Let y = y(x) be the solution... |JEE Main 2026| Differential

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JEE MAIN 2026

21-01-2026 S1

Question

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

Select the correct option:

A

$\frac{2}{e^{\pi / 4}}+\frac{\pi}{4}-1$

B

$\frac{2}{\mathrm{e}^{\pi / 4}}-\frac{\pi}{4}-1$

C

$\frac{4}{e^{\pi / 4}}+\frac{\pi}{2}-1$

D

$\frac{4}{e^{\pi / 4}}-\frac{\pi}{2}-1$

✓ Correct! Well done.

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