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course MTH 279
Query 08 Differential Equations*********************************************
Question: 3.5.6. Solve the equation dP/dt = r ( 1 - P / P_c) P + M with r = 1, P_c = 1 and M = -1/4.
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Your solution:
P’ = rP - rP^2/P_c + M
This is the quadratic form of the logistic equation, which can be put into the form
P’ = ax - bx^2 - k
a = r = 1
b = r/P_c = 1
k = ¼
The stationary solutions of this type equation occur when dx/dt = 0, which are the roots of the equation ax-bx^2-k, so solutions are: x(t) = [a±sqrt(a^2-4bk)]/2b
So the stationary solution to the quadratic is x_1(t) = ½.
The solution to the equation is has the form P(t) = a/(2b) + 1/(b(t-4bk))
So our solution is P(t) = ½ + 1/(t-1)
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Your solution is OK but it sort of mixes up x's and P's.
It's generally preferable to solve the equation directly, from basic principles.
This is a separable equation and for the given values of the constants, is very easy to integrate:
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.
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confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary): OK
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Self-critique rating: OK
3.5.10. Solve dP/dt = k ( N - P) * P with P(0) = 100 000 assuming that P is the number of people, out of a population of N = 500 000, with a disease. Assume that k is not constant, as in the standard logistic model, but that k = 2 e^(-t) - 1. Plot your solution curve and estimate the maximum value of P, and also that value of t when P = 50 000. Interpret all your results in terms of the given situation.
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Your solution:
ʃdP/(P(N-P)) = ʃkdt
1/(P(N-P)) = A/P + B/(N-P)
A = 1/N = B
ʃ1/NP + 1/(N^2-NP) dP = ʃkdt
ln(P/(N-P))/N = -2e^(-t)-t + C
ln(P/(N-P)) = -2Ne^(-t)-Nt+C
P/(N-P) = Ae^(-2Ne^(-t)-Nt)
P = AN/[e^(2Ne^(-t)+Nt) + A]
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
I had a significant amount of trouble with both of these problems, I’m just having trouble visualizing how to solve them.
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You've got a general solution, but you haven't yet applied the initial conditions.
These equations are both solvable by separation of variables, and manipulation of the denominator. In one case the denominator is the perfect square of a linear function, in the other you need to use partial fractions.
Equations of this form very often do require integration by partial fractions.
Check out the following 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
P / (N - P) = e^(-2 N e^(-t) + N t + c) = A e^(-2 N e^(-t) - N t ), A > 0
P = (A e^(-2 N e^(-t) - N t )) ( N - P )
P - P A e^(-2 N e^(-t) - N t ) = N A e^(-2 N e^(-t) - N t )
P ( 1 - A e^(-2 N e^(-t) - N t ) ) = N A e^(-2 N e^(-t) - N t )
P = N A e^(-2 N e^(-t) - N t ) / ( 1 - A e^(-2 N e^(-t) - N t ) )
P = 1 / (1 / (A e^(-2 N e^(-t) - N t )) - 1)
P = 1 / (C e^(2 N e^(-t) + N t ) - 1), where C = 1 / A.
P(0) = P_0 so
1 / (C e^(2 N e^(0) + N * 0 ) - 1) = P_0
C e^(2 N ) - 1 = 1 / P_0
C e^(2 N e^(-t) + N t ) = 1 / P_0 + 1 = (1 + P_0) / P_0
C = (1 + P_0) / P_0 * e^(-2 N)
Thus
P = 1 / ((1 + P_0) / P_0 * e^(-2 N) e^(2 N e^(-t) + N t ) - 1) = 1 / ((1 + P_0) / P_0 * e^(2 N e^(-t) - 2 N + N t) - 1)
Now consider the exponent
2 N e^(-t) - 2 N + N t.
When t = 0 this exponent is 0, and the population is 1 / (1 + P_0) / P_0 - 1) = P_0, as it should be.
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