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Question
The integrating factor of the differential equation `("d"y)/("d"x) - y` = x is ______
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Solution
e–x
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The integrating factor of `("d"y)/("d"x) + y` = e–x is
The rate of growth of population is proportional to the number present. If the population doubled in the last 25 years and present population is 1 lac., when will the city have population 4,00,000?
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 = 100000
∴ log 100000 = k × 0 + c
∴ c = `square`
∴ log p = kt + log 100000
∴ log p – log 100000 = kt
∴ `log ("P"/100000)` = kt ......(i)
Since the number doubled in 25 years, i.e., when t = 25, p = 200000
∴ `log (200000/100000)` = 25k
∴ k = `square`
∴ equation (i) becomes, `log("p"/100000) = square`
When p = 400000, then find t.
∴ `log(400000/100000) = "t"/25 log 2`
∴ `log 4 = "t"/25 log 2`
∴ t = `25 (log 4)/(log 2)`
∴ t = `square` years
Find the population of city at any time t given that rate of increase of population is proportional to the population at that instant and that in a period of 40 years the population increased from 30000 to 40000.
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" prop "p"`
∴ `"dp"/"dt"` = kp, where k is a constant.
∴ `"dp"/"p"` = k dt
On integrating, we get
`int "dp"/"p" = "k" int "dt"`
∴ log p = kt + c
Initially, i.e. when t = 0, let p = 30000
∴ log 30000 = k × 0 + c
∴ c = `square`
∴ log p = kt + log 30000
∴ log p - log 30000 = kt
∴ `log("p"/30000)` = kt .....(1)
when t = 40, p = 40000
∴ `log (40000/30000) = 40"k"`
∴ k = `square`
∴ equation (1) becomes, `log ("p"/30000)` = `square`
∴ `log ("p"/30000) = "t"/40 log (4/3)`
∴ p = `square`
The population of city doubles in 80 years, in how many years will it be triple when the rate of increase is proportional to the number of inhabitants. `("Given" log3/log2 = 1.5894)`
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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Let the population of rabbits surviving at a time t be governed by the differential equation `(dp(t))/dt = 1/2p(t) - 200`. If p(0) = 100, then p(t) equals ______
The rate of increase of bacteria in a certain culture is proportional to the number present. If it doubles in 7 hours, then in 35 hours its number would be ______.
The length of the perimeter of a sector of a circle is 24 cm, the maximum area of the sector is ______.
If a curve y = f(x) passes through the point (1, - 1) and satisfies the differential equation, y (1 + xy) dx = x dy, then `f(-1/2)` is equal to ______
In a certain culture of bacteria, the rate of increase is proportional to the number present. If it is found that the number doubles in 4 hours, find the number of times the bacteria are increased in 12 hours.
Solution:
Let N be the number of bacteria present at time ‘t’.
Since the rate of increase of N is proportional to N, the differential equation can be written as –
`(dN)/dt αN`
∴ `(dN)/dt` = KN, where K is constant of proportionality
∴ `(dN)/N` = k . dt
∴ `int 1/N dN = K int 1 . dt`
∴ log N = `square` + C ...(1)
When t = 0, N = N0 where N0 is initial number of bacteria.
∴ log N0 = K × 0 + C
∴ C = log N0
Also when t = 4, N = 2N0
∴ log (2 N0) = K . 4 + `square` ...[From (1)]
∴ `log((2N_0)/N_0)` = 4K,
∴ log 2 = 4K
∴ K = `square` ...(2)
Now N = ? when t = 12
From (1) and (2)
log N = `1/4 log 2 . (12) + log N_0`
log N – log N0 = 3 log 2
∴ `log(N_0/N_0)` = `square`
∴ N = 8 N0
∴ Bacteria are increased 8 times in 12 hours.
