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I am trying to find the equilibrium values of the system: \begin{align} \dot x &= y-3 \\ \dot y &= (x-1)(y-2) \end{align} I know you set each equation equal to 0, but in the first equation, there are no $x$'s and in the second equation I only found $y = 2$ for one equilibrium value. I am confused on what I should do next. For the phase diagram, would it look like this?: enter image description here

Advent21
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    Equilibrium points are found by setting $\dot x = \dot y = 0$. The first equation requires $y = 3$, but the second can have either $x = 1$ or $y = 2$, hence the only equilibrium value is when both can be satisfied, hence $(1,3)$ is the only value. – Gregory Apr 09 '20 at 01:44
  • @Gregory How would you model this in phase space then? Because I thought phase spaces where only used for either x or y not both values at the same time. – Advent21 Apr 09 '20 at 02:04
  • Your edited post shows one way to look at the phase space, but you did not include the (1,3) point, try expanding the $(x,y)$ range to include this point and you will see the behavior near the point. – Gregory Apr 09 '20 at 03:30
  • As per Gregory's second post: https://www.wolframalpha.com/input/?i=StreamPlot%5B%7By-3%2C%28x-1%29%28y-2%29%7D%2C%7Bx%2C0%2C2%7D%2C%7By%2C2%2C4%7D%5D – Adrian Keister Apr 09 '20 at 16:16

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