Let $y=y\left(x\right)$ be the solution of the differential equation  $\frac{dy}{dx}=\frac{4y^3+2y{x}^2}{3x{y}^2+x^3},y\left(1\right)=1.$ If for some $n\in N,y\left(2\right)\in [n-1,n),$ then $2n$ is equal to ____________ 

Given differential equations is

$\frac{dy}{dx}=\frac{y}{x}\frac{(4y^2+2x^2)}{(3y^2+x^2)}$

Put $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

$\Rightarrow v+x\frac{dv}{dx}=\frac{v(4v^2+2)}{(3v^2+1)}$

$\Rightarrow v+x\frac{dv}{dx}=\frac{v(4v^2+2)}{(3v^2+1)}$

$\Rightarrow x\frac{dv}{dx}=v\left(\frac{(4v^2+2-3v^2-1)}{3v^2+1}\right)$

$\Rightarrow x\frac{dv}{dx}=\frac{v^3+v}{3v^2+1}\Rightarrow \frac{3v^2+1}{v^3+v}dv=\frac{dx}{x}$

Take integral both sides,

$\Rightarrow \int (3v^2+1)\frac{dv}{v^3+v}=\int \frac{dx}{x}$

$\Rightarrow \ln |v^3+v|=\ln x+c$

$\Rightarrow \ln |{\left(\frac{y}{x}\right)}^3+\left(\frac{y}{x}\right)|=\ln x+C$

Here,$y\left(1\right)=1$ and  $C=\ln 2$

Here $y\left(2\right)\in [n-1,n)$

$\ln (\frac{y^3}{8}+\frac{y}{2})=2\ln 2\Rightarrow \frac{y^3}{8}+\frac{y}{2}=4$

$\Rightarrow \left[y\left(2\right)\right]=2$

Therefore,
 $n=3\Rightarrow 2n=6$