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

Here \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}}{\mathrm{x}}+\sec \left(\frac{\mathrm{y}}{\mathrm{x}}\right)\)
Let \(\frac{y}{x}=v \Rightarrow y=v x \Rightarrow \frac{d y}{d x}=v+x \cdot \frac{d v}{d x}\)
\(\begin{aligned} & \Rightarrow v+x \cdot \frac{d v}{d x}=v+\sec v \\ & \Rightarrow x \cdot \frac{d v}{d x}=\operatorname{secv} \\ & \Rightarrow \int \cos v \cdot d v=\int \frac{d x}{x} \\ & \Rightarrow \operatorname{sinv}=\log _e x+C\end{aligned}\)
\(\Rightarrow \sin \left(\frac{y}{x}\right)=\log _e x+C\)
Putting \(\mathrm{x}=1, \mathrm{y}=\frac{\pi}{6} \Rightarrow \mathrm{c}=\frac{1}{2}\)
\(\Rightarrow \sin \left(\frac{y}{x}\right)=\log _e^x+\frac{1}{2}\)