Let $$y = y(x)$$ be the solution of the differential equation $$(x - x^3)dy = (y + yx^2 - 3x^4)dx$$, $$x \gt 2$$. If $$y(3) = 3$$, then $$y(4)$$ is equal to:

We are given the differential equation $$ (x - x^3)dy = (y + yx^2 - 3x^4)dx $$ for $$ x > 2 $$.

Rearranging the terms to write the equation in standard linear form, we get $$ (x - x^3)\frac{dy}{dx} - (1 + x^2)y = -3x^4 $$.

Dividing both sides by $$ x - x^3 $$ or $$ x(1 - x^2) $$, the equation becomes $$ \frac{dy}{dx} - \frac{1 + x^2}{x(1 - x^2)}y = \frac{-3x^4}{x(1 - x^2)} $$.

Simplifying the coefficients gives $$ \frac{dy}{dx} + \frac{1 + x^2}{x(x^2 - 1)}y = \frac{3x^3}{x^2 - 1} $$.

This is a linear differential equation of the first order where $$ P(x) = \frac{1 + x^2}{x(x^2 - 1)} $$ and $$ Q(x) = \frac{3x^3}{x^2 - 1} $$.

To find the integrating factor, we compute $$ \int P(x) dx = \int \left(-\frac{1}{x} + \frac{1}{x - 1} + \frac{1}{x + 1}\right) dx = \ln\left(\frac{x^2 - 1}{x}\right) $$.

Thus, the integrating factor is $$ IF = e^{\ln\left(\frac{x^2 - 1}{x}\right)} = \frac{x^2 - 1}{x} $$.

Multiplying the differential equation by the integrating factor yields the general solution format $$ y \cdot \left(\frac{x^2 - 1}{x}\right) = \int 3x^2 dx $$.

Integrating the right side gives $$ y \cdot \left(\frac{x^2 - 1}{x}\right) = x^3 + C $$.

Using the given condition $$ y(3) = 3 $$, we substitute $$ x = 3 $$ and $$ y = 3 $$ to find the constant $$ C $$:

$$ 3 \cdot \left(\frac{3^2 - 1}{3}\right) = 3^3 + C $$

$$ 8 = 27 + C \implies C = -19 $$.

Thus, the specific solution equation is $$ y \cdot \left(\frac{x^2 - 1}{x}\right) = x^3 - 19 $$.

Now, we substitute $$ x = 4 $$ to find $$ y(4) $$:

$$ y(4) \cdot \left(\frac{4^2 - 1}{4}\right) = 4^3 - 19 $$

$$ y(4) \cdot \left(\frac{15}{4}\right) = 64 - 19 = 45 $$

$$ y(4) = 45 \cdot \frac{4}{15} = 12 $$.

Hence the correct option is B.