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 25,000 in the year 2004, what will be the population of the village in 2009?

Solution: Let the population at any instant be. It is given that the rate of increase of population is proportional to the number of inhabitants at any instant.

$\begin{array}{l} \Rightarrow \frac{{dy}}{{dt}}\alpha y\\ \Rightarrow \frac{{dy}}{{dt}} = ky\\ \Rightarrow \frac{{dy}}{y} = kdt\\\log y = kt + C\end{array}$

In 1999, and y=20000

log20000=C

 In 2004, t=5 and y= 25000

log25000=k.5+C

$\begin{array}{l} \Rightarrow 5k = \log \left( {\frac{{25000}}{{20000}}} \right) = \log \left( {\frac{5}{4}} \right)\\ \Rightarrow k = \frac{1}{5}\log \left( {\frac{5}{4}} \right)\end{array}$

In 2009, t=10 years 

$\begin{array}{l}\log y = 10 \times \frac{1}{5}\log \left( {\frac{5}{4}} \right) + \log 20000\\ \Rightarrow \log y = \log \left[ {20000{{\left( {\frac{5}{4}} \right)}^2}} \right]\\ \Rightarrow y = 20000 \times \frac{5}{4} \times \frac{5}{4} = 31250\end{array}$

Therefore, population of village in 2009 is 31250.