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?

a)

b)

Let's say that

If we multiply the woule equation by k we got:

So then we satisfy that the equation is also logistic since the parameter

c) If we assume that then we have that

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

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

Part a

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

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

And we can write like this:

Part b

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

Let's say that

If we multiply the woule equation by k we got:

So then we satisfy that the equation is also logistic since the parameter

Part c

If we assume that then we have that

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

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

-60.75

step-by-step explanation:

just add the numbers then put the negative sign on it

question: match the eqaution with its graph -8x - 5y = 40

answer: 8 x 5 = 40

x= absiss

y= orderly

so you have a proportional graph

question answered by

(jacemorris04)