A curve passes through the point `(1,(pi)/(6)).` Let the slope of the curve at each point (x,y) be `(y)/(x)+sec((y)/(x)),xgt0.` Then, the equation of the curve is

Options

A. `sin((y)/(x))=logx+(1)/(2)`
B. `cosec(y)/(x)=logx+2`
C. `sec((2y)/(x))=logx+2`
D. `cos((2y)/(x))=logx+(1)/(2)`

Answer

To solve the problem, we need to find the equation of the curve given that the slope of the curve at each point \((x, y)\) is defined by the equation:

\[ \frac{dy}{dx} = \frac{y}{x} + \sec\left(\frac{y}{x}\right) \]

and that the curve passes through the point \((1, \frac{\pi}{6})\).

### Step-by-Step Solution:

1. **Substituting \(y = ux\)**:
We can use the substitution \(y = ux\), where \(u\) is a function of \(x\). This gives us:
\[ \frac{dy}{dx} = u + x\frac{du}{dx} \]

2. **Rewriting the slope equation**:
Substitute \(y = ux\) into the slope equation:
\[ \frac{dy}{dx} = \frac{ux}{x} + \sec\left(\frac{ux}{x}\right) = u + \sec(u) \]
Therefore, we have:
\[ u + x\frac{du}{dx} = u + \sec(u) \]

3. **Simplifying the equation**:
Cancel \(u\) from both sides:
\[ x\frac{du}{dx} = \sec(u) \]

4. **Separating variables**:
We can separate the variables:
\[ \cos(u) du = \frac{dx}{x} \]

5. **Integrating both sides**:
Integrate both sides:
\[ \int \cos(u) du = \int \frac{dx}{x} \]
This gives:
\[ \sin(u) = \log(x) + C \]

6. **Substituting back for \(u\)**:
Recall that \(u = \frac{y}{x}\), so we substitute back:
\[ \sin\left(\frac{y}{x}\right) = \log(x) + C \]

7. **Finding the constant \(C\)**:
We know that the curve passes through the point \((1, \frac{\pi}{6})\). Substitute \(x = 1\) and \(y = \frac{\pi}{6}\):
\[ \sin\left(\frac{\pi}{6}\right) = \log(1) + C \]
Since \(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\) and \(\log(1) = 0\), we have:
\[ \frac{1}{2} = 0 + C \implies C = \frac{1}{2} \]

8. **Final equation of the curve**:
Substitute \(C\) back into the equation:
\[ \sin\left(\frac{y}{x}\right) = \log(x) + \frac{1}{2} \]

Thus, the equation of the curve is:

\[ \sin\left(\frac{y}{x}\right) = \log(x) + \frac{1}{2} \]