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

$(x + 1)\frac{{dy}}{{dx}} = 2{e^{ - y}} - 1$
$\Rightarrow$ $\frac{{dy}}{{2{e^{ - y}} - 1}} = \frac{{dx}}{{x + 1}}$

$\Rightarrow$ $\frac{{{e^y}dy}}{{2 - {e^y}}} = \frac{{dx}}{{x + 1}}$
Integrating both sides,

we get:
$\int {\frac{{{e^y}dy}}{{2 - {e^y}}}} = \log |x + 1| + \log C$

…..(1)
Let $2 - {e^y} = t$.
$\therefore$ $\frac{d}{{dy}}\left( {2 - {e^y}} \right) = \frac{{dt}}{{dy}}$
$\Rightarrow$ $- {e^y} = \frac{{dt}}{{dy}}$
$\Rightarrow$ ${e^y}dt = - dt$

Substituting this value in equation (1),

we get:
$\int {\frac{{ - dt}}{t}} = \log |x + 1| + \log C$

$\Rightarrow$ $- \log |t| = \log |C(x + 1)|$
$\Rightarrow$ $- \log \left| {2 - {e^y}} \right| = \log C(x + 1)$

$\Rightarrow$ $\frac{1}{{2 - {e^y}}} = C(x + 1)$
$\Rightarrow$ $2 - {e^y} = \frac{1}{{C(x + 1)}}$

….(2)
Now, at $x = 0$ and $y = 0$, equation (2) becomes:
$\Rightarrow$ $2 - 1 = \frac{1}{C}$
$\Rightarrow$ ${\rm{C}} = 1$

Substituting ${\rm{C}} = 1$ in equation (2),

we get:
$2 - {e^y} = \frac{1}{{x + 1}}$
$\Rightarrow$ ${e^y} = 2 - \frac{1}{{x + 1}}$

$\Rightarrow$ ${e^y} = \frac{{2x + 2 - 1}}{{x + 1}}$

$\Rightarrow$ ${e^y} = \frac{{2x + 1}}{{x + 1}}$
$\Rightarrow$ $y = \log \left| {\frac{{2x + 1}}{{x + 1}}} \right|,(x \ne - 1)$

This is the required particular solution of the given differential equation.