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Question
Find the particular solution of the following differential equation:
`("x + 2y"^2) "dy"/"dx" = "y",` when x = 2, y = 1
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Solution
`("x + 2y"^2) "dy"/"dx" = "y"`
∴ `"dy"/"dx" = ("x + 2y"^2)/"y" = "x"/"y" + "2y"`
∴ `"dx"/"dy" - 1/"y" * "x" = "2y"` ....(1)
This is the linear differential equation of the form
`"dx"/"dy" + "Px" = "Q"` where P = `- 1/"y"` and Q = 2y.
∴ I.F. = `"e"^(int "P dy") = "e"^(int - 1/"y" "dy")`
`= "e"^(- log "y") = "e"^(log (1/"y")) = 1/"y"`
∴ the solution of (1) is given by
`"x" * ("I.F.") = int "Q" * ("I.F.") "dy" + "c"`
∴ `"x" xx 1/"y" = int "2y" xx 1/"y" "dy" + "c"`
∴ `"x"/"y" = 2 int 1 "dy" + "c"`
∴ `"x"/"y" = 2"y" + "c"`
∴ x = 2y2 + cy
This is the general solution.
When x = 2, y = 1, we have
2 = 2(1)2 + c(1)
∴ c = 0
∴the particular solution is x = 2y2.
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