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Question
Solve the following differential equation:
`("x" + 2"y"^3) "dy"/"dx" = "y"`
Solution
`("x" + 2"y"^3) "dy"/"dx" = "y"`
∴ `("x" + "2y"^3)/"y" = 1/(("dy"/"dx"))`
∴ `"x"/"y" + 2"y"^2 = "dx"/"dy"`
∴ `"dy"/"dx" - 1/"y" * "x" = 2"y"^2` .....(1)
This is the linear differential equation of the form
`"dx"/"dy" + "P"*"x" = "Q"`, where P = `- 1/"y"` and Q = 2y2
∴ I.F. = `"e"^(int "Pdy") = "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"(1/"y") = int 2"y"^2 xx 1/"y" "dy" + "c"`
∴ `"x"/"y" = 2 int "y" "dx" + "c"`
∴ `"x"/"y" = 2 * "y"^2/2 + "c"`
∴ x = y(c + y2)
This is the general solution.
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