Full problem:
Consider the predator-prey model with $\dot{x} = x(1-y)$ and $\dot{y} = (2x-1)y$. Show that every non-constant positive solution is periodic.
My Answer:
First off we can see here that we have two equilibrium points, $E_1 = (0,0)$, and $E_2 = \left( \frac{1}{2},1 \right)$. Since we only care about the non constant positive solutions, we will focus our efforts on $E_2$.
$$Df\left(x , y \right) = \left( \begin{matrix} 1-y & -x \\ 2y & 2x-1 \end{matrix}\right)$$
So,
$$Df\left( \frac{1}{2}, 1 \right) = \left( \begin{matrix} 0 & -\frac{1}{2} \\ 2 & 0\end{matrix}\right) \implies \lambda_{1,2} = \pm i$$
Thus, since we get complex eigenvalues with a zero real part, this implies that we have a center at the point $\left(\frac{1}{2},1 \right)$. Thus, since there are no more positive equilibrium points, all positive nonconstant solutions of the system must be periodic.
I think this is correct, but I'm not sure if there's more I need to say since we are wanting to show this for ALL non-constant positive solutions. Thanks in advance for any advice.
Best Answer
No, the fact that there are no more positive equilibrium points does not show the solutions are periodic.
Hint: find a function $F(x,y)$ that is constant on any solution.
Further hint: the differential equation $\dfrac{dy}{dx} = \dfrac{(2x-1)y}{x(1-y)}$ is separable.