Let $$y = y(x)$$ be the solution of the differential equation $$x\sqrt{1-x^2} \, dy + \left(y\sqrt{1-x^2} - x\cos^{-1}x\right)dx = 0$$, $$x \in (0,1)$$, $$\lim_{x \to 1^-} y(x) = 1$$. Then $$y\left(\frac{1}{2}\right)$$ equals :

This is a first-order linear differential equation. Let's rewrite it in the standard form $$\frac{dy}{dx} + P(x)y = Q(x)$$.

The given equation is:

$$x\sqrt{1-x^2} \, dy + (y\sqrt{1-x^2} - x\cos^{-1}x) \, dx = 0$$

Divide the entire equation by $$dx$$ and rearrange:

$$x\sqrt{1-x^2} \frac{dy}{dx} + y\sqrt{1-x^2} = x\cos^{-1}x$$

Divide by $$x\sqrt{1-x^2}$$ to get the standard form:

$$\frac{dy}{dx} + \frac{1}{x}y = \frac{\cos^{-1}x}{\sqrt{1-x^2}}$$

Here, $$P(x) = \frac{1}{x}$$.

$$\text{I.F.} = e^{\int P(x) \, dx} = e^{\int \frac{1}{x} \, dx} = e^{\ln x} = x$$

The solution is given by $$y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) \, dx$$:

$$yx = \int \frac{\cos^{-1}x}{\sqrt{1-x^2}} \cdot x \, dx$$

To solve the integral on the right, use substitution:

Let $$t = \cos^{-1}x$$. Then $$dt = -\frac{1}{\sqrt{1-x^2}} dx$$ and $$x = \cos t$$.

$$\int \frac{\cos^{-1}x}{\sqrt{1-x^2}} \cdot x \, dx = \int t \cdot \cos t \cdot (-dt) = -\int t \cos t \, dt$$

Using integration by parts ($$\int u \, dv = uv - \int v \, du$$):

$$-\int t \cos t \, dt = -(t \sin t - \int \sin t \, dt) = -(t \sin t + \cos t) + C$$

Substitute back $$t = \cos^{-1}x$$ and $$\sin t = \sqrt{1-x^2}$$:

$$yx = -(\cos^{-1}x \cdot \sqrt{1-x^2} + x) + C$$

We are given $$\lim_{x \to 1^-} y(x) = 1$$.

As $$x \to 1$$:

$$(1)(1) = -(\cos^{-1}(1) \cdot \sqrt{1-1^2} + 1) + C$$

$$1 = -(0 \cdot 0 + 1) + C \implies 1 = -1 + C \implies C = 2$$

So, the particular solution is:

$$yx = 2 - x - \sqrt{1-x^2} \cos^{-1}x$$

$$y = \frac{2}{x} - 1 - \frac{\sqrt{1-x^2}}{x} \cos^{-1}x$$

Substitute $$x = 1/2$$:

$$y(1/2) = \frac{2}{1/2} - 1 - \frac{\sqrt{1-(1/4)}}{1/2} \cos^{-1}(1/2)$$

$$y(1/2) = 4 - 1 - \frac{\sqrt{3}/2}{1/2} \cdot \frac{\pi}{3}$$

$$y(1/2) = 3 - \sqrt{3} \cdot \frac{\pi}{3} = 3 - \frac{\pi}{\sqrt{3}}$$

Correct Option: A