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प्रश्न
Find the particular solution of the following differential equation:
`"dy"/"dx" - 3"y" cot "x" = sin "2x"`, when `"y"(pi/2) = 2`
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उत्तर
`"dy"/"dx" - 3"y" cot "x" = sin "2x"`
∴ `"dy"/"dx" - (3 "cot x")"y" = sin "2x"` ....(1)
This is the linear differential equation of the form
`"dx"/"dy" + "Px" = "Q"` where P = `- 3 cot "x"` and Q = sin 2x.
∴ I.F. = `"e"^(int "P dy") = "e"^(int - 3 cot "x" "dx")`
`= "e"^(- 3 log sin "x") = "e"^(log (sin "x")^-3)`
`= (sin "x")^-3 = 1/(sin^3"x")`
∴ the solution of (1) is given by
`"x" * ("I.F.") = int "Q" * ("I.F.") "dy" + "c"`
∴ `"y" xx 1/(sin^3 "x") = int sin "2x" xx 1/(sin "3x") "dx" + "c"`
∴ y cosec3 x = `int 2 sin "x" cos "x" xx 1/sin^3"x" "dx" + "c"`
∴ y cosec3 x = 2 `int (cos "x")/(sin^2 "x") "dx" + "c"`
Put sin x = t ∴ cos x dx = dt
∴ y cosec3 x = 2`int 1/"t"^2 "dt" + "c"`
∴ y cosec3 x = 2`int "t"^-2 "dt" + "c"`
∴ y cosec3 x = 2`["t"^-1/-1] + "c"`
∴ y cosec3 x = `(-2)/sin "x" + "c"`
∴ y cosec3 x + 2 cosec x = c
This is the general solution.
Now, `"y"(pi/2) = 2`, i.e. y = 2, when x = `pi/2`
∴ `2 "cosec"^3 pi/2 + 2 "cosec" pi/2 = "c"`
∴ 2(1)3 + 2(1) = c
∴ c = 4
∴ the particular solution is
y cosec3 x + 2 cosec x = 4
∴ y cosec2 x + 2 = 4 sin x
Notes
The answer in the textbook is incorrect.
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