Question

Find the general solution of `(x + 2y^3)  "dy"/"dx"` = y

Answer 1

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).