#$&*
course Mth 279
4/1Well 3/31 after midnight
I have started back on these and will be ready for test 1 on Thursday!" "Query 08 Differential Equations
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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:
The Equation: dPdt = r ( 1 - P / P_c) P + M
Becomes:
dP/dt = 1 ( 1 - P / 1) P + (-1/4)
dP/dt = ( 1 - P) P + (-1/4)
dP/dt = P - P^2 + (-1/4)
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You can rewrite this 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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I found that this equation was stated as a ""Riccati Equation""
in the form: y' = p_1(t) + p_2(t)y + p_3(t)y^2
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The equation is of this form, but since in the present case p_1(t), p_2(t) and p_3(t) are all constant, the equation turns out to be separable.
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dP/dt = P - P^2 + (-1/4) or dy/dt = y - y^2 - 1/4
Sustitution of y = y_1 + u
(y_1)' = 1
((y_1)' + u') = (-1/4) + ((y_1) + u) - ((y_1) + u)^2
(1 + u') = (-1/4) + (y_1) + u - ((y_1)^2 + 2(y_1)u + u^2)
u' = -y^2 + y -2yu + u - u^2 -1/2
u' = u - 2yu - u^2
perhaps since u = y - y_1 and u' = 1, y_1 = 0
Which is what I gathered from interpreting the equation that I looked up, and this:
since y' = q_0 + (q_1)(y_1) + (q_2)(y_1)^2
and u' = (q_1)u + 2(q_2)(y_1)u + (q_2)u^2
u' - ((q_1) + 2(q_2)(y_1))u = (q_2)u^2
This is now a bernoulli Equation
Our erquation becomes:
u' = u - 2yu - u^2
u' - u + 2yu = - u^2
u' + (-1 - 2y)u = - u^2
Bernoulli form: y' + p(t)y = q(t)y^n
p(t) = -2y - 1
q(t) = -1
n = 2
v = y^m
y = v^(1/m)
m = (1-n)
y = v^(1/(1-n))
n = 2
y = v^(1/(1-2)) = v^(-1)
This also made the process simpler for me:
y = t, and u = y
Making the equaiton:
y' + (-2t - 1)y = -y^2
dv/dt + (1-n)(-2t-1)v = (1-n)(-1)
dv/dt + (1-2)(-2t-1)v = (1-2)(-1)
dv/dt + (-1)(-2t-1)v = (-1)(-1)
dv/dt + (2t + 1)v = 1
This is now an Equation of the form y' + p(t)y = g(t)
And can be solved by using u(t) = e^(int(p(t))dt)
u(t) = e^(t^2 + t)
u'(t) = (2t + 1)e^(t^2 + t)
v' + (2t + 1)v = 1
v'e^(t^2 + t) + (2t + 1)ve^(t^2 + t) = e^(t^2 + t)
(dv/dt)(ve^(t^2 + t)) = e^(t^2 + t)
Integration of both sides yeilds:
ve^(t^2 + t) = (e^(t^2 + t))/(2t + 1) + c
Division by u(t):
v = 1/(2t + 1) +c
and since y = 1/v
y = (2t + 1) + c
This makes the original equation:
(remember y=t, and u=y)
u = 2y + 1 + c
and the first substitution was:
p = y_1 + u
p = (y_1) + 2(y_1) + 1 + c
p = 3y + 1 + c
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You did some good stuff there. It was unnecessary but it was a good exercise for you, so your effort wasn't wasted.
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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:
Was a bit confused on this one. It was harder to find an good explaination.
dP/dt = k ( N - P) * P
First it wants this in the form
dP/dt = kN ( 1 - P/N) * P
k = 2e^(-t) - 1
I thought the solution might be similar to dp/dt = r(1 - p/p_c)p
but k was making it kinda weird.
dp/(((1-p/N)p) dt) = kN
dp/(((1-p/N)p) dt) = (2e^(-t) - 1)N
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dP/dt = k ( N - P) * P
is separable, and in separated form it's easy to integrate using partial fractions.
This is the approach used and illustrated in the text.
k is a constant and not a function of t.
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Integration yeilds:
Int (2e^(-t) - 1) dt = -2e^(-t) - t + c
Int(1/p - 1/(p-n)) dp = ln(p/(p-n)) Found using partial fractions.
So that
ln(p/(p-n)) = -2e^(-t) - t + c
p/(p-n) = e^(-2e^(-t) - t + c)
p/(p-n) = Ce^(-2e^(-t) - t)
p(o) = 100,000
1/(1-n/p) = Ce^(-2e^(-t) - t)
1/Ce^(-2e^(-t) - t) = (1-n/p)
-N/Ce^(-2e^(-t) - t) = p
and
-N/Ce^(-2) = 100,000
I could find C at this point but only in terms of N
I could not figure out how to solve for N.
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It's not clear how you got
k = 2e^(-t) - 1
and I believe that mislead you. Otherwise you were doing many of the right things.
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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confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:"
&#Good work. See my notes and let me know if you have questions. &#