For $x > 0$, $y$ satisfies $x\frac{dy}{dx} - y = 2\sqrt{x^{2} + y^{2}}$, and $y = 0$ when $x = 1$. What is $y$ when $x = 2$?

Explanation

1. 01

   The general solution is $y + \sqrt{x^{2} + y^{2}} = Cx^{3}$; at $(1, 0)$, $0 + 1 = C$, so $C = 1$.

2. 02

   At $x = 2$: $\sqrt{4 + y^{2}} = 8 - y$, so $4 + y^{2} = 64 - 16y + y^{2}$ and $y = \frac{60}{16} = \frac{15}{4}$.