Limit cycle within $\frac{1}{4}<r<1$

Question

Show that the system \begin{align} x'&=-y+x(1-2x^2-3y^2)\nonumber\\ y'&=x+y(1-2x^2-3y^2)\nonumber \end{align} has a limit cycle in $\frac{1}{4}<r<1$.

Here's what I've done so far: \begin{align} xx'&=-xy+x^2(1-2x^2-3y^2)\nonumber\\ yy'&=xy+y^2(1-2x^2-3y^2)\nonumber \end{align}

\begin{align} xx'+yy'&=-xy+x^2(1-2x^2-3y^2)+xy+y^2(1-2x^2-3y^2)\nonumber\\ \frac{1}{2}(x^2+y^2)'&=x^2(1-2x^2-3y^2)+y^2(1-2x^2-3y^2) \nonumber\\ \frac{1}{2}(x^2+y^2)'&=(x^2+y^2)(1-2x^2-3y^2) \nonumber \end{align} Knowing that $r^2=x^2+y^2$: \begin{align} \frac{1}{2}(x^2+y^2)'&=(x^2+y^2)(1-2x^2-3y^2) \nonumber\\ \frac{1}{2}(r^2)'&=(r^2)(1-2x^2-3y^2) \nonumber \end{align}

I'm now stuck here and I'm not really sure what to do next because of this part $(1-2x^2-3y^2)$.

Answer 1

Noting that on $r=1$, $$ \frac{1}{2}(r^2)'=(r^2)(1-2x^2-3y^2)=r^2(1-2r^2-y^2)<0 $$ and on $r=\frac14$, $$ \frac{1}{2}(r^2)'=(r^2)(1-2x^2-3y^2)=r^2(1-3r^2+y^2)>0 $$ by the Poincare-Bendixson Theorem, there is a closed trajectory lying in $\frac14<r<1$.