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For $a \in \mathbb{R}$ consider $$\begin{cases} y_1'(x) = (ay_1(x) - y_2(x))(1 + y_3(x))\\ y_2'(x) = (-y_1(x) + ay_2(x))(1 + y_1(x))\\ y_3'(x) = ay_3(x) \end{cases}$$ Find all constant solutions.

The case $a = 0$ is fairly easy. So let us consider $a \neq 0$. From the third equation we get $y_3 = 0$. So we have to consider $$ay_1 = y_2 \qquad \text{and} \qquad (-y_1 + ay_2)(1 + y_1) = 0$$ The second equation yields $y_1 = -1$ and so by the first one $y_2 = -a$. Hence a constant solution would be $$(-1,-a,0)$$ Now I would go on with $$ay_2 = y_1$$ Together with above equation, this yields $a = \pm1$. Somehow the solution says, that the case $a = \pm 1$ is degenerate. What does this mean for the constant solutions? Are there more in the case $a \neq 0$?

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    @Moo What should I solve for? The last equation just yields $0 = ay_3$. You mean to solve the ODE $y_3' (x) = ay_3(x)$? – 2017-01-08
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    @Moo Yes, this is easy. But I doubt that solving the resulting system with $y_3$ plugged in is that easy to solve. Maybe some background: this is an exam question and you do not have the time to solve this. Finding the constant solutions is usually much easier than solving the full system. – 2017-01-08
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    The explanation is the following. First, you've overlooked one of the solutions when you were solving $a^2 y_2 = y_2$. There is also a solution $y_2 = 0$, although it will lead to a constant solution $(0, 0, 0)$. What is so degenerate about case $a = \pm 1$? Well, if you look carefully, when $a = \pm 1$ you have a continuum of equilibria with coordinates like $(\gamma, a \gamma, 0)$ where $\gamma$ can take any real value. It is typical to have isolated equilibria, but not the continuum like this, so that's why this situation is degenerate. – 2017-01-09
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    @Evgeny Ah..very nice. Thank you. – 2017-01-09

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