Solve the differential equation dy/dx = e^y.

Step-by-step solution

1. 1

   Separate the variables. All the $y$ dependence is on the right, so divide by $e^{y}$ — legitimate because $e^{y}$ is never zero:

   $e^{-y}\,dy = dx$

2. 2

   Integrate both sides.

   $\int e^{-y}\,dy = \int dx \quad\Longrightarrow\quad -e^{-y} = x + C_0$

   Only one constant is needed; absorbing the sign, write $e^{-y} = C - x$ with $C = -C_0$.

3. 3

   Solve for y explicitly. Taking the natural logarithm requires $C - x > 0$:

   $-y = \ln(C - x) \quad\Longrightarrow\quad y = -\ln(C - x)$

4. 4

   Read off the domain, which is the interesting part. The solution only exists for $x < C$. As $x \to C^{-}$ the argument of the log tends to $0^{+}$ and $y \to +\infty$: the solution blows up in finite $x$, even though the right-hand side $e^{y}$ is smooth everywhere. This is the standard example that local existence does not imply global existence.

5. 5

   Check by substitution. With $C = 5$ and $x = 1$: $y = -\ln 4 \approx -1.386294$, so $e^{y} = \tfrac14 = 0.25$. Differentiating, $\frac{dy}{dx} = \frac{1}{C-x} = \frac14 = 0.25$. The two agree, and a central difference at $x = 1$ with step $10^{-6}$ also returns $0.250000$.