Let y=y(x) be the solution of the differential equation x 1-x^2...

The given differential equation is:

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

Rearranging the terms, we get:

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

Dividing the entire equation by $\sqrt{1-x^2}$:

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

The left side is the exact differential of $xy$:

$d(xy) = \dfrac{x\cos^{-1}x}{\sqrt{1-x^2}}\,dx$

Integrating both sides:

$xy = \int \dfrac{x\cos^{-1}x}{\sqrt{1-x^2}}\,dx + C$

To evaluate the integral, use integration by parts. Let $u = \cos^{-1}x$ and $dv = \dfrac{x}{\sqrt{1-x^2}}\,dx$.

$du = -\dfrac{1}{\sqrt{1-x^2}}\,dx$ and $v = -\sqrt{1-x^2}$

$\int u\,dv = uv - \int v\,du$

$\int \dfrac{x\cos^{-1}x}{\sqrt{1-x^2}}\,dx = -\sqrt{1-x^2}\cos^{-1}x - \int \left(-\sqrt{1-x^2}\right)\left(-\dfrac{1}{\sqrt{1-x^2}}\right)\,dx$

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

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

Substituting this back into the equation:

$xy = -\sqrt{1-x^2}\cos^{-1}x - x + C$

It is given that $\lim_{x\to 1^-} y(x) = 1$. Substituting $x = 1$ and $y = 1$:

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

$1 = 0 - 1 + C \Rightarrow C = 2$

The particular solution is:

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

To find $y\left(\dfrac{1}{2}\right)$, substitute $x = \dfrac{1}{2}$:

$\dfrac{1}{2}y\left(\dfrac{1}{2}\right) = 2 - \dfrac{1}{2} - \sqrt{1 - \left(\dfrac{1}{2}\right)^2}\cos^{-1}\left(\dfrac{1}{2}\right)$

$\dfrac{1}{2}y\left(\dfrac{1}{2}\right) = \dfrac{3}{2} - \dfrac{\sqrt{3}}{2}\left(\dfrac{\pi}{3}\right)$

$\dfrac{1}{2}y\left(\dfrac{1}{2}\right) = \dfrac{3}{2} - \dfrac{\pi}{2\sqrt{3}}$

Multiplying by $2$:

$y\left(\dfrac{1}{2}\right) = 3 - \dfrac{\pi}{\sqrt{3}}$