrac{ x^2}{2} - a^2 x$$
The equilibrium points have $y=0$ and $(x-a)(x^2-a)=0$, so either $x=a$ or (assuming $a > 0$) $x = \pm \sqrt{a}$.
The intuition: think of a particle sliding along the  graph of $V$ under the influence of gravity.  The equilibria correspond to places where the slope of $V$ is $0$.  If there is a direction you can go from the equilibrium where $V$ decreases, that equilibrium is unstable: starting at rest with an arbitrarily small displacement from the equilibrium, you accelerate away from the equilibrium until you escape some neighbourhood of the equilibrium.
If $V$ increases in both directions from the equilibrium, then the equilibrium is stable: if you start near the equilibrium with velocity near $0$, you don't have enough energy to escape a neighbourhood of the equilibrium.
The next step: consider the second derivative of $V$ at an equilibrium.
