Assignment 10

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course Mth 279

6/28

Query 08 Differential Equations*********************************************

Question: 3.5.6. Solve the equation dPdt = r ( 1 - P / P_c) P + M with r = 1, P_c = 1 and M = -1/4.

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Your solution:

dP/dt = r(1 - (P/P_c))P + M

dP/dt = 1(1 - P)P - (1/4)

dP/dt = P - P^2 - (1/4)

separable DE

dP/-(P-.5)^2 = dt then integrate

1/(P-.5) = t + C

Solve for P

1 = (t + C)(P-.5)

1/(t+C) +.5 = P

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Note that as t -> infinity, P approaches 1/2, which is half the 'carrying capacity' P_c = 1 of the system.

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Not sure how this relates to logistics section. This was just a typical separable differential equation right?

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(1 - P / P_c) * P = 1 / P_c ( P_c - P) * P, which is just the product of the constant 1 / P_c with (N-P) * P, where N = P_c.

So this equation is of the form dP/dt = k (N-P)*P + M, a logistic equation with a constant influx of population represented by the term M.

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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 N = 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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Your solution:

dP/dt = k(N-P)P plug in k value

dP/dt = (2e^-t -1)(N-P)P separable DE

dP/((N-P)P) = (2e^-t -1) dt integrate using partial fractions

1/((N-P)P) = A/(N-P) + B/P

1 = AP + BN + BP

A +B = 0 BN = 1

A = -B = -1/N B=1/N

Therefore integrate the following

(-1/N)/(N-P) dP + (1/N)/P dp = (2e^-t -1)

(1/N) ln|P/(N-P)| = -2e^-t + C

Solve for P

ln|P/(P-N)| = -2Ne^-t - Nt + C

P/(P-N) = e^(-2Ne^-t - Nt + C)

P = Ae^(-2Ne^-t - Nt)(P-N)

P(t) = -NAe^-2Ne^-t - Nt)/(1-Ce^(-2Ne^-t - Nt))

find A

A = 100000/(-400000e^-1000000) which is the same as saying

A = P_0/((N-P_0)e^(-2N))

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Given Solution:

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Self-critique (if necessary):

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Self-critique (if necessary):

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&#Good responses. Let me know if you have questions. &#