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

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

The correct answer is Option (2) → $|(x+1) (e^y-2)| = 1$

Given differential equation:

$ (x + 1)\frac{dy}{dx} + 1 - 2e^{-y} = 0 $

Rewriting:

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

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

Separate variables:

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

Integrate both sides:

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

Substitute $u = e^{-y} \Rightarrow dy = -\frac{1}{u} \, du$:

$ -\int \frac{1}{u(2u - 1)} \, du = \ln|x + 1| + C $

Using partial fractions:

$ \frac{1}{u(2u - 1)} = \frac{-1}{u} + \frac{2}{2u - 1} $

Then:

$ -\int \left( \frac{-1}{u} + \frac{2}{2u - 1} \right) du = \ln|x + 1| + C $

$ \int \left( \frac{1}{u} - \frac{2}{2u - 1} \right) du = \ln|x + 1| + C $

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

Back-substitute $u = e^{-y}$:

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

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

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

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

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

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

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

Upon substituting y(0)=0 , C = 1

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