Equilibrium solutions of a differential equation and their nature

Question

I need some help with the following the differential equation $$ \dot{x}=x^{4}-3x^{3}+2 $$ I found that the equilibrium solutions are $x \approx 1, 2.9196, -0.45982+-0.68817i$. If you someone could check that I did it correctly and also if these solutions are stable or unstable, thanks!

Answer 1

The equation $$ \dot x=x^4-3x^3+2 $$ is an autonomous equation, and an equilibrium solution is a constant solution: $x(t)=c$ .

Clearly we have such a solution if $\frac{dx}{dt}=\dot x=0$, so the equilibrium solutions are the roots of the equation $$ x^4-3x^3+2=0 $$ The roots find in OP are correct (see here) and the $x-$ axis of the graph in this link can be used as a phase line for the differential equation. So we see that $x=1$ is a stable solution and $x \approx 2.9$ is unstable.