#$&*
MTH 279
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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3.5.10. Solve dP/dt = k ( N - P) * P with P(0) = 100 000 assuming that P is the number of people, out of a population of 500 000, with a disease. Assume that k is not constant, as in the standard logistic model, but that k = 2 e^(-t) - 1. Plot your solution curve and estimate the maximum value of P, and also that value of t when P = 50 000. Interpret all your results in terms of the given situation.
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I get an impossible result for C when I get to the point of P(0)= 100000. The next box shows my process. The problem comes from the fact that e^(-2e^-t) = e^-2 @ t = 0 meaning that Ce^(N(-2)) which is astronomically small making C astronomically large, specifically 1.516607698401043 × 10^434288. I've retraced the steps again and again and consulted the Schaum's DE book that has a similar problem in it and also found C at other points in the process and still nothing seems to work. What am I missing here?
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Division:
dP/(P(N-P)) = 2e^(-t)-1 dt
Partial Fractions:
1/N (1/P-1/(N-P))
Integrate:
1/N ln (P/(N-P)) = -2e^(-t) -t + C
P/(N-P) = e^[N( -2e^(-t) -t + C)]
Simplify the right side:
e^(NC)= C
Ce^(N(-2e^(-t) -t))
Solve for P:
P = CNe^(N( -2e^(-t) -t)) / (1+Ce^(N(-2e^(-t)-t)))
N = 500 000
P(0)= 100 000
@& Ce^(N(-2e^(-t) -t)) can be expressed as
C e^(-2 N e^-t) e^(-N t) ,
in which form we see that as t increases e^(-N t) approaches zero while e^(-2 N e^(-t)) approaches 1.
Our solution for P is therefore
C N e^(-2 N e^-t) e^(-N t) / (1 + C e^(-2 N e^-t) e^(-N t) ).
Multiplying numerator and denominator by e^(2 N e^(-t)) * e^(N t) we get
C N / ( e^(2 N e^(-t)) * e^(N t) + C)
This is just a rearrangement of the original expression, though, and still leads to the same impossible value of C.
Now, k = 1 at t = 0, and reduces to about .3 by the time t = 1. So that shouldn't make such a big difference at t = 0.
For comparison I'll solve the equation with k = 1.
I get
P = N / (1/C * e^(-N k t) ) + 1)
When t = 0 this is
N / (1/C + 1) = P_0
so
1/C = N / P_0 - 1 and
C = P_0 / (N - P_0),
which is reasonable.
It would be expected that k = 2 e^(-t) - 1, being close to k = 1 when t is near zero, would have little influence on the value of C.
We could replace e^(-t) with its Taylor expansion 1 - t + t^2 / 2 - t^3 / 3 ! etc.. , which would lead us to the expression
e^(N ( t - t^2 + t^3 / 3),
which behaves OK.
I'll need to write this out on the board and compare the terms. I'll get back to you later today.
*@
followup from later:
I've solved the equation with the given function, as a simple logistic equation with k = 1, and replacing the exponential function with the Taylor series. The other two solutions are consistent and behave as expected. The solution with the given function is not consistent with either.
I've reduced the solution to the statement that the Taylor series of the exponential isn't what it is, but there's something I'm not seeing so I'm going to have to look at it again. My alternative approaches should show me what's wrong with my thinking but so far it hasn't.