Stability of a feedback system Take the following feedback system:
$\dot{x} = (\theta - k_1) x - k_2  x^3$
Now my book says:
For $\theta > k_1$, the equilibrium $x = 0$ is unstable. I wonder why...
Furthermore my book indicates that it is easy to see that $x(t)$ will converge to one of the two new equilibria $\pm \sqrt{\frac{\theta-k_1}{k_2}}$. Again, how did they obtain this result?

Suppose $\theta - k_1 = k_3 > 0$.
Then $\dot{x} = k_3 x - k_2  x^3.$
If $k_2 \le 0, \dot{x}$ has the same sign as $x$, so $x(t)$ will tend to go further away from 0.
Even if $k_2 > 0,$ for sufficiently small $x, |k_3 x| > |k_2 x^3|,$ so $\dot{x}$ will have the same sign as $x$, so $x(t)$ will tend to go further away from 0.
I.e. 0 is unstable.
You can get the result of where the equilibria are by solving $$\begin{align}
0 = \dot{x} & = (\theta - k_1) x - k_2  x^3 \\
& = x(\theta - k_1  - k_2  x^2) \\
& = x(\sqrt{\theta-k_1} + \sqrt{k_2} x)(\sqrt{\theta-k_1} - \sqrt{k_2} x).
\end{align}
$$
So the graph of $\dot{x}$ against $x$ is an (anti-symmetric ) cubic. If $k_2 > 0$ the graph slopes down to $-\infty$ as $x \rightarrow + \infty$.  I.e. for the equilibrium at 
$\sqrt{\frac{\theta-k_1}{k_2}}$ the graph is negative for $x$ > equilibrium (leading $x(t)$ down to the equilibrium) and positive for $x$ < equilibrium (leading $x(t)$ up to the equilibrium).  I.e. the equilibrium at $\sqrt{\frac{\theta-k_1}{k_2}}$ is stable.
