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So, I'm supposed to find all the equilibrium points of the following 2D system (and a couple others, but I think just one will give me the hang of it). I'm also supposed to determine the stability. I'm not sure how to do this. Help me!

$\frac{dx}{dt} = -axy$

$\frac{dy}{dt} = axy-by$

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    Where are you stuck? Are you familiar with the relation between stability and differentiation?2011-07-29
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    Well, I thought we just made the equations equal zero to find equilibria, but then I just get y is zero and x could be anything. This seems wrong2011-07-29
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    @rapidash: you're right. There is a set of equilibrium points which is the line $y=0$. For each $x\in\mathbb R$ you should check now if the point $(x,0)$ is a stable equilibrium.2011-07-29
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    I have this $$a_{11}= 0, a_{12}=-ax, a_{21}= 0, a_{22} = ax - b$$ $$p = a_{11} + a_{22} = ax - b$$ $$q = a_{11} a_{22} - a_{12} a_{21} = 0$$ So if $x > \dfrac{b}{a}$, $p>0$, stable center If $x = \dfrac{b}{a}$, $p=0$, stable node? If $x < \dfrac{b}{a}$, $p<0$, unstable saddle point2011-07-29
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    @rapidash: almost. Just check your signs and note that $a$ can vanish as well.2011-07-29
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    @glebovg: I killed the stability and linearisation tags because the terminology is too broad. Both are used in many contexts other than the dynamical systems context that you suggested the tag-wiki for. In any case, those are standard topics under [tag:dynamical-systems] and that tag should be used instead.2013-08-05

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The classification of equilibria is intended for isolated equilibria. Since we have a line of non-isolated equilibria, the classification does not fully apply. We must proceed with caution...

Seeing that the trace of the Jacobian matrix is $ax-b$, we can conclude that the equilibria are stable when $xb/a$. There is no conclusion at $x=b/a$, because adding a nonlinear term to a center can make it an unstable equilibrium.

Fortunately, for this system we can find the equations of trajectories very easily. Divide the second equation by the first to get $$\frac{dy}{dx}=-1+\frac{b}{ax}\tag1$$ Integration yields $$y=-x+\frac{b}{a}\ln |x| +C \tag2$$ as equation of trajectories. One has to be careful in interpreting (2), because trajectories do not actually cross $y=0$: they take infinite time approaching it. If $C$ is such that the curve given by (2) crosses $y=0$, that curve is the union of three trajectories: one above the axis, and two below it.

Consider the case $b/a>0$, when the questionable point falls into the halfplane $x>0$. All trajectories (2) are vertical translates of the same curve, which is concave with maximum at $x=b/a$. Taking $C$ such that the curve is just barely below the $y=0$ line, we conclude that the equilibrium at $x=b/a$ is unstable: there are unbounded trajectories that begin arbitrarily close to it. Here's a picture drawn with Sage for the case $a=b=1$.

orbits

When $b/a<0$, the conclusion is the same: instability at $x=b/a$, but now the relevant unbounded trajectories lie above the line $y=0$.

When $b=0$, we still have instability at $x=0$, which is easier to see since the system becomes essentially one-dimensional ($x+y$ is constant).

All in all, this is pretty hard homework for a typical ODE class.