Question

The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was $20,000$ in $1999$ and $25000$ in the year $2004,$ what will be the population of the village in $2009$?

Answer 1

Let the number of people at a given year '$t$' be $N$
Then,

$\frac{dN}{dt}=kN$ where $k$ is the proportionality constant.


$\frac{dN}{N}=kdt$

${\int }_{N_0}^N\frac{dN}{N}={\int }_0^{t}kdt$

$\ln \left(\frac{N}{N_0}\right)=kt$

$N=N_0{e}^{kt}$ ...(i)
If $N_0=2\times {10}^4$ and $t=2004-1999$

$t=5years$ and
$N=2.5\times {10}^4$

Hence
$\ln \frac{N}{N_0}=kt$

$\ln \left(\frac{2.5\times {10}^4}{2\times {10}^4}\right)=5k$

$\ln \left(1.25\right)=5k$

$k=\frac{1}{5}\ln \left(1.25\right)year{s}^{-1}$

Now
$t=2009-1999$
$=10years$

Hence
$\ln \left(\frac{N}{2\times {10}^4}\right)=\frac{1}{5}\ln \left(1.25\right)t$

$\ln \left(\frac{N}{2\times {10}^4}\right)=\frac{1}{5}\ln \left(1.25\right)\times 10$

$\ln \left(\frac{N}{2\times {10}^4}\right)=\ln \left(1.25\right)\times 2$

$\ln \left(\frac{N}{2\times {10}^4}\right)=\ln \left({1.25}^2\right)$

$\frac{N}{2\times {10}^4}={1.25}^2$

$N=2\times (1.25{)}^2\times {10}^4$

$=3.125\times {10}^4$

$=31250$