## Book: Mathematics Part-II

### Chapter: 9. Differential Equations

#### Subject: Maths - Class 12th

##### Q. No. 15 of Miscellaneous Exercise

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15
##### 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 20, 000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009?

Let the population at any instant (t) be y.

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

Now, integrating both sides, we get,

log y = kt + C ------(1)

According to given conditions,

In the year 1999, t = 0 and y = 20000

log20000 = C ----(2)

Also, in the year 2004, t = 5 and y = 25000

log 25000 = k.5 + C

log 25000 = 5k + log 20000

--------(3)

Also, in the year 2009, t = 10

Now, substituting the values of t, k and c in equation (1), we get

logy =

y = 31250

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

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