Question

The population of a village increases at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20000 in year 1999 and 25000 in the year 2004, what will be the population of the village in 2009?

Answer 1

Let P be the population at time t, then $\frac{dy}{dx}\propto y$
$\frac{dy}{dx}=ky,$ where k is constant $\Rightarrow \frac{dy}{y}=kdt$
On integrating both sides, we get logy=kt+X ....(i)
In the year 1999, t=0, y=20000
$\therefore$From Eq(i) $log20000=k\left(0\right)+C\Rightarrow log20000+C$ .....(ii)
In the year 2004,t=5, y=25000, so from Eq.(i)
$log250000=k5+C\Rightarrow log25000=5k+log20000$ [using ~Eq.~(iii)]
$\Rightarrow 5k=log\left(\frac{25000}{20000}\right)=log\left(\frac{5}{4}\right)\Rightarrow k=\frac{1}{5}log\frac{5}{4}$
For the year 2009, t=10yr
Now, substituting the values of t, k and C in Eq. (i), we get
$logy=10\times \frac{1}{5}log\left(\frac{5}{4}\right)+log\left(20000\right)\Rightarrow logy=log[20000\times {\left(\frac{5}{4}\right)}^2]$ $[\because logm+n=logmn]$
$\Rightarrow y=20000\times \frac{5}{4}\times \frac{5}{4}\Rightarrow y=31250$
Hence, the population of the village in 2009 will be 31250.