JEE Main 2026 — 21 January, Morning Shift — Question 8

Given: $(1+x^2) \, dy + (y - \tan^{-1} x) \, dx = 0$. Rewrite as: $\frac{dy}{dx} + \frac{y}{1+x^2} = \frac{\tan^{-1} x}{1+x^2}$. Integrating factor: $\text{I.F.} = e^{\int \frac{1}{1+x^2} \, dx} = e^{\tan^{-1} x}$. Multiply both sides: $e^{\tan^{-1} x} \frac{dy}{dx} + \frac{y e^{\tan^{-1} x}}{1+x^2} = \frac{e^{\tan^{-1} x} \tan^{-1} x}{1+x^2}$. Left side is $\frac{d}{dx} \left( y e^{\tan^{-1} x} \right)$. Integrate: $y e^{\tan^{-1} x} = \int \frac{e^{\tan^{-1} x} \tan^{-1} x}{1+x^2} \, dx$. Let $u = \tan^{-1} x$, then $du = \frac{dx}{1+x^2}$, so RHS = $\int u e^u \, du = u e^u - e^u + C = e^{\tan^{-1} x} (\tan^{-1} x - 1) + C$. Thus $y e^{\tan^{-1} x} = e^{\tan^{-1} x} (\tan^{-1} x - 1) + C$. Using $y(0)=1$: $1 \cdot e^{0} = e^{0}(0-1)+C \Rightarrow 1 = -1 + C \Rightarrow C = 2$. Hence $y = \tan^{-1} x - 1 + 2 e^{-\tan^{-1} x}$. At $x=1$: $y(1) = \frac{\pi}{4} - 1 + 2 e^{-\pi/4} = 2 e^{-\pi/4} + \frac{\pi}{4} - 1$.