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प्रश्न
Show that the general solution of differential equation `"dy"/"dx" + ("y"^2 + "y" + 1)/("x"^2 + "x" + 1) = 0` is given by (x + y + 1) = (1 - x - y - 2xy).
उत्तर
`"dy"/"dx" + ("y"^2 + "y" + 1)/("x"^2 + "x" + 1) = 0`
∴ `"dy"/"dx" = - (("y"^2 + "y" + 1)/("x"^2 + "x" + 1))`
∴ `1/("y"^2 + "y" + 1)"dy" = - 1/("x"^2 + "x" + 1)"dx"`
Integrating both sides, we get
`int 1/("y"^2 + "y" + 1)"dy" = - int 1/("x"^2 + "x" + 1)"dx"`
∴ `int 1/(("y"^2 + "y" + 1/4) + 3/4) "dy" = - int 1/(("x"^2 + "x" + 1/4) + 3/4)"dx"`
∴ `int 1/(("y" + 1/2)^2 + (sqrt3/2)^2)"dy" = - int 1/(("x" + 1/2)^2 + (sqrt3/2)^2)"dx"`
∴ `1/(sqrt3/2) tan^-1 [("y" + 1/2)/(sqrt3/2)] = - 1/((sqrt3/2)) tan^-1 [("x" + 1/2)/(sqrt3/2)] + "c"_1`
∴ `2/sqrt3 tan^-1 (("2y" + 1)/sqrt3) + 2/sqrt3 tan^-1 (("2x" + 1)/sqrt3) = "c"_1`
∴ `2/sqrt3 tan^-1 [((("2y" + 1)/sqrt3) + (("2x" + 1)/sqrt3))/(1 - (("2y" + 3)/sqrt3)(("2x" + 1)/sqrt3))] = "c"_1`
∴ `tan^-1 ((("2y" + 1 + 2"x" + 1)/sqrt3))/(((3 - 4"xy" - 2"y" - 2"x" - 1)/3)) = sqrt3/2 "c"_1`
∴ `(("2y" + "2x" + 2)sqrt3)/(2 - "2x" - 2"y" - 4"xy") = tan (sqrt3/2 "c"_1)`
∴ `("x + y + z")/(1 - "x" - "y" - 2"xy") = 1/sqrt3 tan (sqrt3/2 "c"_1) = "c"`, where c = `1/sqrt3 tan (sqrt3/2 "c"_1)`
∴ (x + y + 1) = c(1 - x - y - 2xy)
This is the general solution.
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∴ `"dx"/"dt"` = kx, where k is a constant
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∴ log x = kt + log x0
∴ log x - log x0 = kt
∴ `log ("x"/"x"_0)`= kt ......(1)
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x = 2x0
∴ `log ((2"x"_0)/"x"_0)` = 4k
∴ k = `square`
∴ equation (1) becomes, `log ("x"/"x"_0) = "t"/4` log 2
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`log ("x"/"x"_0) = 12/4` log 2 = 3 log 2
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∴ `"x"/"x"_0 = 8`
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∴ number of bacteria will be 8 times the original number in 12 hours.
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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
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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.
Bacteria increase at the rate proportional to the number of bacteria present. If the original number N doubles in 3 hours, find in how many hours the number of bacteria will be 4N?
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
