Question

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?

Answer 1

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
dy
dt

is proportional to y
dy
dt

=ky, where k is a proportional constant.
seperating the variables we get,
dy
y

=kdt
now integrating on both sides we get,
logy=kt+C-------(1)
Step 2:
In the year 1999, when t = 0 and y = 20,000
substituting this in equ(1)
log20,000=C--------(2)
In the year 2004, when t= 5 y = 25,000
substituing this in equ(1) we get,
log25,000=5k+C
Substituting for C we get
log25,000=5k+log20,000
log
25000
20,000

=5k
log
5
4

=5k
Or k=(
1
5

)log(
5
4

)
Step 3:
In the year 2009,when t=10
Substituting for t,k and C equ(1) becomes,
logy=10(
1
5

)log(
5
4

)+log20,000
logy=2log(5/4)+log20,000=log(
5
4

)2+log20,000
logy=log20,000×(5/4)2
y=20,000×
5
4

×
5
4

=31250
Hence the population in the year 2009 is 31250.

i hope this will help.