JEE Main 2026 — Jan 21, Shift 1 — Mathematics solution

JEE Main2026Jan 21, Shift 1Mathematics

Let $y = y (x)$ be the solution curve of the differential equation $(1 + x^{2}) d y + (y - tan^{- 1} x) d x = 0$ , $y (0) = 1$ . Then the value of $y (1)$ is :

A
$\frac{2}{e ^{π /4}} + \frac{π}{4} - 1$
B
$\frac{2}{e ^{π /4}} - \frac{π}{4} - 1$
C
$\frac{4}{e ^{π /4}} + \frac{π}{2} - 1$
D
$\frac{4}{e ^{π /4}} - \frac{π}{2} - 1$

Solution

Rewriting in standard linear form: $\frac{d y}{d x} + \frac{1}{1 + x ^{2}} y = \frac{tan ^{- 1} x}{1 + x ^{2}}$ .

Integrating factor $= e^{t a n^{- 1} x}$ . Then $y \cdot e^{t a n^{- 1} x} = \int e^{t a n^{- 1} x} \cdot \frac{tan ^{- 1} x}{1 + x ^{2}} d x$ . With $u = tan^{- 1} x$ the integral becomes $\int u e^{u} d u = (u - 1) e^{u} + C$ , giving $y = (tan^{- 1} x - 1) + C e^{- t a n^{- 1} x}$ .

$y (0) = 1$ gives $C = 2$ . At $x = 1$ : $y (1) = \frac{π}{4} - 1 + \frac{2}{e ^{π /4}}$ .

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