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

Explanation

1. 01

   $y = vx$: $x\frac{dv}{dx} = \cos^{2}v$, so $\sec^{2}v\,dv = \frac{dx}{x}$ and $\tan\frac{y}{x} = \log x + C$; $y(1) = 0$ gives $C = 0$.

2. 02

   At $x = e$: $\tan\frac{y}{e} = 1$, so $\frac{y}{e} = \frac{\pi}{4}$ and $y = \frac{e\pi}{4}$.