**Find the particular solution of the following differential equation:**

`"dy"/"dx" - 3"y" cot "x" = sin "2x"`, when `"y"(pi/2) = 2`

#### Solution

`"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 cosec^{3} x = `int 2 sin "x" cos "x" xx 1/sin^3"x" "dx" + "c"`

∴ y cosec^{3} x = 2 `int (cos "x")/(sin^2 "x") "dx" + "c"`

Put sin x = t ∴ cos x dx = dt

∴ y cosec^{3} x = 2`int 1/"t"^2 "dt" + "c"`

∴ y cosec^{3} x = 2`int "t"^-2 "dt" + "c"`

∴ y cosec^{3} x = 2`["t"^-1/-1] + "c"`

∴ y cosec^{3} x = `(-2)/sin "x" + "c"`

∴ y cosec^{3} 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 cosec^{3} x + 2 cosec x = 4

∴ y cosec^{2} x + 2 = 4 sin x

#### Notes

[**Note:** Answer in the textbook is incorrect.]