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(A) The given differential equation is $x\sqrt{1-x^2} dy + (y\sqrt{1-x^2} - x\cos^{-1}x) dx = 0$.Dividing by $dx$ and $x\sqrt{1-x^2}$, we get $\frac{d

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$3 - \frac{\pi}{\sqrt{3}}$

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(A) The given differential equation is $x\sqrt{1-x^2} dy + (y\sqrt{1-x^2} - x\cos^{-1}x) dx = 0$.Dividing by $dx$ and $x\sqrt{1-x^2}$, we get $\frac{dy}{dx} + \frac{y}{x} = \frac{\cos^{-1}x}{\sqrt{1-x^2}}$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$, where $P = \frac{1}{x}$ and $Q = \frac{\cos^{-1}x}{\sqrt{1-x^2}}$.The Integrating Factor ($I$.$F$.) is $e^{\int P dx} = e^{\int (1/x) dx} = e^{\ln x} = x$.The solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$, so $yx = \int \frac{x\cos^{-1}x}{\sqrt{1-x^2}} dx$.Let $t = \cos^{-1}x$, then $dt = -\frac{dx}{\sqrt{1-x^2}}$ and $x = \cos t$.Substituting these, $yx = -\int t \cos t dt = -(t \sin t + \cos t) + C = -(\cos^{-1}x \sqrt{1-x^2} + x) + C$.Given $\lim_{x	o 1^-} y(x) = 1$, as $x 	o 1$, $yx 	o 1$. Thus, $1 = -(\cos^{-1}(1) \cdot 0 + 1) + C$, which gives $1 = -1 + C$, so $C = 2$.Therefore, $yx = 2 - x - \sqrt{1-x^2} \cos^{-1}x$.At $x = \frac{1}{2}$, $y(\frac{1}{2}) \cdot \frac{1}{2} = 2 - \frac{1}{2} - \sqrt{1 - (1/2)^2} \cos^{-1}(1/2) = \frac{3}{2} - \frac{\sqrt{3}}{2} \cdot \frac{\pi}{3} = \frac{3}{2} - \frac{\pi}{2\sqrt{3}}$.Multiplying by $2$, we get $y(\frac{1}{2}) = 3 - \frac{\pi}{\sqrt{3}}$.

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

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