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Question
Solve the following differential equation:
y dx + (x - y2) dy = 0
Solution
y dx + (x - y2) dy = 0
∴ y dx = - (x - y2) dy
∴ `"dx"/"dy" = - (("x - y"^2))/"y" = - "x"/"y" + "y"`
∴ `"dx"/"dy" + (1/"y") * "x" = "y"` ....(1)
This is the linear differential equation of the form
`"dx"/"dy" + "P" * "x" = "Q"`, where P = `1/"y"` and Q = y
∴ I.F. = `"e"^(int "P dy") = "e"^(int 1/"y" "dy") = "e"^(log "y") = "y"`
∴ the solution of (1) is given by
`"x" * ("I.F.") = int "Q" * (I.F.) "dy" + "c"_1`
∴ `"xy" = int "y" * "y" "dy" + "c"_1`
∴ `"xy" = int "y"^2 "dy" + "c"_1`
∴ `"xy" = "y"^3/3 + "c"_1`
∴ `"y"^3/3 = "xy" + "c"`, where c = -c1
This is the general solution.
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