The general solution of the differential equation \((1 + e^x)dy + ye^x dx = 0\), where \(y \gt 0\), is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(C = y(1 + e^x)\), C is an arbitrary constant
The variables can be separated (variables separable form):
\((1 + e^x)\,dy = -ye^x\,dx\)
\( \displaystyle \frac{dy}{y} = -\frac{e^x}{1 + e^x}\,dx \)
Integrating both sides (the numerator \(e^x\) is the derivative of \(1 + e^x\)):
\( \displaystyle \int \frac{dy}{y} = -\int \frac{e^x}{1 + e^x}\,dx \)
\(\log y = -\log(1 + e^x) + \log C\)
Since \(y \gt 0\), \(\log y\) needs no modulus. Combining the logarithms:
\(\log y + \log(1 + e^x) = \log C\)
\(y(1 + e^x) = C\)