Find the particular solution of the differential equation
$\left( {1 + {e^{2x}}} \right)dy + \left( {1 + {y^2}} \right){e^x}dx = 0$,
given that $y = 1$ when $x = 0$

$\left( {1 + {e^{2x}}} \right)dy + \left( {1 + {y^2}} \right){e^x}dx = 0$
$\Rightarrow$ $\frac{{dy}}{{1 + {y^2}}} + \frac{{{e^x}dx}}{{1 + {e^{2x}}}} = 0$
Integrating both sides,

we get:
${\tan ^{ - 1}}y + \int {\frac{{{e^x}dx}}{{1 + {e^{2x}}}}} = C$

….(1)
Let ${e^x} = t \Rightarrow {e^{2x}} = {t^2}$.
$\Rightarrow$ $\frac{d}{{dx}}\left( {{e^x}} \right) = \frac{{dt}}{{dx}}$

$\Rightarrow$ ${e^x} = \frac{{dt}}{{dx}}$
$\Rightarrow$ ${e^x}dx = dt$

Substituting these values in equation (1),

we get:
${\tan ^{ - 1}}y + \int {\frac{{dt}}{{1 + {t^2}}}} = C$

$\Rightarrow$ ${\tan ^{ - 1}}y + {\tan ^{ - 1}}t = {\rm{C}}$

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

….(2)
Now, $y = 1$ at $x = 0$.
Therefore, equation (2) becomes:
${\tan ^{ - 1}}1 + {\tan ^{ - 1}}1 = C$
$\Rightarrow$ $\frac{\pi }{4} + \frac{\pi }{4} = {\rm{C}}$

$\Rightarrow$ ${\rm{C}} = \frac{\pi }{2}$
Substituting ${\rm{C}} = \frac{\pi }{2}$ in equation (2),

we get:
${\tan ^{ - 1}}y + {\tan ^{ - 1}}\left( {{e^x}} \right) = \frac{\pi }{2}$
This is the required particular solution of the given differential equation.