QJEEM24Ma29ab: JEE Main 2024 Mathematics - Class 12 JEEM24Ma29ab Solution: Let be the solution of the differential equation . Then is…

$\begin{aligned}& \left(1+y^2\right) e^{\tan x} d x+\cos ^2 x\left(1+e^{2 \tan x}\right) d y=0 \\& \frac{d y}{1+y^2}=-\frac{e^{\tan x} \cdot \sec ^2 x d x}{1+e^{2 \tan x}} \\& \int \frac{d y}{1+y^2}=-\int \frac{e^{\tan x} \cdot \sec ^2 x d x}{1+e^{2 \tan x}}\end{aligned}$

Let $e^{\tan x}=t$

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

$\begin{aligned}& \tan ^{-1} y=-\tan ^{-1} t+c \\& \tan ^{-1} y=-\tan ^{-1}\left(e^{\tan x}\right)+c \\& \text { at } x=0, y=1 \\& \tan ^{-1}(1)=-\tan ^{-1}(1)+c \\& \frac{\pi}{4}=-\frac{\pi}{4}+c \\& c=\frac{\pi}{2} \\& \tan ^{-1} y=-\tan ^{-1}\left(e^{\tan x}\right)+\frac{\pi}{2}\end{aligned}$

Now, at $x=\frac{\pi}{4}$

$\begin{aligned}& \tan ^{-1} y=-\tan ^{-1}(e)+\frac{\pi}{2} \\& \tan ^{-1} y=\cot ^{-1} e=\tan ^{-1} \frac{1}{e} \\& \Rightarrow y=\frac{1}{e}\end{aligned}$