The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time.

Solution:

Let the population at any instant (t) be y.

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

$\therefore \frac{d y}{d t} \propto y$

$\Rightarrow \frac{d y}{d t}=k y$                              ( $k$ is a constant)

$\Rightarrow \frac{d y}{y}=k d t$

Integrating both sides, we get:

$\log y=k t+C \ldots(1)$

In the year $1999, t=0$ and $y=20000$.

Therefore, we get:

$\log 20000=C \ldots$ (2)

In the year $2004, t=5$ and $y=25000$.

Therefore, we get:

$\log 25000=k \cdot 5+\mathrm{C}$

$\Rightarrow \log 25000=5 k+\log 20000$

$\Rightarrow 5 k=\log \left(\frac{25000}{20000}\right)=\log \left(\frac{5}{4}\right)$

$\Rightarrow k=\frac{1}{5} \log \left(\frac{5}{4}\right)$                     ...(3)

In the year 2009, t = 10 years.

Now, on substituting the values of t, k, and C in equation (1), we get:

$\log y=10 \times \frac{1}{5} \log \left(\frac{5}{4}\right)+\log (20000)$

$\Rightarrow \log y=\log \left[20000 \times\left(\frac{5}{4}\right)^{2}\right]$

$\Rightarrow y=20000 \times \frac{5}{4} \times \frac{5}{4}$

$\Rightarrow y=31250$

Hence, the population of the village in 2009 will be 31250.