Which of the following is the solution to the differential equation \(\dfrac{dy}{dx}=e^{x+y}\) with the initial condition \(y(0)=-\ln 5 \) ?

Summary Submit

$$ \frac{dy}{dx}=e^{x+y} $$ $$ \frac{dy}{dx}=e^x\cdot e^y $$ $$ \frac{1}{e^y}  dy = e^x  dx $$ $$ e^{-y}  dy = e^x   dx $$ $$ \int e^{-y}  dy = \int e^x   dx $$ $$ -e^{-y}=e^x+C $$ $$ e^{-y}=-e^x+C $$ $$ -y = \ln (-e^x+C) $$ $$ y = -\ln(-e^x+C) $$

Using the initial condition:

$$ -\ln 5 = -\ln(-e^0+C) $$ $$ -\ln 5 = -\ln(-1+C) $$ $$ C=6 $$ $$ \boxed{y = -\ln(-e^x+6)} $$