A biologist is tracking Lyme Disease cases in a deer population of 8500 in a state forest in Wisconsin. Suppose at time t = 0 the number D(t) of deer with Lyme Disease is 2250 and is increasing at a rate of 110 deer per month. Assume that D'(t) is proportional to the product of the number of deer who have caught Lyme's disease and of those who have not. (a) Write a differential equation to represent this scenario. Use k for your proportionality constant and D(t) as your function (not D). D'(t) kD(t) (8500 – D(t)) ✓ os = می (b) If you did part (a) correctly, the general solution to your differential equation is D(t) = 8500 1+ Ce-8500kt Find C (keep C exact). C = 25 9 من (c) Find k. (Hint: you do NOT need the function in part (b) to find k). Keep k exact. k 1 125000 (d) How long will it take for 50% of the deer population to develop Lyme Disease? Keep k and C exact until your final calculation. Round your final answer to two decimal places. 13.59 ✓ months
A biologist is tracking Lyme Disease cases in a deer population of 8500 in a state forest in Wisconsin. Suppose at time t = 0 the number D(t) of deer with Lyme Disease is 2250 and is increasing at a rate of 110 deer per month. Assume that D'(t) is proportional to the product of the number of deer who have caught Lyme's disease and of those who have not. (a) Write a differential equation to represent this scenario. Use k for your proportionality constant and D(t) as your function (not D). D'(t) kD(t) (8500 – D(t)) ✓ os = می (b) If you did part (a) correctly, the general solution to your differential equation is D(t) = 8500 1+ Ce-8500kt Find C (keep C exact). C = 25 9 من (c) Find k. (Hint: you do NOT need the function in part (b) to find k). Keep k exact. k 1 125000 (d) How long will it take for 50% of the deer population to develop Lyme Disease? Keep k and C exact until your final calculation. Round your final answer to two decimal places. 13.59 ✓ months
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:A biologist is tracking Lyme Disease cases in a deer
population of 8500 in a state forest in Wisconsin.
Suppose at time t = 0 the number D(t) of deer with
Lyme Disease is 2250 and is increasing at a rate of 110
deer per month. Assume that D'(t) is proportional to
the product of the number of deer who have caught
Lyme's disease and of those who have not.
(a) Write a differential equation to represent this
scenario. Use k for your proportionality constant and
D(t) as your function (not D).
D'(t) kD(t) (8500 – D(t)) ✓ os
=
می
(b) If you did part (a) correctly, the general solution
to your differential equation is
D(t)
=
8500
1+ Ce-8500kt
Find C (keep C exact).
C
=
25
9
من
(c) Find k. (Hint: you do NOT need the function in
part (b) to find k). Keep k exact.
k
1
125000
(d) How long will it take for 50% of the deer
population to develop Lyme Disease? Keep k and C
exact until your final calculation. Round your final
answer to two decimal places.
13.59
✓ months
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