Annette Arroyo

2021-08-29

Find all of the equilibrium points for two systems. Explain the significance of these points in terms of the predator and prey populations

(i)

(ii)

Aamina Herring

For equilibrium point, $\frac{dx}{dt}=0$ and $\frac{dy}{dt}=0$

For system (i):
$\frac{dx}{dt}=0$

$10x\left(1-\frac{x}{10}\right)-20xy=0$

$x\left(10-x-20y\right)=0$ .....(1)
$\frac{dy}{dt}=0$

$-5y+\frac{xy}{20}=0$

$y\left(-5+\frac{x}{20}\right)=0$......(2)

Solution of the equations (1) and (2) will have 3 cases.
Case 1:

$x=0,y=0$

This will represent 0 prey and 0 predator.

Case 2:

$-5+\frac{x}{20}=0$

$x=100$

$10-x-20y=0$

$10-100-20y=0$

$y=-4.5$

As y cannot be negative, the solution set is rejected.

Case 3:

when $y=0$

$10-x-20\left(0\right)=0$

$10-x=0$

$x=10$

For system (ii):

$\frac{dx}{dt}=0$

$0.3x-\frac{xy}{100}=0$

$x\left(0.3-\frac{y}{100}\right)=0$ .....(3)

$\frac{dy}{dt}=0$

$15y\left(1-\frac{y}{15}\right)+25xy=0$

$y\left(15-y+25x\right)=0$ ...(4)

Solution of the equations (3) and (4) will have 3 cases.

Case 1:

$x=0,y=0$

This will represent 0 prey and 0 predator.

Case 2:

$0.3-\frac{y}{100}=0$

$y=30$

$15-y+25x=0$

$15-30+25x=0$

$x=\frac{3}{5}$

So, $\left(\frac{3}{5},30\right)$ will represent a stable population.

Case 3:

when $y=0$

$15-\left(0\right)+25x=0$

$x=-\frac{3}{5}$

As x cannot be negative, the solution set is rejected.

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