Question

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

• A. \(3 - \frac{\pi}{\sqrt{3}}\)
• B. \(4 - \sqrt{3} \pi\)
• C. \(4 - \frac{2\pi}{\sqrt{3}}\)
• D. \(3 - \frac{\pi}{2\sqrt{3}}\)

Answer 1

The Correct Option is C

Step 1: Convert into linear differential equation Divide by \(x\sqrt{1-x^2}\,dx\): \[ \frac{dy}{dx}+\frac{1}{x}y=\frac{\cos^{-1}x}{\sqrt{1-x^2}} \] This is of the form \[ \frac{dy}{dx}+P(x)y=Q(x) \] where \[ P(x)=\frac{1}{x} \] Step 2: Find integrating factor \[ \text{I.F.}=e^{\int \frac1x dx}=e^{\ln x}=x \] Multiply throughout by \(x\): \[ x\frac{dy}{dx}+y=\frac{x\cos^{-1}x}{\sqrt{1-x^2}} \] \[ \frac{d}{dx}(xy)=\frac{x\cos^{-1}x}{\sqrt{1-x^2}} \] Step 3: Integrate both sides \[ xy=\int \frac{x\cos^{-1}x}{\sqrt{1-x^2}}dx+C \] Let \[ t=\cos^{-1}x \] Then \[ x=\cos t,\qquad dx=-\sin t\,dt \] and \[ \sqrt{1-x^2}=\sin t \] So, \[ \int \frac{x\cos^{-1}x}{\sqrt{1-x^2}}dx = -\int t\cos t\,dt \] Using integration by parts, \[ \int t\cos t\,dt=t\sin t+\cos t \] Hence, \[ \int \frac{x\cos^{-1}x}{\sqrt{1-x^2}}dx = -\left(t\sin t+\cos t\right) \] Substituting back, \[ = -\cos^{-1}x\sqrt{1-x^2}-x \] Therefore, \[ xy=-\cos^{-1}x\sqrt{1-x^2}-x+C \] \[ y=-\frac{\cos^{-1}x\sqrt{1-x^2}}{x}-1+\frac{C}{x} \] Step 4: Apply boundary condition \[ \lim_{x\to1^-}y(x)=1 \] As \(x\to1^-\), \[ \cos^{-1}x\to0,\qquad \sqrt{1-x^2}\to0 \] Thus, \[ 1=-1+C \] \[ C=2 \] So, \[ y(x)=\frac{2}{x}-1-\frac{\cos^{-1}x\sqrt{1-x^2}}{x} \] Step 5: Find \(y(1/2)\) \[ y\left(\frac12\right)=4-1-\frac{\frac{\pi}{3}\cdot\frac{\sqrt3}{2}}{\frac12} \] \[ =3-\frac{\pi\sqrt3}{3} \] \[ =3-\frac{\pi}{\sqrt3} \] \[ \boxed{y\left(\frac12\right)=3-\frac{\pi}{\sqrt3}} \]