Chapter 1 - First-Order Differential Equations - 1.4 Separable Differential Equations - Problems - Page 43: 3

$y=ln(C-e^x)$

Split $e^{x+y}$ into $e^xe^y$ using exponent rules. $$e^xe^ydy-dx=0$$ Multiply the equation by $\frac{1}{e^x}$ to get $$e^y dy-e^{-x}dx=0$$ You can integrate each of the $dx$ and $dy$ terms since they are in terms of one variable. $$\int e^y dy-\int e^{-x} dx=\int 0dx$$ $$e^y+e^x=C$$ Solve for $y$. $$e^y=C-e^x$$ $$y=ln(C-e^x)$$