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?

Starting with:
$$
\dot{x} = (\theta - k_{1})x - k_{2}x^{3} = V(x).
$$
The equilibrium occurs at $\dot{x}=0$ i.e.
$$
(\theta - k_{1})x - k_{2}x^{2} = 0 = x\left[(\theta - k_{1}) - k_{2}x^{2}\right]
$$
so clearly the stationary points occur,
$
x_{0} = 0
$
and when 
$$
(\theta - k_{1}) - k_{2}x^{2} = 0
$$
To compute the other fixed point we utilise $x^{2} =\frac{(\theta - k_{1})}{k_{2}}$ which you can probably carry from here with. 
To prove the stability criteria of the points i.e. the nature of the points you linearise about the stationary points $x_{0}$'s using $x = x_{0} + \delta x$, and inserting into your original ode as follows,
$$
\dot{\delta x} = \frac{\partial V}{\partial x}
$$
where the last term is evaluated at the fixed points. This leads to
$$
\dot{\delta x} = (\theta - k_{1}) + 3k_{2}x_{0}^{2}\delta x
$$
Solving this for $\delta x$ will show how perturbations of the fixed points behave. 
