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:

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