Question

Consider a rabbits population p(t) of a country satisfies the differential equations dp/dt=kp(400-p) with k constant, if population in 1960 was 100 million and was then growing at the rate of 2 million per year. Predict this country's population for the year 2010 ?

Answer 1

Explanation:

Given the differentials equation

dp/dt=kp(400-p)

dp/p(400-p) = kdt

Integrating both sides

1/200(lnp - ln(400-p)) = kt+C

(lnp - ln(400-p)) = 200kt+C

ln(p/400-p) = 200kt+C

p/400-p = Ce^200kt

At t = 0, P(t) = 100

On substituting

100/400-100 = Ce^200k(0)

100/300 = C

C = 1/3

Since the population is growing at 2 million per year, then at t = 1, p(t) = 102

1 = kp(400-p)

1 = k(102)(400-102)

1 = k(102)(298)

1 = 30396k

k = 1/30396

Next is to predict the population by 2010.

From 1960 to 2010, there are 50years

Using the expression

p/400-p = Ce^200kt.

p(50)/400-p(50) = 1/3e^200(50)/30396

p(50)/400-p(50) = 1/3e^10000/30396.

p(50)/400-p(50) = 1/3e^0.3289

p(50)/400-p(50) = 0.4631

P = (400-p)0.4631

p = 185.26-0.4631p

1.4631p = 185.26

p = 185.26/1.463

p = 126.63

Hence the country's population by 2010 is approximately 127million