Q42 / 75 · JEE Main 2026 (21 January Shift 1) · Mathematics
Let $y=y(x)$ be the solution curve of the differential equation $\left(1+x^{2}\right) \mathrm{d} y+\left(y-\tan ^{-1} x\right) d x=0, y(0)=1$. Then the value of $y(1)$ is :
- A$\frac{4}{\mathrm{e}^{\pi / 4}}+\frac{\pi}{2}-1$
- Correct answerB$\frac{2}{\mathrm{e}^{\pi / 4}}+\frac{\pi}{4}-1$
- C$\frac{2}{\mathrm{e}^{\pi / 4}}-\frac{\pi}{4}-1$
- D$\frac{4}{e^{\pi / 4}}-\frac{\pi}{2}-1$
Solution
Rewrite as $\frac{dy}{dx} + \frac{y}{1+x^2} = \frac{\tan^{-1}x}{1+x^2}$.
IF $= e^{\tan^{-1}x}$.
$y \cdot e^{\tan^{-1}x} = \int \frac{\tan^{-1}x}{1+x^2} \cdot e^{\tan^{-1}x} dx$.
Let $t = \tan^{-1}x$: $= \int te^t dt = e^t(t-1) + C$.
$y = \tan^{-1}x - 1 + Ce^{-\tan^{-1}x}$.
Using $y(0) = 1$: $C = 2$.
$y(1) = \frac{\pi}{4} - 1 + \frac{2}{e^{\pi/4}}$.
IF $= e^{\tan^{-1}x}$.
$y \cdot e^{\tan^{-1}x} = \int \frac{\tan^{-1}x}{1+x^2} \cdot e^{\tan^{-1}x} dx$.
Let $t = \tan^{-1}x$: $= \int te^t dt = e^t(t-1) + C$.
$y = \tan^{-1}x - 1 + Ce^{-\tan^{-1}x}$.
Using $y(0) = 1$: $C = 2$.
$y(1) = \frac{\pi}{4} - 1 + \frac{2}{e^{\pi/4}}$.
Hence, the correct answer is option: 'B'.