Question

Start with the logistic equation 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 (Similar to earlier example with h replaced with hx).

a) Construct the differential equation.
b) Show that if kM > h, then the equation is still logistic.
c) What happens when kM < h?

Answer 1

Answer:

a) $\frac{dx}{dt}= kx (M-x) -hx$

$\frac{dx}{dt}= kx (M -x- \frac{h}{k})$

b) $M -\frac{h}{k}>0$

Let's say that $a=M -\frac{h}{k}>0$

If we multiply the woule equation by k we got:

$kM -h >0$

So then we satisfy that the equation is also logistic since the parameter $a>0$

c) If we assume that $kM <h$ then we have that $a<0$

And then [tec] kx (a -x) <0[/tex] for any value of $x>0$

And if that hhapens then the population will tend to 0 for any initial condition established/

Step-by-step explanation:

For this case we have the following logistic equation $\frac{dx}{dt}= kx (M-x)$

Part a

We want to modify our harvesting for this case, so we harvest hx per unit of time for some $h>0$

So then the model with harvesting who is proportional is given by:

$\frac{dx}{dt}= kx (M-x) -hx$

And we can write like this:

$\frac{dx}{dt}= kx (M -x- \frac{h}{k})$

Part b

For this case we assume that $kM>h$and we need to show that the equation is still logistic. So we need that the sollowing quantity higher than 0

$M -\frac{h}{k}>0$

Let's say that $a=M -\frac{h}{k}>0$

If we multiply the woule equation by k we got:

$kM -h >0$

So then we satisfy that the equation is also logistic since the parameter $a>0$

Part c

If we assume that $kM <h$ then we have that $a<0$

And then [tec] kx (a -x) <0[/tex] for any value of $x>0$

And if that hhapens then the population will tend to 0 for any initial condition established/