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zenterix

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- Homework Statement
- The following two problems are from chapter 8.7 "Introduction to Differential Equations" of Apostol's Calculus Volume I.

Problem 13. (Prelude to Problem 14, about which my question is)

Express ##x## as a function of ##t## for the growth law

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

with ##k## and ##M## both constant. Show that

$$x(t)=\frac{M}{1+e^{-kM(t-t_1)}}\tag{8.23}$$

where ##t_1## is the time at which ##x=M/2##.

Problem 14.

Assume the growth law in formula 8.23 of exercise 13, and suppose a census is taken at three equally spaced times ##t_1,t_2,t_3##, the resulting numbers being ##x_1,x_2,x_3##. Show that this suffices to determine ##M## and that, in fact, we have

$$M=x_2\frac{x_3(x_2-x_1)-x_1(x_3-x_2)}{x_2^2-x_1x_3}\tag{8.24}$$

- Relevant Equations
- My question is strictly about exercise 14.

Exercise 13 is a relatively simple matter (and I include at the end my solution to it).

First of all, a few observations

1) It is not clear if the ##t_1## used in problem 14 is the same ##t_1## from problem 13 where ##x(t_1)=\frac{M}{2}##.

However, if it were, then the problem seems like it wouldn't make too much sense because we'd have ##M=2x_1## and that'd be it (though this wouldn't match the expression the problem has for ##M##).

We would be able to solve for ##k## as well in this case with just one census reading.

2) From (8.23) it seems that the population never actually reaches ##M##. It just gets really close from below.

Now let me rewrite the equation for ##x## as follows

$$x(t)=\frac{M}{1+e^{-kM(t-t_h)}}$$

where now ##t_h## is such that ##x(t_h)=\frac{M}{2}##.

I've tried a few things but am a bit stuck.

Each attempt is a lot to type so I will have to screenshot them.

First I tried to just plug the values in

But this doesn't seem to tell me much.

Well, I will post now, as I continue to think about this problem.Since it will be too much work to write out all my work for problem 13 in equations, here is a screenshot of the work

Note that I used partial fractions to solve the integral in purple, and the calculations were as follows

1) It is not clear if the ##t_1## used in problem 14 is the same ##t_1## from problem 13 where ##x(t_1)=\frac{M}{2}##.

However, if it were, then the problem seems like it wouldn't make too much sense because we'd have ##M=2x_1## and that'd be it (though this wouldn't match the expression the problem has for ##M##).

We would be able to solve for ##k## as well in this case with just one census reading.

2) From (8.23) it seems that the population never actually reaches ##M##. It just gets really close from below.

Now let me rewrite the equation for ##x## as follows

$$x(t)=\frac{M}{1+e^{-kM(t-t_h)}}$$

where now ##t_h## is such that ##x(t_h)=\frac{M}{2}##.

I've tried a few things but am a bit stuck.

Each attempt is a lot to type so I will have to screenshot them.

First I tried to just plug the values in

But this doesn't seem to tell me much.

Well, I will post now, as I continue to think about this problem.Since it will be too much work to write out all my work for problem 13 in equations, here is a screenshot of the work

Note that I used partial fractions to solve the integral in purple, and the calculations were as follows

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