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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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∴ `("dp")/"dt" prop "p"`
∴ Differential equation can be written as `("dp")/"dt" = "kp"`
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∴ `("dp")/"p" = "k.dt"`
On integrating we get
`square` = kt + c ...(i)
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∴ 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
