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I am given the following two equations as models for male and female population

$\dfrac {df} {dt}$ = $\dfrac {mf} {m+f}$$B_f-fD_f$

$\dfrac {dm} {dt}$ = $\dfrac {mf} {m+f}$$B_m-mD_m$

assume $D_m$=$D_f$=$D$

The equilibrium points two the previous two equations form a line in the $f-m$ plane and I am trying to find the equilibrium ratio of $m/f$

I have tried moving everything to one side for both of them and then integrating and solving for $m$ and $f$ respectively but it seems to just get more complicated

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2 Answers 2

up vote 1 down vote accepted

(For simplicity I will assume that $B_m\neq 0 \neq B_f$).


An equilibrium point must satisfy $\dfrac {df} {dt}=0$ and $\dfrac {dm} {dt}=0$, i.e.,

$\dfrac {mf} {m+f}$$B_f$ = $fD$

$\dfrac {mf} {m+f}$$B_m$ = $mD$

Notice that this is trivially satisfied by $m=f=0$ meaning the origin is always an equilibrium point. Moreover, if $D=0$ then it is necessary that either $m=0$ or $f=0$. In other words, the $m$ and $f$ axes are the set of equilibrium points.

Now the only case left to worry about is when $D\neq 0$. Since we already know that the origin is an equilibrium point, let us ignore it for now and we'll be able to divide the second equation by the first and yield,

$\dfrac{m}{f}=\dfrac{B_m}{B_f}$

which is the only other family of equilibrium points.

So in summary,

  • If $D=0$, then the axes are the equilibrium points.
  • If $D\neq 0$, then the line $f=\dfrac{B_f}{B_m}m$ are the equilibrium points.
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Equilibrium occurs when $\dfrac{\mathrm{d}f}{\mathrm{d}t}=\dfrac{\mathrm{d}m}{\mathrm{d}t}=0$.

$\dfrac{f}{B_f}=\dfrac{m}{B_m}=\dfrac{mf}{D(m+f)}\\ \dfrac{m}{f}=\dfrac{B_m}{B_f}$

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