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प्रश्न
Solve the following differential equation:
`(1 + "x"^2) "dy"/"dx" + "y" = "e"^(tan^-1 "x")`
उत्तर
`(1 + "x"^2) "dy"/"dx" + "y" = "e"^(tan^-1 "x")`
∴ `"dy"/"dx" + 1/(1 + "x"^2) * "y" = "e"^(tan^-1 "x")/(1 + "x"^2)` ....(1)
This is the linear differential equation of the form
`"dy"/"dx" + "P" * "y" = "Q",` where P = `1/(1 + "x"^2)` and Q = `"e"^(tan^-1 "x")/(1 + "x"^2)`
∴ I.F. = `"e"^(int "P dx") = "e"^(int 1/"1 + x"^2"dx")`
`= "e"^(tan^-1 "x")`
∴ the solution of (1) is given by
`"y" * ("I.F.") = int "Q" * ("I.F.") "dx" + "c"`
∴ `"y" * "e"^(tan^-1"x") = int "e"^(tan^-1 "x")/(1 + "x"^2) * "e"^(tan^-1"x") "dx" + "c"`
∴ `"y" * "e"^(tan^-1"x") = int ("e"^(tan^-1 "x")) * ("e"^(tan^-1"x")/(1 + "x"^2)) "dx" + "c"`
Put `"e"^(tan^-1"x") = "t"`
∴ `"e"^(tan^-1"x")/(1 + "x"^2) "dx" = "dt"`
∴ `"y" * "e"^(tan^-1"x") = int "t dt" + "c"`
∴ `"y" * "e"^(tan^-1"x") = "t"^2/2 + "c"`
∴ `"y" * "e"^(tan^-1"x") = 1/2 ("e"^(tan^-1"x"))^2 + "c"`
∴ y = `1/2 "e"^(tan^-1"x") + "ce"^(- tan^-1 "x")`
This is the general solution.
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