The general solution of the differential equation $(1 + e^x)dy + ye^x dx = 0$, where $y > 0$, is

$C=y(1+e^x)$, C is an arbitary constant

The correct answer is Option (3) → $C=y(1+e^x)$, C is an arbitary constant

$ (1 + e^x) \, dy + y e^x \, dx = 0 $

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

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

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

$ \Rightarrow \ln y = -\ln(1 + e^x) + C $

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

$ \Rightarrow y = A \cdot \frac{1}{1 + e^x} = \frac{A}{1 + e^x} $

General solution: $ y = \frac{A}{1 + e^x} $, where $A > 0$