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I am looking at the equilibrium points of the following system:

$$ X' = a_1 X - a_2 X^2 -a_3 XY \\ Y' = \beta a_3YX-a_4Y-a_5Y^2 $$

I found the following $(X,Y)$ equilibrium points

$(0,0); \ (\frac {a_1}{a_2},0 ); \ (0,-\frac{a_4}{a_5})$ but I can not find the 4th equilibrium point when both $X\ne 0$ and $Y \ne 0$. I get lost in computation and cant find the solution.

Bernard
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Frostic
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1 Answers1

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From $0 = a_1 X - a_2 X^2 -a_3 XY \\ 0 = \beta a_3XY-a_4Y-a_5Y^2$

we get $a_3XY=a_1X-a_2X^2$. Use the second equation to derive

$0=X(\beta a_1- \beta a_2 X)-Y(a_4+a_5Y)$

Now it is easy to see that $(\frac {a_1}{a_2},- \frac {a_4}{a_5})$ is an equilibrium point.

Fred
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