Start with the logistic equation dx dt kx(M−x). Suppose 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 (Similar to earlier example with h replaced with hx).

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

the perimeter is increased by a factor of 2