Calculus (3rd Edition)

The general solution is $$y=\ln \frac{1}{c-e^x}.$$
We have $$y'-e^{x+y}=0\Longrightarrow e^{-y}dy=e^{x}dx\\ \Longrightarrow\int e^{-y}dy=\int e^{x}dx+c\\ \Longrightarrow -e^{-y} =e^{x}+c\\ \Longrightarrow e^{-y} =c-e^{x}\Longrightarrow -y =\ln(c-e^x).$$ Hence, the general solution is $$y=\ln \frac{1}{c-e^x}.$$