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Let $y=y(x)$ be the solution of the differential equation $\left(1+y^2\right) e^{\tan x} d x+\cos ^2 x\left(1+e^{2 \tan x}\right) d y=0, y(0)=1$. Then $y\left(\frac{\pi}{4}\right)$ is equal to
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The correct answer is:
$\frac{1}{e}$
$\begin{aligned} & \left(1+y^2\right) e^{\tan x} d x+\cos ^2 x\left(1+e^{2 \tan x}\right) d y=0 \\ & \int \frac{\sec ^2 x e^{\tan x}}{1+e^{2 \tan x}} d x+\int \frac{d y}{1+y^2}=C \\ & \Rightarrow \tan ^{-1}\left(e^{\tan x}\right)+\tan ^{-1} y=C \\ & \text { for } x=0, y=1, \tan ^{-1}(1)+\tan ^{-1} 1=C\end{aligned}$
$\mathrm{C}=\frac{\pi}{2}$
$\tan ^{-1}\left(e^{\tan x}\right)+\tan ^{-1} y=\frac{\pi}{2}$
Put $x=\pi, \tan ^{-1} e+\tan ^{-1} y=\frac{\pi}{2}$
$\tan ^{-1} y=\cot ^{-1} e$
$y=\frac{1}{e}$
$\mathrm{C}=\frac{\pi}{2}$
$\tan ^{-1}\left(e^{\tan x}\right)+\tan ^{-1} y=\frac{\pi}{2}$
Put $x=\pi, \tan ^{-1} e+\tan ^{-1} y=\frac{\pi}{2}$
$\tan ^{-1} y=\cot ^{-1} e$
$y=\frac{1}{e}$
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