Query 8

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course MTH 279

2/13 12I have reviewed your notes on the query over the mixing and cooling problems. I hope to have time to rework those in the near future. Thank you for the feedback.

Query 08 Differential Equations

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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:

P/(1-P) = Ce^(-.25t)

or explicitly: P= [Ce^(-.25t)] / [1 + Ce^(-.25t)]

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

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

dP/[(1-P)P] = -1/4 dt integrate left side using partial fractions

ln(abs(P) ) - ln(abs(1-P) ) = -1/4t + C Simplify using properties of logs and exponents.

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If r = 1, P_c = 1 and M = -1/4, our equation becomes

dP/dt = (1 - P) * P + 1/4,

which we can rewrite as

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

The denominator factors into -(P - 1/2) ^ 2 so we have

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

which is easily integrated to obtain

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

so that

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

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

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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:

P/(500,000 - P) = (1/4e^1,000,000)e^(-1,000,000e^(-t) - 500,000t)

The maximum value of P appears to be 100,000 (at t=0), because the curve begins to slope down from this point as time elapses.

The value of t when the population would be 50,000 would be about 3.5.

Based on these answers, it appears as if the number of people infected is decreasing as time elapses.

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The steps by which you obtained your solution are not clear, and you haven't solved for P, but your solution is in the riight general form. I believe the correct solution for P is

P = 500 000 / ( 4 e^(1 000 000 (e^(-t) - 1) + 500 000 t) + 1).

or equivalently

P = 125 000 / ( e^(1 000 000 (e^(-t) - 1) + 500 000 t) + 1/4).

At t = 0 this yields population 100 000, and as t -> infinity the limiting population is 0, as expected.

I agree that this solution appears to decrease throughout its domain.

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): I think I went about it okay, but the numbers were a little messy, so i may've made some minor miscalculations.

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Self-critique rating: 3"

Self-critique (if necessary):

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Self-critique rating:

&#Good responses. See my notes and let me know if you have questions. &#