Differential Equations — Class 12 Maths Solution

Let the population at any instant $({\rm{t}})$ be ${\rm{y}}$.

It is given that the rate of increase of population is proportional to the number of inhabitants at any instant.

$\therefore \frac{{dy}}{{dt}} \propto y$
$\Rightarrow$ $\frac{{dy}}{{dt}} = ky$

($k$is a constant)
$\Rightarrow$ $\frac{{dy}}{y} = kdt$
Integrating both sides,

we get:
$\log y = kt + C$ $\ldots (1)$
In the year 1999,${\rm{t}} = 0$ and ${\rm{y}} = 20000$.
Therefore,

we get:
$\log 20000 = C$ $\ldots (2)$
In the year 2004, ${\rm{t}} = 5$ and ${\rm{y}} = 25000$.
Therefore,

we get:
$\log 25000 = k \cdot 5 + {\rm{C}}$
$\Rightarrow$ $\log 25000 = 5k + \log 20000$
$\Rightarrow$ $5k = \log \left( {\frac{{25000}}{{20000}}} \right) = \log \left( {\frac{5}{4}} \right)$

$\Rightarrow$ $k = \frac{1}{5}\log \left( {\frac{5}{4}} \right)$

……(3)
In the year $2009,$ ${\rm{t}} = 10$ years.
Now, on substituting the values of $t$, $k$, and $C$ in equation (1),

we get:
$\log y = 10 \times \frac{1}{5}\log \left( {\frac{5}{4}} \right) + \log (20000)$
$\Rightarrow$ $\log y = \log \left[ {20000 \times {{\left( {\frac{5}{4}} \right)}^2}} \right]$

$\Rightarrow$ $y = 20000 \times \frac{5}{4} \times \frac{5}{4}$
$\Rightarrow$ $y = 31250$

Hence, the population of the village in 2009 will be 31250.