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प्रश्न
Solve the following differential equation:
`"dy"/"dx" + "y cot x" = "x"^2 "cot x" + "2x"`
उत्तर
`"dy"/"dx" + "y cot x" = "x"^2 "cot x" + "2x"` ....(1)
This is the linear differential equation of the form
`"dy"/"dx" + "Py" = "Q",` where P = cot x and Q = x2 cot x + 2x
∴ I.F. = `"e"^(int "P dx") = "e"^(int "cot x" "dx")`
`= "e"^(log ("sin x")) = sin "x"`.
∴ the solution of (1) is given by
`"y" * ("I.F.") = int "Q" * ("I.F.")"dx" + "c"`
∴ y sin x = `int ("x"^2 "cot x" + "2x") "sin x dx" + "c"`
∴ y sin x = `int ("x"^2 cot "x" * sin "x" + "2x" sin "x") "dx" + "c"`
∴ y sin x = `int "x"^2 cos "x" "dx" + 2 int "x" sin "x" "dx" + "c"`
∴ y sin x = `"x"^2 int "cos x dx" - int["d"/"dx" ("x"^2) int "cos x dx"] "dx" + 2 int "x sin x dx" + "c"`
`= "x"^2 (sin "x") - int "2x" (sin "x")"dx" + 2 int "x sin x dx" + "c"`
`= "x"^2 sin x - 2 int "x" sin "x" "dx" + 2 int "x sin x dx" + "c"`
∴ y sin x = x2 sin x + c
∴ y = x2 + c cosec x
This is the general solution.
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