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Question
Find the general solution of `(x + 2y^3) "dy"/"dx"` = y
Solution
Given equation is `(x + 2y^3) "dy"/"dx"` = y
⇒ `"dy"/"dx" = y/(x + 2y^3)`
⇒ `"dx"/"dy" = (x + 2y^3)/y`
⇒ `"dx"/"dy" = x/y + (2y^3)/y`
⇒ `"dx"/"dy" - x/y` = 2y3
Here P = `- 1/y` and Q = 2y2.
∴ Integrating factor I.F. = `"e"^(intPdy)`
= `"e"^(int 1/y dy)`
= `"e"^(-log y)`
= `"e"^(log 1/y)`
= `1/y`.
So the solution of the equation is
x.I.F. = `int "Q"."I"."F". "d"y + "c"`
`x . 1/y = int 2y^2 . 1/y "d"y + "c"`
⇒ `x/y = 2 int y "d"y + "c"`
⇒ `x/y = 2. y^2/2 + "c"`
⇒ `x/y = y^2 + "c"`
So x = y3 + cy = y(y2 + c)
Hence, the required solution is x = y(y2 + c).
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