(x + 1) $\frac{dy}{dx}$ = $2e^{-y}$ - 1 ⇒ $\frac{dy}{2e^{-y}-1}$ = $\frac{dx}{x+1}$ ⇒ $\frac{e^y}{2-e^y}$ = $\frac{dx}{x+1}$ Integrating both sides, we get: ∫ $\frac{e^y dy}{2-e^y}$ = log |x + 1| + log C ... (1) Let 2 - $e^y$ = t ∴ $\frac{d}{dy}(2-e^y) =${dt}/{dy}$$\Rightarrow -$e^y$=${dt}/{dy}$\Rightarrow$e^y$dy = - dt \text{ Substituting this value in equation 1 , we get: } \int -${dt}/t$= \log |x + 1| + \log C \Rightarrow - \log |t| = \log |C (x + 1)| \Rightarrow - \log |2 -$e^y$,| = \log |C (x + 1)| \Rightarrow$1/{2-e^y}$, = C (x + 1) \Rightarrow 2 -$e^y$, =$1/{C(x+1)}$, \ldots (2) \text{ Now, at } x = 0 \text{ and } y = 0, \text{ equation 2 becomes: } \Rightarrow 2 - 1 =$1/C$,$e^y$\Rightarrow C = 1 \text{ Substituting } C = 1 \text{ in equation 2, we get: } 2 -$1/{x+1}$, =$e^y$, \Rightarrow$1/{x+1}$, = 2 -$e^y$, \Rightarrow${2x+2-1}/{x+1}$, =$e^y$, \Rightarrow${2x+1}/{x+1}$, =$|{2x+1}/{x+1}|$ , (x ≠ - 1) This is the required particular solution of the given differential equation.