Solve differential equation xy'= 6y+12x^4 y^(2/3)

$dy/dx+P(x)y=Q(x)y^n$
$n=0$ or $n=1$
Substituting $v=y^1-n$ transforms the above equation into the linear equation
$(dv)/dx+(1-n)P(x)v=(1-n)Q(x)$
$(dv)/dx+P(x)y=Q(x)$
$y(x)=e^{(\int P(x)dx)}[\int (Q(x)e^{(\int P(x)dx))}dx+C]$
The differential equation is,
$x{y}^{\prime }=6y+12x^4{y}^{(2/3)}$
The above equation can be written as follows
$y^{\prime }-6/xy=12x^3{y}^{(2/3)}$
Substituting $v=y^1-2/3$ or y $1/3$ transforms the above Bernoulli equation into the linear equation
$(dv)/dx-2/xv=4x^3$
$v(x)=e^{(-\int (2/x)dx)}[\int (4x^3{e}^{(\int (-2/x)dx))}dx+C^{\prime }]$
$=e^{(2\ln x)}[\int (4x^3{x}^{-}2\ln x)dx+C^{\prime }]$
$=x^2[4x^2/x+C^{\prime }]$
$v(x)=2x^4+C{x}^2$
$y1/3=2x^4+C{x}^2$
$y(x)=(2x^4+C{x}^2{)}^3$
where C is constant