$y$ satisfies $\frac{dy}{dx} + 2y = e^{x}$, and $y = 1$ when $x = 0$. What is $y$ when $x = \log 2$?

Explanation

1. 01

   The factor is $e^{2x}$: $ye^{2x} = \int e^{3x}\,dx = \frac{e^{3x}}{3} + C$, so $y = \frac{e^{x}}{3} + Ce^{-2x}$.

2. 02

   $y(0) = \frac{1}{3} + C = 1$ gives $C = \frac{2}{3}$.

3. 03

   At $x = \log 2$: $\frac{2}{3} + \frac{2}{3} \cdot \frac{1}{4} = \frac{5}{6}$.