Question

A model for the population p(t) in a suburb of a large city is given by the initial-value problem dp dt p(101 2 107p), p(0) 5000, where t is measured in months. what is the limiting value of the population? at what time will the population be equal to one-half of this limiting value?

Answer 1

Given - dp/dt = P(10^-1 - 10^-7P), P(0) = 5000, Solution - a = 10^-1, b = 10^-7,
dp/dt = P(a-bP),
P(t) = aP/{bP+(a-bP)e^-at}
P(t) = 500/{0.0005+0.0995e^-0.1t}
on limiting t--infinite -
limiting value P = 1000000
P(t) = 500000 = 500/{0.0005+0.0995e^-0.1t}
on solving... t = 52.9 months.