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Find the fixed points of the following dynamical system\begin{align}\frac{dx}{dt}&= (a_1 -b_1x - c_1y)x \\ \frac{dy}{dt} &= (-a_2 +c_2x)y\end{align} Note that ALL the parameters are positive.

I have found the following fixed points so far

\begin{align*}(x,y) &= (0,0) \\ & \text{and} \\ (x,y) &= \bigg(\frac{a_1}{b_1},0\bigg)\end{align*}

Are there any other points that I am missing?

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Observe that if $(x,y)$ is a fixed point then $$\frac{dx}{dt}=0\qquad\text{and}\qquad\frac{dy}{dt}=0$$

We have $$\frac{dy}{dt}=0 \implies -a_2+c_2x=0 \;\text{ or }\;y=0$$ I see you have worked the case when $y=0$. The other case, $-a_2+c_2x=0$ implies $x=a_2/c_2$, and $$\frac{dx}{dt}=0\iff \left(a_1-\frac{b_1a_2}{c_2}-c_1y\right)\frac{a_2}{c_2}=0$$ Since $a_2/c_2>0$, the last equation imply $$a_1-\frac{b_1a_2}{c_2}-c_1y=0\iff y=\frac{a_1c_2-a_2b_1}{c_1c_2}$$


There is another one fixed point:

\begin{align*} x&=\frac{a_2}{c_2}\quad\text{and}\quad y=\frac{a_1c_2-a_2b_1}{c_1c_2} \end{align*}

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  • $\begingroup$ Thank you! :). Can you please show me how you got this one? :). $\endgroup$ – user860374 Aug 16 '15 at 18:39

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