Showing an equilibrium point with an inequality

**FoxyGrandma3000** · March 4th 2010, 02:30 PM

Given $\frac{dy}{dt}=r(1-\frac{y}{k})y-Ey$

if $E<r$ show that there are two equilibrium points given by $y_{1}=0$ and $y_{2}=k(1-\frac{E}{r})>0$

The demonstration of $y_{1}$ is plain to see, but how do you go about proving the equilibrium point at $y_{2}$ ?

**Danny** · March 4th 2010, 03:43 PM

Originally Posted by FoxyGrandma3000

Given $\frac{dy}{dt}=r(1-\frac{y}{k})y-Ey$

if $E<r$ show that there are two equilibrium points given by $y_{1}=0$ and $y_{2}=k(1-\frac{E}{r})>0$

The demonstration of $y_{1}$ is plain to see, but how do you go about proving the equilibrium point at $y_{2}$ ?

I believe that you'll have two distinct eq. points provided that $E \ne r$ . From your ODE

$<br /> \frac{dy}{dx} = y\left(r - \frac{ry}{k} - E \right) <br />$

$y = 0$ is one, the other is found by setting $r - \frac{ry}{k} - E =$ 0 and solving for $y$ . If you want the second eq. point to be positive then you would need $E < r$ .

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