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

Explanation

1. 01

   The general solution is $y = x\log x + Cx$; $y(1) = 2$ gives $C = 2$.

2. 02

   At $x = e$: $e \cdot 1 + 2e = 3e$.