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

Given $$\frac{dy}{dx} = \frac{4y^3 + 2yx^2}{3xy^2 + x^3}$$ with $$y(1) = 1$$.

This is a homogeneous ODE, so set $$y = vx$$ giving $$\frac{dy}{dx} = v + x\frac{dv}{dx}$$.

$$v + xv' = \frac{4v^3x^3 + 2vx^3}{3xv^2x^2 + x^3} = \frac{4v^3 + 2v}{3v^2 + 1}$$ $$xv' = \frac{4v^3 + 2v}{3v^2 + 1} - v = \frac{4v^3 + 2v - 3v^3 - v}{3v^2 + 1} = \frac{v^3 + v}{3v^2 + 1} = \frac{v(v^2+1)}{3v^2+1}$$

Separating variables yields

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

Using partial fractions gives

$$\frac{3v^2+1}{v(v^2+1)} = \frac{1}{v} + \frac{2v}{v^2+1}$$

Integrating both sides leads to

$$\ln|v| + \ln(v^2+1) = \ln|x| + C$$ $$v(v^2+1) = kx$$

Substituting back $$v = \frac{y}{x}$$ yields

$$\frac{y}{x}\Bigl(\frac{y^2}{x^2} + 1\Bigr) = kx \implies y(y^2 + x^2) = kx^4$$

Applying $$y(1) = 1$$ gives $$1(1+1) = k$$ so $$k = 2$$, and hence

$$y(y^2 + x^2) = 2x^4$$

For $$x = 2$$ we have

$$y(y^2 + 4) = 32$$

Testing $$y = 2$$ gives $$2(4+4) = 16 < 32$$ and testing $$y = 3$$ gives $$3(9+4) = 39 > 32$$, so $$y(2) \in [2,3)$$, meaning $$n = 3$$.

The answer is $$\boxed{3}$$.