Question

Solve the following differential equation:

`cos^2 "x" * "dy"/"dx" + "y" = tan "x"`

Answer 1

`cos^2 "x" * "dy"/"dx" + "y" = tan "x"`

`∴ "dy"/"dx" + 1/cos^2"x" * "y" = (tan "x")/(cos^2"x")`

∴ `"dy"/"dx" + sec^2"x" * "y" = tan "x" * sec^2 "x"`       ....(1)

This is the linear differential equation of the form

`"dy"/"dx" + "P"*"y" = "Q"`, where P = sec²x and Q = `tan "x" * sec^2 "x"`

∴ I.F. = `"e"^(int "Pdx") = "e"^(int sec^2"x"  "dx") = "e"^(tan"x")`

∴ the solution of (1) is given by

`"y" * ("I.F.") = int "Q" (I.F.) "dx" + "c"`

∴ `"y"*"e"^"tan x" = int "tan x" * sec^2"x" * "e"^(tan"x") "dx" + "c"`

Put tan x = t

∴ `sec^2"x"  "dx" = "dt"`

∴ `"y" * "e"^"tan x" = int "t" * "e"^"t" "dt" + "c"`

∴ `"y" * "e"^"tan x" = "t" int "e"^"t" "dt" - int["d"/"dt" ("t") int "e"^"t" "dt"] "dt" + "c"`

`= "t" * "e"^"t" - int 1 * "e"^"t" "dt" + "c"`

`= "t" * "e"^"t" - "e"^"t" + "c"`

`= "e"^"t" ("t - 1") + "c"`

∴ `"y" * "e"^"tan x" = "e"^"tan x" (tan"x" - 1) + "c"`

This is the general solution.