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प्रश्न
Solve the following differential equation:
`cos^2 "x" * "dy"/"dx" + "y" = tan "x"`
उत्तर
`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.
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