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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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