Suppose a population growth is modeled by the logistic differential equation with the carrying capacity 2000 and the relative growth rate k = 0.06 per year. If the initial population is P(0) = 500, and P(10).
A)309
B)756
C)310
D)757
E)308
F)755
G)307
H)800

Question

Quizplus · Accepted Answer

Using the logistic differential equation, we have:
dP/dt = 0.06P(1 - P/2000)
Separating variables and integrating, we get:
∫dP/(P(1 - P/2000)) = ∫0.06dt
∫[1/P + 1/(2000 - P)]dP = 0.06t + C
ln|P| - ln|2000 - P| = 0.06t + C
ln|P/(2000 - P)| = 0.06t + C
Taking exponential of both sides:
P/(2000 - P) = e^(0.06t+C) = Ce^(0.06t)
Using the initial condition P(0) = 500, we have:
500/(2000 - 500) = Ce^0
C = 1/3
Therefore:
P/(2000 - P) = (1/3)e^(0.06t)
Solving for P when t = 10, we get:
P/(2000 - P) = (1/3)e^(0.06*10)
P/(2000 - P) = 0.756
P = 2000(0.756)/(1 + 0.756) = 756.16
Rounding to the nearest whole number, we get:
P(10) = 756. Therefore, the best choice is B.

Suppose a Population Growth Is Modeled by the Logistic Differential