For $y > 0$, $x$ satisfies $\frac{dx}{dy} + \frac{2x}{y} = 4y$, and $x = 2$ when $y = 1$. What is $x$ when $y = 2$?

Explanation

1. 01

   Linear in $x$ with $P_{1} = \frac{2}{y}$: the factor is $y^{2}$, and $xy^{2} = \int 4y^{3}\,dy = y^{4} + C$.

2. 02

   At $y = 1$: $2 = 1 + C$, so $C = 1$ and $x = y^{2} + \frac{1}{y^{2}}$.

3. 03

   At $y = 2$: $x = 4 + \frac{1}{4} = \frac{17}{4}$.