<s>[INST] Finding Equilibrium Points of Dynamical System I am wondering how to find the equilibrium points from these $2$ equations
\begin{align}
\frac{\mathrm{d}x}{\mathrm{d}t} & =
0.4x\left(1 - \frac{x}{1000} + \frac{y}{1000}\right)
\\[2mm]
\frac{\mathrm{d}y}{\mathrm{d}t} & =
0.1y\left(1 + \frac{x}{1000} - \frac{y}{500}\right)
\end{align}

*

*From the $2$nd equation, the $y$ value I got is $y = 0$ and
$x = -1,000$

*and from the $1$st equation I got $x = 0$ and
$y = -1,000$
So I am wondering is it use the $1$st equation to get the
$\left(x,y\right)$ value or use the $2$nd equation to get the $\left(x,y\right)$ value ?
 [/INST] We want to simultaneously find the zeros of
$$0.4x\left(1-\frac{x}{1000}+\frac{y}{1000}\right) = 0\\ 0.1y\left(1+\frac{x}{1000}-\frac{y}{500}\right) = 0$$
We can see that each equation is of the form $a b = 0$, so each of those being zero, will satisfy the equation and we just need for them to do that simultaneously for both equations.
We can set $x = 0$ in the first equation and substitute that in the second equation
$$0.1y\left(1+\frac{0}{1000}-\frac{y}{500}\right) = 0$$
From this we get two $y$ values of
$$y = 0, 500$$
We can set $y = 0$ in the second equation and substitute that in the first equation
$$0.1x\left(1-\frac{x}{1000}+\frac{0}{1000}\right) = 0$$
From this we get two $x$ values of
$$x = 0, 1000$$
So that takes care of the $x$ and $y$ terms by themselves. Now, ignoring that term, we have
$$0.4\left(1-\frac{x}{1000}+\frac{y}{1000}\right) = 0\\ 0.1\left(1+\frac{x}{1000}-\frac{y}{500}\right) = 0$$
Solving the first for $y$
$$y = x - 1000$$
Note: We could have solved for $x$ or also used the 2nd equation - try it.
Substituting that into the second
$$\frac{1}{10} \left(-\frac{1}{500} (x-1000)+\frac{x}{1000}+1\right)=0$$
Solving this gives $x = 3000$, which means $y = 2000$.
So, our solution set is
$$(x, y) = (0,0), (0,500), (1000, 0), (3000, 2000)$$
We can use a contour plot to see these points of intersection

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