Q.
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 20000 in 1999 and 25000 in the year 2004 , the population of the village in 2009 is
Let P be the population at time t, then dtdy∝y ⇒dtdy=ky, where k is constant ⇒ydy=kdt
On integrating both sides, we get logy=kt+C.....(i)
In the year 1999,t=0,y=20000 ∴ From equation (i), log20000=k(0)+C ⇒log20000=C....(ii)
In the year 2004,t=5,y=25000, so from Eq. (i) ⇒log25000=k5+C ⇒log25000=5k+log20000 [using Eq. (iii)] ⇒5k=log(2000025000) =log(45) ⇒k=51log45
For year 2009,t=10yrs
Now, substituting the values of t,k and C in Eq. (i), we get logy=10×51log(45)+log(20000) ⇒logy=log[20000×(45)2](∵logm+logn=logmn) ⇒y=20000×45×45 ⇒y=31250
Hence, the population of the village in 2009 will be 31250 .