The general solution of the differential equation \((e^{x} + 2)\,y\,dy = (y + 2)\,e^{x}\,dx\), \(y \gt 0\), is (C is an arbitrary constant)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(y - 2\log (y + 2) = \log (e^{x} + 2) + C\)
Dividing by \((y + 2)(e^{x} + 2)\) separates the variables:
\( \displaystyle \frac{y}{y + 2}\,dy = \frac{e^{x}}{e^{x} + 2}\,dx \)
On the left, write
\( \displaystyle \frac{y}{y + 2} = 1 - \frac{2}{y + 2} \)
On the right, the numerator \(e^{x}\) is the derivative of the denominator \(e^{x} + 2\).
Integrating both sides:
\( y - 2\log (y + 2) = \log (e^{x} + 2) + C \)
Why the other options are incorrect
Option (1) \(y + 2\log (y + 2) = \log (e^{x} + 2) + C\) — Incorrect. This integrates \(1 + \frac{2}{y + 2}\); the fraction \(\frac{y}{y + 2}\) is \(1 - \frac{2}{y + 2}\).
Option (2) \(\log (y + 2) = \log (e^{x} + 2) + C\) — Incorrect. This drops the factor y on the left, so it solves \((e^{x} + 2)\,dy = (y + 2)e^{x}\,dx\) instead.
Option (4) \(y - 2\log (y + 2) = e^{x} + C\) — Incorrect. The right side integrates \(e^{x}\) alone, forgetting the denominator \(e^{x} + 2\).