JEE Main 2025 — 22 January, Evening Shift — Question 21

$\frac{dy}{dx} + \frac{xy}{x^2 - 1} = \frac{x^6 + 4x}{\sqrt{1 - x^2}}, \; -1 < x < 1$

$\text{Integrating\; factor:\; } \mu = e^{\int \frac{x}{x^2 - 1} dx} = \sqrt{|x^2 - 1|} = \sqrt{1 - x^2}$

$\Rightarrow \frac{d}{dx}(\mu y) = \sqrt{1 - x^2} \cdot \frac{x^6 + 4x}{\sqrt{1 - x^2}} = x^6 + 4x$

$\int \frac{d}{dx}(\mu y) dx = \int (x^6 + 4x) dx$

$\mu y = \frac{x^7}{7} + 2x^2 + C$

$\text{Given\; } f(0) = 0 \Rightarrow C = 0$

$\therefore f(x) = \frac{\tfrac{x^7}{7} + 2x^2}{\sqrt{1 - x^2}}$

$6 \int_{-\frac{1}{2}}^{\frac{1}{2}} f(x) \, dx = 6 \left( 2 \int_{0}^{\frac{1}{2}} f(x) \, dx \right)$

$= 12 \int_{0}^{\frac{1}{2}} \frac{\tfrac{x^7}{7} + 2x^2}{\sqrt{1 - x^2}} \, dx$

$\text{The\; first\; term\; } \int \frac{x^7}{7\sqrt{1 - x^2}} dx \text{ is\; odd,\; so\; it\; vanishes\; from\; } -\tfrac{1}{2} \text{ to\; } \tfrac{1}{2}$

$\Rightarrow I = 4 \int_{0}^{\frac{1}{2}} \frac{x^2}{\sqrt{1 - x^2}} \, dx$

$\text{Put } x = \sin t, \; dx = \cos t \, dt, \; t \in [0, \tfrac{\pi}{6}]$

$I = 4 \int_{0}^{\frac{\pi}{6}} \sin^2 t \, dt$

$I = 4 \left( \frac{t}{2} - \frac{\sin 2t}{4} \right)_{0}^{\frac{\pi}{6}}$

$I = \frac{\pi}{3} - \frac{\sqrt{3}}{2}$

$6I = 6 \left( \frac{\pi}{3} - \frac{\sqrt{3}}{2} \right) = 2\pi - 3\sqrt{3}$

$\text{Given\; } 6I = 2\pi - \alpha \Rightarrow \alpha = 3\sqrt{3}$

$\therefore \alpha^2 = 27$

$\boxed{27}$