Step 1:

Given the population of the village increases at the rate proportional to its number of its inhabitants. Let y be the number of inhabitants of the village.

Hence $\large\frac{dy}{dt}$ is proportional to $y$

$\large\frac{dy}{dt}$$= ky$, where k is a proportional constant.

seperating the variables we get,

$\large\frac{dy}{y}$$ = kdt$

now integrating on both sides we get,

$\log y = kt + C$-------(1)

Step 2:

In the year 1999, when t = 0 and y = 20,000

substituting this in equ(1)

$\log 20,000 = C$--------(2)

In the year 2004, when t= 5 y = 25,000

substituing this in equ(1) we get,

$\log 25,000 = 5k + C$

Substituting for C we get

$\log 25,000 = 5k +\log 20,000$

$\log \large\frac{25000}{20,000} $$= 5k$

$\log \large\frac{5}{4} $$= 5k$

Or $k=\big(\large\frac{1}{5}\big)\log\big(\large\frac{5}{4}\big)$

Step 3:

In the year 2009,when $t=10$

Substituting for t,k and C equ(1) becomes,

$\log y = 10 \big(\large\frac{1}{5}\big)$$\log\big(\large\frac{5}{4}\big)$$+\log 20,000$

$\log y = 2\log(5/4) + \log 20,000 = \log\big(\large\frac{5}{4}\big)^2 +$$ \log 20,000$

$\log y = \log 20,000 \times (5/4)^2$

$y=20,000 \times\large\frac{5}{4}\times\large\frac{5}{4}$$ = 31250$

Hence the population in the year 2009 is 31250.