The solution of the differential equation \( (1+x)\displaystyle\frac{dy}{dx}=e^{-y},y(0)=0 \) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( e^y=log_e|1+x|+1 \)
Since \( \frac{1}{e^{-y}}=e^{y} \), the variables separate:
\( \displaystyle e^{y}\,dy=\frac{dx}{1+x} \)
Integrating both sides:
\( \displaystyle e^{y}=\log_e|1+x|+C \)
Using the initial condition y(0) = 0, put x = 0 and y = 0:
\( \displaystyle e^{0}=\log_e 1+C \)
\( \displaystyle 1=0+C \)
\( \displaystyle C=1 \)
\( \displaystyle e^{y}=\log_e|1+x|+1 \)
Shortcut: Put x = 0, y = 0 into each option:
\( \displaystyle \text{Option (1)}: \log_e|0|\ \text{is not defined} \)
\( \displaystyle \text{Option (2)}: 0=1+3\ \text{is false} \)
\( \displaystyle \text{Option (3)}: 0=0+1\ \text{is false} \)
\( \displaystyle \text{Option (4)}: e^0=1=0+1\ \text{is true} \)