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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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