Equation of Curve from Given Point and Homogeneous Slope

To find the equation of the curve, we start from the given differential equation for the slope: $\frac{dy}{dx} = \frac{y}{x} + \sec\left(\frac{y}{x}\right), \quad x > 0$

This is a standard homogeneous differential equation. Let: $y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$

Substitute $y$ and $\frac{dy}{dx}$ into the differential equation: $v + x\frac{dv}{dx} = v + \sec(v)$

Subtracting $v$ from both sides gives: $x\frac{dv}{dx} = \sec(v)$

Separating the variables: $\frac{dv}{\sec(v)} = \frac{dx}{x}$ $\cos(v) \, dv = \frac{dx}{x}$

Integrating both sides: $\int \cos(v) \, dv = \int \frac{1}{x} \, dx$ $\sin(v) = \log x + C$

Substituting back $v = \frac{y}{x}$: $\sin\left(\frac{y}{x}\right) = \log x + C$

Since the curve passes through the point $\left(1, \frac{\pi}{6}\right)$, we substitute $x = 1$ and $y = \frac{\pi}{6}$: $\sin\left(\frac{\pi}{6}\right) = \log(1) + C$ $\frac{1}{2} = 0 + C \implies C = \frac{1}{2}$

Therefore, the equation of the curve is: $\sin\left(\frac{y}{x}\right) = \log x + \frac{1}{2}$

This corresponds to Option (A).