Question

Sol. [Key Concept: This is a linear ODE $\frac{dy}{dx} + \frac{y}{x} = \frac{\cos^{- 1}x}{\sqrt{1 - x^{2}}}$ (I.F. $x$); integrate by parts, apply $\lim_{x \rightarrow 1^{-}}y = 1$, then $y(\frac{1}{2}) = 3 - \frac{\pi}{\sqrt{3}}$.]

$${x\sqrt{1 - x^{2}}dy + \left( y\sqrt{1 - x^{2}} - x\cos^{- 1}x \right)dx = 0 }{\Rightarrow \frac{dy}{dx} + \frac{y}{x} = \frac{\cos^{- 1}x}{\sqrt{1 - x^{2}}} }$$it is L.D.E.

∴ I.F. $= e^{\int\frac{1}{x}dx} = e^{lnx} = x$

∴ solution is $y \cdot x = \int\frac{x\cos^{- 1}x}{\sqrt{1 - x^{2}}}dx + C$

$$\begin{array}{r} y \cdot x = I_{1} + C\#(1) \end{array}$$

$$I_{1} = \int\frac{x\cos^{- 1}x}{\sqrt{1 - x^{2}}}dx $$Let $\cos^{- 1}x = t \Rightarrow x = cost$

$${- \frac{1}{\sqrt{1 - x^{2}}}dx = dt }{I_{1} = \int - t \cdot cotdt }{= - (tsint + cost)}$$

$$\begin{array}{r} I_{1} = - \left( \sqrt{1 - x^{2}} \cdot \cos^{- 1}x + x \right)\#(2) \end{array}$$

$${\therefore y \cdot x = - \left( \sqrt{1 - x^{2}} \cdot \cos^{- 1}x + x \right) + C }{\because\lim_{x \rightarrow 1^{-}}\mspace{2mu} y(x) = 1 }{\therefore 1 = - (0 + 1) + C }{\Rightarrow C = 2 }{\therefore y \cdot x = - \sqrt{1 - x^{2}} \cdot \cos^{- 1}x - x + 2 }$$$\left( \right.\ $ put $\left. \ x = \frac{1}{2} \right)$

$${y\left( \frac{1}{2} \right) \times \frac{1}{2} = - \frac{\sqrt{3}}{2} \times \frac{\pi}{3} - \frac{1}{2} + 2 }{\Rightarrow y\left( \frac{1}{2} \right) = 3 - \frac{\pi}{\sqrt{3}} }$$