Let y=y(x) be the solution of the differential equation (d y)/(d x)=(4 y^3+2 y x^2)/(3 x y^2+x^3), y(1)=1. If for some n ∈ N, y(2) ∈[n-1, n), then n…

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