How do I find the equilibria of this autonomous system of ODEs?

Question

I have the following system of ODEs: $$x'(t)=x(x^3-2y^3)$$ $$y'(t)=y(2x^3-y^3)$$ and I am asked to find the equilibria. I've equated both to zero and solved, but this just gives me one answer of $(0,0)$ and I'm not sure how else to find them - every example I can find uses much simpler equations that give you the answers by just equating to zero. Can anyone suggest methods?

Answer 1

Let $u(x,y)=x(x^3-2y^3)$ and $v(x,y)=y(2x^3-y^3)$.

• If $x=0$ and $v(x,y)=0$ then $-y^4=0$ hence $y=0$.
• If $y=0$ and $u(x,y)=0$ then $x^4=0$ hence $x=0$.
• If $x\ne0$, $y\ne0$ and $u(x,y)=v(x,y)=0$ then $x^3=2y^3$ and $y^3=2x^3$ hence $x^3=4x^3$, thus $x=0$ and $y=0$.

Finally, the only equilibrium is at $(x,y)=(0,0)$.