$y$ satisfies $x\frac{dy}{dx} = x + 2y$, and $y = 0$ when $x = 1$. What is $y$ when $x = 3$?

Explanation

1. 01

   The general solution is $y = Cx^{2} - x$; $y(1) = 0$ gives $C = 1$.

2. 02

   $y = x^{2} - x$, so at $x = 3$: $9 - 3 = 6$.