Question

Solve the following differential equation:

`(1 + "x"^2) "dy"/"dx" + "y" = "e"^(tan^-1 "x")`

Answer 1

`(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.