Question

If \((x\sqrt{1-x^2}) \, dy - (y\sqrt{1-x^2} - x^2 \cos^{-1} x) \, dx = 0\) and \(\lim_{x \to 1^-} y(x) = 1\), then \(y\left(\frac{1}{2}\right)\) is

• A. \(\frac{\pi^2}{36}\)
• B. \(\frac{\pi^2 + 18}{36}\)
• C. \(\frac{\pi^2 - 18}{36}\)
• D. \(\frac{\pi^2}{18}\)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We rearrange the differential equation into the linear form \(\frac{dy}{dx} + P(x)y = Q(x)\), find the integrating factor, and solve.

Step 2: Key Formula or Approach:
\[ \frac{dy}{dx} - \frac{1}{x} y = -\frac{x \cos^{-1} x}{\sqrt{1-x^2}} \] Integrating Factor (I.F.) = \( e^{\int -1/x dx} = 1/x \).

Step 3: Detailed Explanation:
1. Multiply by I.F.: \[ \frac{y}{x} = \int -\frac{\cos^{-1} x}{\sqrt{1-x^2}} dx \] 2. Let \( t = \cos^{-1} x \), \( dt = -\frac{1}{\sqrt{1-x^2}} dx \): \[ \frac{y}{x} = \int t \, dt = \frac{t^2}{2} + C = \frac{(\cos^{-1} x)^2}{2} + C \] 3. Using the condition \( \lim_{x \to 1^-} y = 1 \): As \( x \to 1, \cos^{-1} x \to 0 \). \[ \frac{1}{1} = 0 + C \implies C = 1 \] 4. Equation: \( y = \frac{x(\cos^{-1} x)^2}{2} + x \). 5. Substitute \( x = 1/2 \): \[ y\left(\frac{1}{2}\right) = \frac{1}{2} \cdot \frac{(\pi/3)^2}{2} + \frac{1}{2} = \frac{\pi^2}{36} + \frac{1}{2} = \frac{\pi^2 + 18}{36} \]

Step 4: Final Answer:
The value is \(\frac{\pi^2 + 18}{36}\).