Let $$y = y(x)$$ be the solution of the differential equation $$(1 - x^2)dy = \left[xy + (x^3 + 2)\sqrt{3(1 - x^2)}\right]dx$$, $$-1 < x < 1$$, $$y(0) = 0$$. If $$y\left(\frac{1}{2}\right) = \frac{m}{n}$$, $$m$$ and $$n$$ are coprime numbers, then $$m + n$$ is equal to __________.

We begin by solving the differential equation $$(1-x^2)\,dy = \bigl[xy + (x^3+2)\sqrt{3(1-x^2)}\bigr]\,dx$$ with the initial condition $$y(0)=0$$. Dividing both sides by $$(1-x^2)$$ gives $$\frac{dy}{dx} = \frac{xy}{1-x^2} + \frac{(x^3+2)\sqrt{3(1-x^2)}}{1-x^2} = \frac{xy}{1-x^2} + \frac{(x^3+2)\sqrt{3}}{\sqrt{1-x^2}}\,. $$

This is a linear first-order equation of the form $$\frac{dy}{dx} - \frac{x}{1-x^2}\,y = \frac{(x^3+2)\sqrt{3}}{\sqrt{1-x^2}}\,. $$ The integrating factor is $$e^{-\int \frac{x}{1-x^2}\,dx} = e^{\frac12\ln(1-x^2)} = \sqrt{1-x^2}\,. $$

Multiplying through by this integrating factor yields $$\frac{d}{dx}\bigl[\sqrt{1-x^2}\,y\bigr] = (x^3+2)\sqrt{3}\,. $$ Integrating both sides with respect to $$x$$ gives $$\sqrt{1-x^2}\,y = \sqrt{3}\Bigl(\frac{x^4}{4} + 2x\Bigr) + C\,. $$

Applying the initial condition $$y(0)=0$$ shows that at $$x=0$$, $$1\cdot 0 = 0 + C$$, so $$C=0$$. Hence $$y = \frac{\sqrt{3}\bigl(\tfrac{x^4}{4} + 2x\bigr)}{\sqrt{1-x^2}}\,. $$

Evaluating this expression at $$x=\tfrac12$$ gives $$y\bigl(\tfrac12\bigr) = \frac{\sqrt{3}\bigl(\tfrac{1}{64} + 1\bigr)}{\sqrt{3/4}} = \frac{\sqrt{3}\cdot \tfrac{65}{64}}{\tfrac{\sqrt{3}}{2}} = \frac{65/64}{1/2} = \frac{65}{32}\,. $$ Thus $$m=65$$, $$n=32$$ and $$m+n=97$$.

The correct answer is $$\boxed{97}$$.