The question is about the system of equations

$$\frac{dx}{dt} = y - 1, \frac{dy}{dt} = -xy$$

and I'm trying to find the periodic solutions by finding closed orbits. I'm not sure if orbit is the correct term, but I mean the following: if $\gamma: t \mapsto (x(t), y(t))$ is a solution of the system with initial values $x(t_0) = x_0, y(t_0) = y_0$ for $t \in I$ for some interval $I$, then the orbit is $\gamma[I]$. The solution is given globally (because $x', y'$ are Lipschitz continious in the second variable) by solving $\frac{dy}{dx} = \frac{-xy}{y - 1}$, which gives $x = \pm \sqrt{\log|y| - y + C}$ where $C = y_0 - \log|y_0| + \frac{1}{2}x_0^2$ if $y(t) \neq 1$ $\forall t$. I have difficulty finding out when this is closed, because when $y \to -\infty$, $x \to \infty$ so it can't be closed and there are also some problems when $y(t) = 1$. Can I just say $y > 1$?

The stationary points are $(0, 1)$ because that is the solution of $y - 1 = 0$ and $-xy = 0$.

  • $\begingroup$ Can $y,x$ be complex? $\endgroup$ – Yuriy S Feb 10 '18 at 14:11
  • $\begingroup$ @YuriyS they must be real $\endgroup$ – Pel de Pinda Feb 10 '18 at 14:42

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