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Just a quick question I have the following system: $$\frac{dx}{dt}=x\left(2-x-\frac{y}{1+x}\right),\qquad \frac{dy}{dt}=3y\left(\frac{x}{1+x}-4y\right)$$ It asks which variable represents the predator and which the prey, write down the long term size of the prey in the absence of predators.

I approached this by considering that $\frac{dy}{dx}= \frac{y(3x-12y-12yx)}{x(2+x-x^{2}-y)}$, hence when y=o we have that $\frac{dy}{dx}$=0 and therefore we have that our predator is represented by $\frac{dx}{dt}$ and prey by $\frac{dy}{dt}$. Additionally in the absence of prey our prey will tend to $\infty$.

Many thanks in advance.

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Made some corrections to your post. I hope this is indeed the system you are interested in. – Did Oct 20 '12 at 13:49
up vote 1 down vote accepted

If $x(0)=0$ and $y(0)\gt0$, then $x(t)=0$ for every $t$ and $y(t)\to0$ when $t\to\infty$. If $x(0)\gt0$ and $y(0)=0$, then $y(t)=0$ for every $t$ and $x(t)\to2$ when $t\to\infty$. Hence $x(t)$ describes the amount of preys (which can survive without predators) and $y(t)$ describes the amount of predators (which die out without preys).

To show that $x(t)\to2$ when $t\to\infty$ when $y(0)=0$, one can use a phase diagram: since $x'(t)=x(t)(2-x(t))$, the function $t\mapsto x(t)$ decreases at times $t$ such that $x(t)\gt2$ and increases at times $t$ such that $0\lt x(t)\lt2$.

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Note that the amount of preys does not grow to infinity in the absence of predators, rather it converges to 2. – Did Oct 20 '12 at 13:54

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