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:
This is a first-order linear ordinary differential equation (ODE) of the form $\dfrac{dy}{dx} + P(x)\,y = Q(x)$.
We will rearrange the given equation into this standard linear form, find the integrating factor (I.F.), and then solve using the formula: \[ y \cdot (\text{I.F.}) = \int Q(x)\cdot(\text{I.F.})\,dx + C \] Step 2: Rearranging into Standard Linear Form:
Dividing the original equation by $x\sqrt{1-x^2}\,dx$: \[ \frac{dy}{dx} - \frac{y}{x} = \frac{-x\cos^{-1}x}{\sqrt{1-x^2}} \] So we have $P(x) = -\dfrac{1}{x}$ and $Q(x) = \dfrac{-x\cos^{-1}x}{\sqrt{1-x^2}}$.
Step 3: Finding the Integrating Factor:
\[ \text{I.F.} = e^{\int P(x)\,dx} = e^{\int -\frac{1}{x}\,dx} = e^{-\ln x} = \frac{1}{x} \] Step 4: Multiplying through and Integrating:
Multiplying both sides by $\dfrac{1}{x}$: \[ \frac{d}{dx}\!\left(\frac{y}{x}\right) = \frac{-\cos^{-1}x}{\sqrt{1-x^2}} \] Integrating both sides: \[ \frac{y}{x} = -\int \frac{\cos^{-1}x}{\sqrt{1-x^2}}\,dx + C \] Let $u = \cos^{-1}x$, then $du = \dfrac{-1}{\sqrt{1-x^2}}\,dx$, so $\dfrac{dx}{\sqrt{1-x^2}} = -du$.
\[ -\int \frac{\cos^{-1}x}{\sqrt{1-x^2}}\,dx = -\int u\cdot(-du) = \int u\,du = \frac{u^2}{2} = \frac{(\cos^{-1}x)^2}{2} \] Therefore: \[ \frac{y}{x} = \frac{(\cos^{-1}x)^2}{2} + C \] Step 5: Applying the Initial Condition:
We are given $\displaystyle\lim_{x\to 1^-} y(x) = 1$.
As $x \to 1^-$: $\cos^{-1}(1) = 0$, and $y \to 1$.
\[ \frac{1}{1} = \frac{(0)^2}{2} + C \implies C = 1 \] So the solution is: \[ \frac{y}{x} = \frac{(\cos^{-1}x)^2}{2} + 1 \implies y = \frac{x(\cos^{-1}x)^2}{2} + x \] Step 6: Computing $y\!\left(\dfrac{1}{2}\right)$:
\[ y\!\left(\frac{1}{2}\right) = \frac{\frac{1}{2}\cdot\left(\cos^{-1}\frac{1}{2}\right)^2}{2} + \frac{1}{2} \] Since $\cos^{-1}\!\left(\dfrac{1}{2}\right) = \dfrac{\pi}{3}$: \[ = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{\pi^2}{9} + \frac{1}{2} = \frac{\pi^2}{36} + \frac{18}{36} = \frac{\pi^2 + 18}{36} \] Step 7: Final Answer:
\[ y\!\left(\frac{1}{2}\right) = \frac{\pi^2 + 18}{36} \] The answer is Option (2).