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Question
The population of a town grows at the rate of 10% per year. Using differential equation, find how long will it take for the population to grow 4 times.
Solution
Her `dx/dy prop x` [Since the increase in population speeds up with the increase in population] and let x be the population at any time t.
`:. dx/dy = rx` (where r is proportionality constant)
`:. dx/ x = r.dt`
integrating both sides
In x = rt + c, (where c is the integration constant)
`:. x = e^(rt + c)`
`x = Ke^(rt)` where `K = e^c`
Here r is the rate of increase and K is the initial population let x0 then t = 0
`x_0 = ke^o => k = x_0`
Given to find the time t taken to attain 4 times population, so `x = 4x_0`
So, `x = Ke^(rt)`
`=> 4x_0 = x_0e^0.10t`
`2 = e^0.05t`
Taking log on both sides
In 2 = In `e^(0.1t)`
0.69314 = 0.1t t = 6.9314
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