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Question
Choose the correct alternative:
The integrating factor of `("d"y)/("d"x) + y` = e–x is
Options
x
– x
ex
y = e–x
Solution
ex
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Solution: Let p be the population at time t.
Then the rate of increase of p is `"dp"/"dt"` which is proportional to p.
∴ `"dp"/"dt" ∝ "p"`
∴ `"dp"/"dt"` = kp, where k is a constant
∴ `"dp"/"p"` = kdt
On integrating, we get
`int "dp"/"p" = "k" int "dt"`
∴ log p = kt + c
Initially, i.e., when t = 0, let p = N
∴ log N = k × 0 + c
∴ c = `square`
When t = 80, p = 2N
∴ log 2N = 80k + log N
∴ log 2N – log N = 80k
∴ `log ((2"N")/"N")` = 80k
∴ log (2) = 80k
∴ k = `square`
∴ p = 3N, then t = ?
∴ log p = `log2/80 "t" + log "N"`
∴ log 3N – log N = `square`
∴ t = `square` = `square` years
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The rate of growth of population is proportional to the number present. If the population doubled in the last 25 years and the present population is 1,00,000, when will the city have population 4,00,000?
Let ‘p’ be the population at time ‘t’ years.
∴ `("dp")/"dt" prop "p"`
∴ Differential equation can be written as `("dp")/"dt" = "kp"`
where k is constant of proportionality.
∴ `("dp")/"p" = "k.dt"`
On integrating we get
`square` = kt + c ...(i)
(i) Where t = 0, p = 1,00,000
∴ from (i)
log 1,00,000 = k(0) + c
∴ c = `square`
∴ log `("p"/(1,00,000)) = "kt"` ...(ii)
(ii) When t = 25, p = 2,00,000
as population doubles in 25 years
∴ from (ii) log2 = 25k
∴ k = `square`
∴ log`("p"/(1,00,000)) = (1/25log2).t`
(iii) ∴ when p = 4,00,000
`log ((4,00,000)/(1,00,000)) = (1/25log2).t`
∴ `log 4 = (1/25 log2).t`
∴ t = `square ` years
