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 $2009$?

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

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


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

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

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

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

$t=5years$ and 
$N=2.5\times 10^4$

Hence 
$\ln \frac{N}{N_0}=kt$

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

$\ln (1.25)=5k$

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

Now 
$t=2009-1999$
$=10years$

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

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

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

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

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

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

$=3.125\times 10^4$

$=31250$