The solution of the differential equation $(x + 1)\frac{dy}{dx} + 1 - 2e^{-y} = 0$, $y(0) = 0$ is

Given differential equation: $(x + 1)\frac{dy}{dx} + 1 - 2e^{-y} = 0$ with initial condition $y(0) = 0$

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

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

Multiply numerator and denominator by $e^y$:

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

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Let $u = e^y$

Then $du = e^y \, dy$

The equation becomes:

$\frac{du}{2 - u} = \frac{dx}{x + 1}$

$-\frac{du}{u - 2} = \frac{dx}{x + 1}$

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Integrating both sides:

$\int -\frac{du}{u - 2} = \int \frac{dx}{x + 1}$

$-\ln|u - 2| = \ln|x + 1| + C$

Substituting back $u = e^y$:

$-\ln|e^y - 2| = \ln|x + 1| + C$

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Using the initial condition $y(0) = 0$:

$-\ln|e^0 - 2| = \ln|0 + 1| + C$

$-\ln|1 - 2| = \ln(1) + C$

$-\ln(1) = 0 + C$

$C = 0$

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With $C = 0$:

$-\ln|e^y - 2| = \ln|x + 1|$

$\ln|e^y - 2|^{-1} = \ln|x + 1|$

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

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

$1 = |x + 1| \cdot |e^y - 2|$

$|(x + 1)(e^y - 2)| = 1$