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# The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time.If the population was 20,000 in 1999 and 25000 in the year of 2004,what will be the population of the village in 2009?

Toolbox:
• Word problems related to differential equations can be solved by framing a linear differential equation and then solve by seperating the variables.
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.