Find the general solution of
$\left( {x + 2{y^3}} \right)\frac{{dy}}{{dx}} = y$.

Given that, $\left( {x + 2{y^3}} \right)\frac{{dy}}{{dx}} = y$

$\Rightarrow$ $y \cdot \frac{{dx}}{{dy}} = x + 2{y^3}$

$\Rightarrow$ $\frac{{dx}}{{dy}} = \frac{x}{y} + 2{y^2}$

[dividing throughout by y]

$\Rightarrow$ $\frac{{dx}}{{dy}} - \frac{x}{y} = 2{y^2}$

which is a linear differential equation.
On comparing it with

$\frac{{dx}}{{dy}} + Px = Q,$

we get
$P = - \frac{1}{y},Q = 2{y^2}$

${\rm{IF}} = {e^{\int - \frac{1}{y}dy}} = {e^{ - \int {\frac{1}{y}} dy}}$

$\therefore = {e^{ - \log y}} = \frac{1}{y}$

The general solution is
$x \cdot \frac{1}{y} = \int 2 {y^2} \cdot \frac{1}{y}dy + C$

$\Rightarrow$ $\frac{x}{y} = \frac{{2{y^2}}}{2} + C$

$\Rightarrow$ $\frac{x}{y} = {y^2} + C$
$\Rightarrow$ $x = {y^3} + Cy$