The general solution of the differential equation \( (x+1)\displaystyle\frac{dy}{dx}=2e^{-y}-1 \) is:
(where C is an arbitrary constant)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( (x+1)(e^y-2)=C \)
Over a common denominator the right side is
\( \displaystyle 2e^{-y}-1=\frac{2-e^y}{e^y} \)
so the variables separate:
\( \displaystyle \frac{e^y}{2-e^y}\,dy=\frac{dx}{x+1} \)
On the left put \( t=2-e^y \), so \( dt=-e^y\,dy \):
\( -\log|2-e^y|=\log|x+1|+\log|C_1| \)
\( \log|x+1|+\log|2-e^y|=-\log|C_1| \)
\( \displaystyle (x+1)(2-e^y)=\pm\frac{1}{C_1} \)
Changing the sign of the constant:
\( (x+1)(e^y-2)=C \)
Shortcut: Differentiate Option (3):
\( \displaystyle (e^y-2)+(x+1)e^y\frac{dy}{dx}=0 \)
\( \displaystyle (x+1)\frac{dy}{dx}=\frac{2-e^y}{e^y}=2e^{-y}-1 \)