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 $n$ is equal to _______.

Given,

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

Let $y=vx$

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

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

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

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

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

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

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

Given, $y\left(1\right)=1$

$\Rightarrow c=\ln 2$

So the equation becomes $\ln \left|{\left(\frac{y}{x}\right)}^3+\left(\frac{y}{x}\right)\right|=\ln x+\ln 2$

Now for $y\left(2\right)$

So putting $x=2$ in $\ln \left|{\left(\frac{y}{x}\right)}^3+\left(\frac{y}{x}\right)\right|=\ln x+\ln 2$

 We get, $\ln \left(\frac{y^3}{8}+\frac{y}{2}\right)=2\ln 2\Rightarrow \frac{y^3}{8}+\frac{y}{2}=4$

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

So, $y\left(2\right)\in [2,3)$

So on comparing with $y\left(2\right)\in [n-1,n)$

$\Rightarrow n=3$