Question

The rate of inceease of population of city is 10% per annum.if in the year 2005,its population were 80000.what will be its population in 2008

Answer 1

Since the population increases at a rate proportional to the population,

$\frac{dP}{dt} =kP,k>0$

Solving the above differential equation,

$\frac{dP}{dt} =kP\\ \frac{dP}{P} =kdt\\ \int \frac{dP}{P} =\int kdt\\ \ln P=kt+C$

Let the initial population be $P_0$. Then at $t=0$,

$\ln P_0=0+C ,C=\ln P_0$

$\ln P=kt+\ln P_0\\ \ln (\frac{P}{P_0})=kt\\ \frac{P}{P_0}=e^{kt}\\ P=P_0e^{kt}$

Here $k=10\%=0.1.$. At $t=2008-2005=3$, the population is

$P=P_0e^{kt}\\ P=80000e^{0.3}\\ P=107,988.7$