A curve passes through the point $\left(1, \frac{\pi}{6}\right)$. Let the slope of the curve at each point $(x, y)$ be $\frac{y}{x} + \sec\left(\frac{y}{x}\right)$, $x > 0$. Then the equation of the curve is

$\text{curve passes through } \left(1, \tfrac{\pi}{6}\right)$

$\frac{dy}{dx} = \frac{y}{x} + \sec\!\left(\frac{y}{x}\right),\quad x>0$

$\text{Find the equation of the curve}$

$\text{(choose the correct option)}$

$1.\ \text{homogeneous: put } v=y/x$

$2.\ \text{separate and integrate}$

$3.\ \text{use the point to fix } C$

$\frac{dy}{dx} = \frac{y}{x}+\sec\!\left(\frac{y}{x}\right) \ \text{is homogeneous}$

$v=\frac{y}{x} \Rightarrow y=vx$

$\frac{dy}{dx} = v + x\frac{dv}{dx}$

$v + x\frac{dv}{dx} = v + \sec v \Rightarrow x\frac{dv}{dx} = \sec v$

$\cos v\,dv = \frac{dx}{x}$

$\sin v = \log x + C$

$\Rightarrow \sin\!\left(\frac{y}{x}\right) = \log x + C$

$x=1,\ \tfrac{y}{x}=\tfrac{\pi}{6}:\ \sin\tfrac{\pi}{6} = \log 1 + C$

$\frac{1}{2} = 0 + C \Rightarrow C = \frac{1}{2}$

$\sin\!\left(\frac{y}{x}\right) = \log x + \frac{1}{2} \ \Rightarrow\ \text{option (1)}$

$\boxed{\sin\!\left(\frac{y}{x}\right) = \log x + \frac{1}{2} \Rightarrow (1)}$