JEE Main 2024 — 8 April, Shift 1 — Question 19

$\left(1+y^{2}\right) e^{\tan x} d x+\cos ^{2} x\left(1+e^{2 \tan x}\right) d y=0$

$$\begin{aligned} & \int \frac{\sec ^{2} x e^{\tan x}}{1+e^{2 \tan x}} d x+\int \frac{d y}{1+y^{2}}=C & \end{aligned}$$

$\Rightarrow \tan ^{-1}\left(e^{\tan x}\right)+\tan ^{-1} y=C$

$$\begin{array}{r} \text { for } \mathrm{x}=0, \mathrm{y}=1, \tan ^{-1}(1)+\tan ^{-1} 1=\mathrm{C} \qquad \end{array}$$

$\mathrm{C}=\frac{\pi}{2}$

$\tan ^{-1}\left(e^{\tan x}\right)+\tan ^{-1} y=\frac{\pi}{2}$

Put $x=\pi, \tan ^{-1} e+\tan ^{-1} y=\frac{\pi}{2}$

$$\tan ^{-1} y=\cot ^{-1} e$$ $$y=\frac{1}{e}$$