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