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Question
Solve the following differential equation:
x dy = (x + y + 1) dx
Solution
x dy = (x + y + 1) dx
∴ `"dy"/"dx" = ("x + y + 1")/"x" = ("x + 1")/"x" + "y"/"x"`
∴ `"dy"/"dx" - 1/"x" * "y" = ("x + 1")/"x"` ....(1)
This is the linear differential equation of the form
`"dy"/"dx" + "Py" = "Q",` where P = `- 1/"x" and "Q" = ("x + 1")/"x"`
∴ I.F. = `"e"^(int "P dx") = "e"^(int - 1/"x" "dx")`
`= "e"^(- log "x") = "e"^(log (1/"x")) = 1/"x"`
∴ the solution of (1) is given by
`"y" * ("I.F.") = int "Q" * ("I.F.")"dx" + "c"`
∴ `"y"*1/"x" = int ("x + 1")/"x" xx 1/"x" "dx" + "c"`
∴ `"y"/"x" = int ("x + 1")/"x"^2 "dx" + "c"`
∴ `"y"/"x" = int (1/"x" + 1/"x"^2) "dx" + "c"`
∴ `"y"/"x" = int 1/"x" "dx" + int "x"^-2 "dx" + "c"`
∴ `"y"/"x" = log |"x"| + "x"^-1/-1 + "c"`
∴ y = x log x - 1 + cx
This is the general solution.
Notes
The answer in the textbook is incorrect.
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