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The particular solution of the differential equation $\left(1+e^{2 x}\right) d y+e^x\left(1+y^2\right) d x=0$ at $x=0$ and $y=1$ is
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$\tan ^{-1} \mathrm{e}^{\mathrm{x}}+\tan ^{-1} \mathrm{y}=\frac{\pi}{2}$
$\begin{aligned}
& \left(1+\mathrm{e}^{2 \mathrm{x}}\right) \mathrm{dy}+\mathrm{e}^{\mathrm{x}}\left(1+\mathrm{y}^2\right) \mathrm{dx}=0 \\
& \therefore \frac{\mathrm{dy}}{1+\mathrm{y}^2}+\frac{\mathrm{e}^{\mathrm{x}}}{\left(1+\mathrm{e}^{2 \mathrm{x}}\right)} \mathrm{dx}=0 \\
& \therefore \int \frac{\mathrm{dy}}{1+\mathrm{y}^2}=-\int \frac{\mathrm{e}^{\mathrm{x}}}{1+\mathrm{e}^{2 \mathrm{x}}} d \mathrm{x}
\end{aligned}$
Put $\mathrm{e}^{\mathrm{x}}=\mathrm{t} \Rightarrow \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{dt}$
$\therefore \int \frac{\mathrm{dy}}{1+\mathrm{y}^2}=-\int \frac{\mathrm{dt}}{1+\mathrm{t}^2} \Rightarrow \tan ^{-1}(\mathrm{y})=-\tan ^{-1}(\mathrm{t})+\mathrm{c}$
$\therefore \tan ^{-1}(\mathrm{y})+\tan ^{-1}\left(\mathrm{e}^{\mathrm{x}}\right)=\mathrm{c}$
We have $\mathrm{x}=0, \mathrm{y}=1$
$\therefore \tan ^{-1}(1)+\tan ^{-1}\left(\mathrm{e}^{\circ}\right)=\mathrm{c} \Rightarrow \mathrm{c}=2 \tan ^{-1}(1)=\frac{\pi}{2}$
$\therefore \tan ^{-1} \mathrm{y}+\tan ^{-1} \mathrm{e}^{\mathrm{x}}=\frac{\pi}{2}$
& \left(1+\mathrm{e}^{2 \mathrm{x}}\right) \mathrm{dy}+\mathrm{e}^{\mathrm{x}}\left(1+\mathrm{y}^2\right) \mathrm{dx}=0 \\
& \therefore \frac{\mathrm{dy}}{1+\mathrm{y}^2}+\frac{\mathrm{e}^{\mathrm{x}}}{\left(1+\mathrm{e}^{2 \mathrm{x}}\right)} \mathrm{dx}=0 \\
& \therefore \int \frac{\mathrm{dy}}{1+\mathrm{y}^2}=-\int \frac{\mathrm{e}^{\mathrm{x}}}{1+\mathrm{e}^{2 \mathrm{x}}} d \mathrm{x}
\end{aligned}$
Put $\mathrm{e}^{\mathrm{x}}=\mathrm{t} \Rightarrow \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{dt}$
$\therefore \int \frac{\mathrm{dy}}{1+\mathrm{y}^2}=-\int \frac{\mathrm{dt}}{1+\mathrm{t}^2} \Rightarrow \tan ^{-1}(\mathrm{y})=-\tan ^{-1}(\mathrm{t})+\mathrm{c}$
$\therefore \tan ^{-1}(\mathrm{y})+\tan ^{-1}\left(\mathrm{e}^{\mathrm{x}}\right)=\mathrm{c}$
We have $\mathrm{x}=0, \mathrm{y}=1$
$\therefore \tan ^{-1}(1)+\tan ^{-1}\left(\mathrm{e}^{\circ}\right)=\mathrm{c} \Rightarrow \mathrm{c}=2 \tan ^{-1}(1)=\frac{\pi}{2}$
$\therefore \tan ^{-1} \mathrm{y}+\tan ^{-1} \mathrm{e}^{\mathrm{x}}=\frac{\pi}{2}$
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