Sketch a phase portrait for the non-linear system $\dot x =(1+x-2y)x$, $\dot y = (x-1)y$

Question

Example 3.10 Sketch a phase portrait for the non-linear system \begin{aligned} \dot x &=(1+x-2y)x \\ \dot y &= (x-1)y \end{aligned}

To do this we need to identity the critical points which occurs when $\dot x = \dot y =0$. The answers says that this occurs at when $(x,y) =(0,0)$, $(1,1)$, $(-1,0)$. What method do you use to find these points?

Answer 1

We have

$$x(1+x-2y) = 0 \\ y(x-1) =0$$

From the second equation, we see that $y = 0$ and $x = 1$ are two possible solutions.

Now substitute "each" of those in the first and solve for the remaining variables.

You get the critical points

$$(x, y) = (-1, 0), (0, 0) , (1, 1)$$

Using these, you should get a phase portrait