Answer:
a) Prey: [tex]\frac{dy}{dt}=y(1-y)-\frac{xy}{2+y}[/tex]
Predator: [tex]\frac{dx}{dt}=\frac{xy}{2+y}-\frac{x}{2}[/tex]
b) they grow as a function y(1-y).
c) It will grow faster and probably exponentially,
Step-by-step explanation:
a)
Prey
For prey population the rate of change is given by its own growth rate, it would be a constant times the variable, minus the rate at which it is preyed upon, usually it is represent a parameter times x and y.
So the differential equation could be:
[tex]\frac{dy}{dt}=y(1-y)-\frac{xy}{2+y}[/tex]
Predator
By the other hand, the rate of change of the predator's population depends only the rate at which the predators eat preys, minus predator's intrinsic death rate.
[tex]\frac{dx}{dt}=\frac{xy}{2+y}-\frac{x}{2}[/tex]
b) If there are no predators present the differential equation that models prey population will be:
[tex]\frac{dy}{dt}=y(1-y)[/tex]
So they grow as a function y(1-y).
c) In the case of the population of prey is much grader than predator, it will grow faster and exponentially, because it we assume that preys have unlimited food supply.
I hope it helps you!
