Answer:
a) dx/dt = kx*(M - h/k - x)
Step-by-step explanation:
Given:
- The harvest differential Equation is:
dx/dt = kx*(M-x)
Suppose that we modify our harvesting. That is we will only harvest an amount proportional to current population.In other words we harvest hx per unit of time for some h > 0
Find:
a) Construct the differential equation.
b) Show that if kM > h, then the equation is still logistic.
c) What happens when kM < h?
Solution:
- The logistic equation with harvesting that is proportional to population is:
dx/dt = kx*(M-x) hx
It can be simplified to:
dx/dt = kx*(M - h/k - x)
- If kM > h, then we can introduce M_n=M -h/k >0, so that:
dx/dt = kx*(M_n - x)
Hence, This equation is logistic because M_n >0
- If kM < h, then M_n <0. There are two critical points x= 0 and x = M_n < 0. Since, kx*(M_n - x) < 0 for all x<0 then the population will tend to zero for all initial conditions
Uh 10... because 10 is 10
First, let's input "a" and "b" into the equation:
(Don't forget to put brackets around the inputted numbers to show you are inputting "a" and "b")
<span>4 ÷ a × b = </span>4 ÷ (444) x (213)
Now we can solve!
= 1.918
= ~ 2
Hope this helps!
Feel free to message me if you have any questions :)
