I'm new to differential equations as well as to controls so I'd like to show my thought process for a problem.
Let
$$ \begin{align} \dot x &= z \\ \dot z &= -\frac{d}{dx}V_\alpha(x) \end{align} $$ where $V_\alpha(x) = x^4 - \alpha x^2$
We want to show that $(z,x) = (0,0)$ is an equilibrium point or all $-1 \leq \alpha \leq 1$
I started with substituting in our equation and evaluating what $\dot z$ is.
$$ \begin{align} -\frac{d}{dx} (x^4 - \alpha x^2) &= -(4x^3 - 2\alpha x) \\ &= -2(2x^2 - \alpha x) \end{align} $$
I know that I'm supposed to show that at $(z,x)=(0,0)$ I'm supposed to see that when $\alpha$ is on the interval $[-1,1]$ that I have a real component but $z$ isn't in the equation...
Another aspect that I'm familiar with but still not comfortable is that I can put the equations in matrix form but this doesn't make sense to me in this context since it's just one variable.
I know I'm probably pretty far off the mark but some guidance would be well appreciated.
For equilibrium all derivatives must be zero. So you have to find $x$, $z$ such that
$$ \begin{align} \dot{z} &= 0 \\ \dot{x} &= 0 \end{align} $$
You have the system
$$ \begin{align} \dot{z} &= z \\ \dot{x} &= - \frac{d}{d x} V_{\alpha}(x) \end{align} $$
and $V_{\alpha}(x) = x^4 - \alpha x^2$. Your derivative is wrong, because $\frac{d}{d x} V_{\alpha}(x) = 4 x^3 - 2 \alpha x$ and so your system is
$$ \begin{align} \dot{z} &= z \\ \dot{x} &= -4 x^3 + 2 \alpha x \end{align} $$
First look at the first equation: You get zero derivative only when
$$ z = 0 \tag{1} $$
Now look at the second equation: You want $-4 x^3 + 2 \alpha x = 0$. Factor out $x$ you get:
$$ x(\alpha - 2 x^2) = 0 \tag{2} $$
This is true if $x = 0$ for all $\alpha$ so you have a equilibrium at $(z,x)=(0,0)$.
For this equilibrium the value of $\alpha$ does not matter. However you have two more equilibrium points, one at $(z,x) = (0,\frac{\sqrt{2}\sqrt{\alpha}}{2})$ and one at $(z,x) = (0,-\frac{\sqrt{2}\sqrt{\alpha}}{2})$ because there $(1)$ and $(2)$ are also zero.
Some more notes: