#$&*
course Phy 279
3/6
Query 08 Differential Equations*********************************************
Question: 3.5.6. Solve the equation dPdt = r ( 1 - P / P_c) P + M with r = 1, P_c = 1 and M = -1/4.
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Your solution:
dPdt = r ( 1 - P / P_c) P + M dP/dt = (1 - P)P - (1/4)
@& This doesn't work. a = b + c doesn't rearrange into a / b = c.*@
dP/P(1-P) = -(1/4) dt
1/P(1-P) = A/P + B/(1-P) = A(1-P) + BP = A - AP + BP = A - P(A-B), A = 1, B= 1
Int(1/P(1-P) dP) = Int( 1/P + 1/(1-P) dt = ln|P| + ln|(1-P)| = ln|P(1-P)|
ln|P(1-P)| = -(1/4)t + C
P(1-P) = e^(-(1/4)t + C) = Ae^(-(1/4)t)
P = Ae^(-(1/4)t)* (1-P) = Ae^(-(1/4)t) - APe ^(-(1/4)t)
P + APe ^(-(1/4)t) = Ae ^(-(1/4)t)
P(1 + Ae ^(-(1/4)t)) = Ae ^(-(1/4)t)
P = Ae ^(-(1/4)t)/( 1 + Ae ^(-(1/4)t))
@& If r = 1, P_c = 1 and M = -1/4, our equation becomes
dP/dt = (1 - P) * P + M,
which we can rewrite as
dP / ( -P^2 + P - 1/4) = dt.
The denominator factors into -(P - 1/2) ^ 2 so we have
-dP / (P - 1/2)^2 = dt,
which is easily integrated to obtain
1 / (P - 1/2) = t + c
so that
P = 1 / (t + c) + 1/2.
At t -> infinity, P approaches 1/2, which is half the 'carrying capacity' P_c = 1 of the system.*@
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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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 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:
I’m confused is N the max number in the population? I’m just not being able to put this question together?
@& You need to check the book for the complete statement of the problem, but N is the limiting population.
The logistic equation is
dP/dt = k P ( N - P).
Now if k is replaced by the given expression we get
dP/dt = (2 e^-t - 1) ( N - P) * P.
This equation is separable, and the integration is straightforward. See what you get, and see how far you can go from there.
Jonathan and I independently got the same solution, which led to the same ridiculous value of the integration constant. The solution up to that point is fairly straightforward, so see if you can break through our logjam.
I've made some progress by keeping things symbolic, but I'm still not getting a completely sensible interpretation.
Don't get too bogged down on this, but work it out until you get to that point.*@