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?

Let us Assume the population is (y) at any instant (t)

Since Rate of increase of population with respect to any time is

And, the rate of change of population is proportional to the inhabitants ,

So,

where, k is the proportionality constant

On integrating both sides, we get

log y = kt + C …(i)

Now, According to question,

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

On putting these values, in equation (i) , we get

log 20000 = C …(ii)

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

on putting these values in equation (i), we get

log 25000 = 5k + C …(iii)

Now, from the equation (ii) put the value of C in equation (iii)

log 25000 = 5k + log 20000

5k = log 25000 - log 20000

…(iv)

And, In the year 2009, t = 10 years

Now, putting the value of k, t and C in equation (i), we get

y = 31250

Hence, The population of the village in the year 2009 will be 31250