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course Mth 279
6/30 8:58 pm
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:
P' - (1-P) * P = -1/4
p(t) = -(1-P)
P(t) = 1/2 P^2 - P
u = e^(1/2 P^2 - P)
e^(1/2 P^2 - P)P' - (1-P) * e^(1/2 P^2 - P) * P = -1/4 e^(1/2 P^2 - P)
integrate
e^(1/2 P^2 - P) P = -1/4 * 1/(P-1) e^(1/2 P^2 - P) + C
P = (-1/4) (1/(P-1)) + Ce^(1/2 P^2 - P)
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Nice try, but this equation is not first-order linear. p(t) would have to be a function of t, not a function of P.
This equation is solved by separation of variables. It can be rearranged to the form
dP / (r(1 - P / P_c)* P + M) = dt
then both sides can be integrated.
The left-hand side becomes
dP / ( (1 - P) * P - 1/4) =
- dP / (P^2 - P + 1/4).
The denominator factors into a perfect square, and the integral can then be done by a straightforward substitution.
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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 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:
What does N stand for???
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N is the limiting population, which would be the total population of 500 000.
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P' - kP(N-P) = 0
Is this a Bernoulli equation ???
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This equation is a Bernoulli equation, but it is also separable. You get
dP/dt = k P ( N - P )
so that
dP / ( P(N-P) ) = k dt
The left-hand side is integrated using partial fractions.
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P' - kPN + kP^2 = 0
Multiplying out kP doesn't help me solve.
Treating it like a Bernoulli equation
P' -kPN = -kP^2
v = y^(1-n) = y^(-1)
v' + (1-n)Pv = (1-n)(-k)
v' -Pv = -k
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We need to avoid confusing all those p's and P's, so we're going to change the notation slightly:
The basic form of a Bernoulli equation is
y ' + p(t) y = q(t) y^n.
Using y instead of P, simply because P could easily be confused with p and its antiderivative P, the equation is
y ' + (-k N) y = k y^2
so that p(t) = -k N and q(t) = k.
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p(t) = -P
P(t) = -Pt
u = e^(-Pt)
e^(-Pt)v' - e^(-Pt)Pv = -ke^(-Pt)
integrate
e^(-Pt)v = -kP * e^(-Pt) + C
v = -kP + Ce^(Pt)
I think I should plug in values here, but I don't think I solved right because it doesn't make sense
P(0) = 100,000
v = P^(-1)
v(0) = 1/100000
plugging in t = 0
1/100000 = -k(100,000) + C
and k = 2e^(-t) - 1
1/100000 =- (200000)e^(-t)-1 + C
C = -200000.999
v = [-2e^(-t) - 1]P + -200000.999e^(Pt)
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You did well to recognize this as a Bernoulli equation.
Check my note on notation, which might change some of your solution.
You should solve this as a separable equation as well.
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Given Solution:
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#*&!
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The main method used with these equations is separation of variables, which often leads to integration by partial fractions. That method is unavoidable in this course, so if necessary be sure you review it.
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