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Question
Solve the following differential equation:
`("x + y") "dy"/"dx" = 1`
Solution
`("x + y") "dy"/"dx" = 1`
∴ `"dx"/"dy" = "x + y"`
∴ `"dx"/"dy" - "x" = "y"`
∴ `"dx"/"dy" + (- 1)"x" = "y"` ....(1)
This is the linear differential equation of the form
`"dx"/"dy" + "P"*"x" = "Q",` where P = - 1 and Q = y
∴ I.F. = `"e"^(int "P dy") = "e"^(int - 1 "dy") = "e"^-"y"`
∴ the solution of (1) is given by
x.(I.F.) = `int "Q" * ("I.F.") "dy" + "c"`
∴ `"x" * "e"^-"y" = int "y" * "e"^-"y" "dy" + "c"`
∴ `"e"^-"y" * "x" = "y" int "e"^-"y" "dy" - int ["d"/"dx" ("y") int "e"^-"y" "dy"] "dy" + "c"`
`= "y" * ("e"^-"y")/-1 - int 1 * ("e"^-"y")/-1 "dy" + "c"`
`= - "ye"^-"y" + int "e"^-"y" "dy" + "c"`
∴ `"e"^-"y" * "x" = - "ye"^-"y" + "e"^-"y"/-1 + "c"`
∴ `"e"^-"y" * "x" + "ye"^-"y" + "e"^-"y" = "c"`
∴ `"e"^"-y" ("x + y + 1") = "c"`
∴ x + y + 1 = cey
This is the general solution.
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