I was wondering how we would show that the system: $$\frac{dx}{dt}=-x^3+2x-4y \\ \frac{dy}{dt}=-y^3+2y+4x$$ has only one equilibrium point.
I have seen cases where the system is, for example:
$$\frac{dx}{dt}=x(2x-y)-x \\ \frac{dy}{dt}=2y(1-x)+y$$
which I understand more how to calculate the equilibria, as you can simplify it better. But in the case above I am stuck.
I am also unsure how to show it has a periodic solution. Any help would be appreciated!
Equilibrium point is where both $x'(t)$ and $y'(t)$ are equal to zero. $4y=2x-x^3$, $64y^3-128y-256x=0\Rightarrow$ $(2x-x^3)^3-32(2x-x^3)-256x=0\Rightarrow$ $x(x^8-6x^6+12x^4-40x^2+2720)=0$. Let $x^2=t> 0$, then $x^8-6x^6+12x^4-40x^2+2720=$ $t^4-6t^3+12t^2-40t+2720=$ $(t^2-3t)^2+3(t-\frac{20}{3})^2+\frac{7760}{3} > 0$. Then $x=0$ is the only equilibrium point. $4y=2x-x^3=0 \Rightarrow y=0$.