If $\frac{dy}{dx} = e^{-y}$ and $y = 0$ when $x = 0$, what is $y$ when $x = e - 1$?

Explanation

1. 01

   Separate: $e^{y}\,dy = dx$, so $e^{y} = x + C$; $y(0) = 0$ gives $C = 1$.

2. 02

   $y = \log(x + 1)$, and at $x = e - 1$, $y = \log e = 1$.