Q42 / 75 · JEE Main 2026 (08 April Shift 2) · Mathematics
Let $y=y(x)$ be the solution of the differential equation $x\sqrt{1-x^2}\,dy + \left(y\sqrt{1-x^2} - x\cos^{-1}x\right)dx = 0$, $x \in (0, 1)$, $\displaystyle\lim_{x\to 1^-} y(x) = 1$. Then $y\left(\dfrac{1}{2}\right)$ equals:
- Correct answerA$3 - \dfrac{\pi}{\sqrt{3}}$
- B$4 - \sqrt{3}\pi$
- C$4 - \dfrac{2\pi}{\sqrt{3}}$
- D$3 - \dfrac{\pi}{2\sqrt{3}}$
Solution
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 $2$:
$y\left(\dfrac{1}{2}\right) = 3 - \dfrac{\pi}{\sqrt{3}}$
$x\sqrt{1-x^2}\,dy + \left(y\sqrt{1-x^2} - x\cos^{-1}x\right)dx = 0$
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 $2$:
$y\left(\dfrac{1}{2}\right) = 3 - \dfrac{\pi}{\sqrt{3}}$
Hence, the correct answer is option: 'A'.