Query_08

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course MTH-279

06/25 around 11:30AM. Pretty good amount of confusion on this assignment as far as whether or not I am in the right direction/what comes next.

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.YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Rewritten

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

Factor the denominator

-(P - 1/2) ^ 2

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

Integrate

1 / (P - 1/2) = t + c

So,

P = 1 / (t + c) + 1/2

???? I am lost after this point, but suppose we use a form of the Riccati Equation ????

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This is a good solution.

Note that as t -> infinity, P approaches 1/2, meaning that P approaches half the carrying capacity of the system.
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Given Solution:

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Self-critique (if necessary):

Self-critique rating:

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 = (2 e^-t - 1) ( N - P) * P

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

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

ln( P / (N - P) ) = -2 N e^(-t) - N t + c

???? Not sure if I started this correctly, and if so I’m lost on the next steps????

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You are trying to solve for P, so you would apply the inverse of the ln function to get

e^(ln( P / (N - P) )) = -e^(2 N e^(-t) - N t + c).

The resulting equation has P / (N - P) on the left-hand side, and can be solved for P (multiply both sides by N - ), collect all P terms on one side and factor out P, divide both sides by the rest).
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