Population Growth

Let the population at any instant $t$ be $y$ . According to the question,
$\dfrac{dy}{dt} \propto y$ .
$\Rightarrow \dfrac{dy}{dt} = k y$ , [ where $k$ is the proportionality constant. ]
$\Rightarrow \dfrac{dy}y = k \, dt$ .

Integrating both sides, we get
$\ln y = kt + c$ , [ where $c$ is a constant to be determined ]

In the year 1999, $t = 0$ and $y =20000$ ,
therefore, $\ln(20000) = 0 + c$ .

In the year 2004, $t = 5$ and $y =25000$ ,
therefore, we get $\ln(25000) = 5k + c$ $\Rightarrow \ln(25000) = 5k + \ln(20000)$
$\Rightarrow k = \dfrac15 \ln \left( \dfrac54 \right)$ .

Now, in the year 2009, $t=10$ years.
$\ln y = 10 \cdot \dfrac15 \ln \left( \dfrac54 \right) + \ln(20000)$
$\Rightarrow \ln y = \ln \left(20000\cdot \dfrac54 \cdot \dfrac54\right)$ .
$\Rightarrow y = 20000 \cdot \dfrac54 \cdot \dfrac54$ .
$\Rightarrow y = 31250$ .

Hence, the population of the village in 2009 will be $\boxed{31250}$ .