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I'm suppose to find equilibrium solutions that satisfy the following system of differential equations:

$$\frac{dx}{dt} = -3x+2y$$ $$\frac{dy}{dt} = 6x+y$$

This may be a simple question, but I feel like my instructor didn't explain how to go about this. All I know, if I'm even correct, is that we need to make sure that $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} = 0$. I am not interested in a linear algebra approach for this problem. I'm just looking for a simple explanation as to how to approach problems similar to the one I asked.

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As you wrote, you need $\frac{dx}{dt}$ and $\frac{dy}{dt}$ to both be zero. That means the right sides of the differential equations must be $0$ as well. So you have a system of two equations in two unknowns to solve: $$ \eqalign{-3x+2y &= 0\cr 6x+y &= 0\cr}$$

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  • $\begingroup$ So $x=0 , y=0$ is our one and only equilibrium solution then. $\endgroup$
    – Kyogre
    Nov 28, 2017 at 19:14
  • $\begingroup$ Yes, that's right. $\endgroup$
    – Robert Israel
    Nov 28, 2017 at 23:09

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