Let $$y = y(x)$$ be the solution of the differential equation $$x(1 - x^2)\frac{dy}{dx} + (3x^2y - y - 4x^3) = 0, x > 1$$ with $$y(2) = -2$$. Then $$y(3)$$ is equal to

We begin with the differential equation $$x(1 - x^2)\frac{dy}{dx} + (3x^2 - 1)y = 4x^3$$. Dividing by $$x(1 - x^2)$$ rewrites it in standard linear form as $$\frac{dy}{dx} + \frac{3x^2 - 1}{x(1 - x^2)} \cdot y = \frac{4x^2}{1 - x^2}$$.

To find the integrating factor, we set $$P(x) = \frac{3x^2 - 1}{x(1-x^2)}$$ and decompose it into partial fractions. Since $$1 - x^2 = -(x-1)(x+1)$$, we write $$\frac{-(3x^2-1)}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}$$. Setting $$x = 0$$ gives $$\frac{1}{(-1)(1)} = A \implies A = -1$$, setting $$x = 1$$ gives $$\frac{-2}{1 \cdot 2} = B \implies B = -1$$, and setting $$x = -1$$ gives $$\frac{-2}{-1 \cdot (-2)} = C \implies C = -1$$. Thus $$P(x) = -\frac{1}{x} - \frac{1}{x-1} - \frac{1}{x+1}$$.

Integrating, we get $$\int P(x) \, dx = -\ln|x| - \ln|x-1| - \ln|x+1| = -\ln|x(x^2-1)|$$, so the integrating factor is $$\mathrm{IF} = e^{\int P \, dx} = \frac{1}{x(x^2-1)}$$.

Multiplying the standard form by the integrating factor, the left side becomes $$\frac{d}{dx}\left(\frac{y}{x(x^2-1)}\right)$$, while the right side simplifies as $$\frac{4x^2}{1-x^2} \cdot \frac{1}{x(x^2-1)} = \frac{4x^2}{-(x^2-1)} \cdot \frac{1}{x(x^2-1)} = \frac{-4x}{(x^2-1)^2}$$.

To integrate the right side, we use the substitution $$u = x^2 - 1$$ and $$du = 2x \, dx$$, yielding $$\int \frac{-4x}{(x^2-1)^2} \, dx = -2 \int \frac{du}{u^2} = \frac{2}{u} = \frac{2}{x^2-1} + C$$. Hence $$\frac{y}{x(x^2-1)} = \frac{2}{x^2-1} + C$$, which leads to $$y = 2x + Cx(x^2-1)$$.

Applying the initial condition $$y(2) = -2$$ gives $$-2 = 2(2) + C(2)(4-1) = 4 + 6C$$, so $$6C = -6 \implies C = -1$$. Therefore $$y = 2x - x(x^2 - 1) = 2x - x^3 + x = 3x - x^3$$.

Finally, evaluating at $$x = 3$$ yields $$y(3) = 3(3) - 3^3 = 9 - 27 = -18$$, so the correct answer is Option A: $$-18$$.