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JEE MAIN 2026 · Differential Equations

JEE Main 2026 Session 2 Question Paper: 8 Apr, Shift 2 (B.E./B.Tech), question 17

Updated
Single correct+4 / -1Answer matches NTA’s final key

Let y=y(x)y = y(x) be the solution of the differential equation

x1−x2 dy+(y1−x2−xcos⁡−1x)dx=0,x∈(0,1),lim⁡x→1−y(x)=1.x\sqrt{1 - x^2}\,dy + \left(y\sqrt{1 - x^2} - x\cos^{-1}x\right)dx = 0,\quad x \in (0, 1),\quad \lim_{x \to 1^-} y(x) = 1.

Then y(12)y\left(\frac12\right) equals:

Choose one

Solution

Solution

Spot the exact derivative. Divide by 1−x2\sqrt{1 - x^2}:

x dy+y dx=xcos⁡−1x1−x2 dx⟹d(xy)=xcos⁡−1x1−x2 dx.x\,dy + y\,dx = \frac{x\cos^{-1}x}{\sqrt{1 - x^2}}\,dx \quad\Longrightarrow\quad d(xy) = \frac{x\cos^{-1}x}{\sqrt{1 - x^2}}\,dx.

Integrate the right side by parts, with u=cos⁡−1xu = \cos^{-1}x and dv=x1−x2dxdv = \frac{x}{\sqrt{1 - x^2}}dx, so v=−1−x2v = -\sqrt{1 - x^2} and du=−dx1−x2du = -\frac{dx}{\sqrt{1 - x^2}}:

∫xcos⁡−1x1−x2 dx=−1−x2cos⁡−1x−x+C.\int \frac{x\cos^{-1}x}{\sqrt{1 - x^2}}\,dx = -\sqrt{1 - x^2}\cos^{-1}x - x + C.

So xy=−1−x2cos⁡−1x−x+Cxy = -\sqrt{1 - x^2}\cos^{-1}x - x + C.

Use the condition. As x→1−x \to 1^-: 1⋅1=0−1+C1 \cdot 1 = 0 - 1 + C, so C=2C = 2.

Evaluate at x=12x = \frac12. 1−14=32\sqrt{1 - \frac14} = \frac{\sqrt3}{2} and cos⁡−112=π3\cos^{-1}\frac12 = \frac{\pi}{3}:

12y=−32⋅π3−12+2  ⇒  y=3−3 π3=3−π3.\tfrac12 y = -\tfrac{\sqrt3}{2}\cdot\tfrac{\pi}{3} - \tfrac12 + 2 \;\Rightarrow\; y = 3 - \frac{\sqrt3\,\pi}{3} = 3 - \frac{\pi}{\sqrt3}.

Answer: 3−π33 - \frac{\pi}{\sqrt3}.

Answer checked against the official answer key. The solution is our own working.

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