$\left( {x + 3{y^2}} \right)\cfrac{{dy}}{{dx}} = y\left( {y > 0} \right)$

: The given equation is $(x + 3{y^2})\cfrac{{dy}}{{dx}} = y$

$\Rightarrow x + 3{y^2} = y\cfrac{{dx}}{{dy}} \Rightarrow \cfrac{{dx}}{{dy}} - \cfrac{x}{y} = 3y$

… (1)
which is a linear equation of the type $\cfrac{{dy}}{{dx}} + Rx = S$

Where, $R = - \cfrac{1}{y}$ and $S = 3y$

$\therefore {\rm{I}}.{\rm{F}}. = {e^{\int {\cfrac{1}{y}dy} }} = {e^{\log {y^{ - 1}}}} = {y^{ - 1}} = \cfrac{1}{y}$

The solution is $x \cdot ({\rm{I}}.{\rm{F}}.) = \int S \cdot ({\rm{I}}.{\rm{F}}.)dy + C$

$\Rightarrow x \cdot \cfrac{1}{y} = \int {\cfrac{{3y}}{y}dy} + C \Rightarrow x \cdot \cfrac{1}{y} = 3\int {1 \cdot dy} + C \Rightarrow \cfrac{x}{y} = 3y + C$

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

which is the required solution.