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The solution to quiz 2 problem of math 320, lecture 2 - spring 2009. The problem deals with the population dynamics of rabbits during an epidemic, where the death rate is proportional to the population. The solution involves writing the differential equation, solving the initial value problem, and determining the number of rabbits alive after 12 months.
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Math 320, Lecture 2 - Spring 2009
To get the most of this quiz, allow yourself no more than 20 minutes to completely answer it, and do not use any notes or outside help. I will grade it if you hand it in during discussion on February 16 (for Monday sections), February 18 (for Wednesday sections).
This problem was meant to use remind you of the logistic equation problems we did where dP dt = (β^ −^ δ)P^ , where^ β^ and^ δ^ are the rate of births and of deathes, respectively.^ Recall homework problem # 17 from Section 2.2.
(a) The general equation is dPdt = (β − δ)P , where β and δ are the birth and death rates, respectively. In our problem, β = 0 and δ = kP for some positive constant k, because there are no births and the death rate is propornal to P. So the differential equation is dP dt =^ −kP^
(b) We can solve this by using the separation of variables technique.
dP dt
= −kP 2 1 P 2
dP = −k dt ∫ 1 P 2
dP =
(−k) dt
= −kt + C 1 P
= kt + C′
P =
kt + C′
At time t = 0, we have P (0) = P 0. We must use this information in order to solve for our constant C′.
P (t) =
kt + C′ P 0 =
k(0) + C′
P 0 =
Thus, we get
P (t) =
kt + (^) P^10
=
P 0 kt + 1
=
1 + P 0 kt
(c) We see from the given information that P 0 = 1500. Also, P (6) = 500. Using this piece of information, we can solve for the constant k.
P (t) =
1 + 1500kt 500 =
1 + 1500k(6) 1 + 9000k = 3 9000 k = 2
k =
Thus,
P (t) =
4500
t
=
1 + 13 t
=
3 + t
We can now easily see that after 12 months, there will be P (12) = 300 bunnies remaining.