solving coupled equation

Question

$$\begin{aligned}\dot{x}&=x\left ( 3-x-2y \right )\\ \dot{y}&=y\left ( 2-x-y \right )\end{aligned}$$

The above is a coupled equation.

The fixed point condition requires all $\;x,y\;$ for which $x^{\ast} =0\;$ and $\; y^{\ast}=0.$
Solving, I arrive at $\left ( x^*=0,y^*=0 \right )\;$ and $\;\left ( x^*=1,y^*=1 \right ).$
But there are $2$ other solutions: $\left ( x^*=0,y^*=2 \right )\;$ and $\;\left ( x^*=3,y^*=0 \right ).$

I am unable to determine the last $2$ ordered pairs of fixed points. Any help would be good.

Answer 1

Analytical solution for the third and fourth fixed points will be a bit tricky (messy and cumbersome). However, you can use substitution in any one of the equations (replace $y$ in $\dot x$ or $x$ in $\dot y$) and solve the quadratic equation to get your other two fixed points.

Answer 2

Consider first solving the $\dot{x}$ equation. You should conclude that either $x=0$ or $x=3-2y$.

Now solve the $\dot{y}$ equation in each of these cases i.e. first suppose $x=0$, then substitute this into the $\dot{y}$ equation to get $0=y(2-y)$, so this is satisfied when $y=0$ and when $y=2$.

So you will have found the points $(0,0)$ and $(0,2)$. Repeat this process for the other relation found on $x$ (i.e. $x=3-2y$) to get the other points.

Answer 3

make the equations quadratic:

$y^2+(x-2)y=0$

solving for this we get $y=0,-(x-2)$ so when you substitute $y=0$ in the above equation you will get $x=3~x=0$, now when you substitute $x=0$ in the second equation you get $y=2,0$ when you solve for the equation on the whole, you get $(1,1),~(0,0)$ as the fixed points. Now from the above obtained fixed point pairs you can see that the fixed points are $(0,2),~(3,0)$.