Question

The population of the Mysore city of a country was 30 lakhs in the year 2000 and 60 lakhs in the year 2010. What is the estimated population in the year 1990 with the help of the population model?

(a) 18 lakhs

(b) 15 lakhs

(c) 20 lakhs

(d) 12 lakhs

This question was addressed to me in an interview.

My question comes from Law of Natural Growth and Decay in portion Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Right answer is (b) 15 lakhs

Easiest explanation: Here rate of change of population is proportional to the population present at that instant of time i.e \(\frac{dp}{dt} ∝ P\) i.e \(\frac{dp}{dt}=kP\), k is a constant of proportionality.

solving using the variable separable integral form we get \(\int \frac{1}{P} \,dP = \int K \,dt + a\)

log P=kt+c –> e^kt+a = P or P=ce^kt…where e^a is a constant let the year 2000 be the initial reference year –> P(0)=30=ce^k(0) = c i.e c = 30 & P(10) = ce^k10

=30e^k10 = 60 –> log 60 = log 30 + 10k –> log ^6⁄3 =log 2 = 10k –> k = 0.1*0.693 = 0.0693

to find P(-10) = ce^kt = 30*e^-0.693=15 lakhs.