A curve passes through the point (1 , (π)/(6)) . Let the slope of...

Given slope at (x, y) is

$\dfrac{\text{dy}}{\text{dx}} = \dfrac{\text{y}}{\text{x}} + \text{sec}\left(\dfrac{y}{x}\right)$......\left(i\right)$<br />let$\dfrac{\text{y}}{\text{x}} = \text{t} ⇒ \text{y} = \text{xt}$</p><p>Differentiating this wrt$x$</p><p>$⇒ \dfrac{\text{dy}}{\text{dx}} = \text{t} + \text{x}\dfrac{\text{dt}}{\text{dx}}$......\left(ii\right)$

Now using equation \left(i\right)$and \left(ii\right)$, we have
$\text{t} + \text{x} \dfrac{\text{dt}}{\text{dx}} = \text{t} + \text{sec} \left(\text{t}\right)$

$⇒ \dfrac{dt}{\sec \left(t\right)} = \dfrac{dx}{x}$

Integrating both sides.
$∫ \cos \left(t\right)\text{dt} = ∫ \dfrac{1}{\text{x}}\text{dx}$
$\sin \left(t\right) = \ln \left(x\right) + \text{c}$

Now Put value of $t$ in above equation.
\text\sin\left(\dfrac{y}{x}\right) = ln\left(x\right) + \text{c}$.....\left(iii\right)$
Given that curve passes through \left(1\text{,}\dfrac{\pi}{6}\right)$so put in above equation.<br />$\text\sin\left(\dfrac{\pi}{6}\right) = \ln \left(1\right) + \text{c} ⇒ \text{c} = \dfrac{1}{2}$</p><p>Now, put the value of$c$in equation \left(iii\right)$
\text\sin\left(\dfrac{y}{x}\right) = \ln \left(x\right) + \dfrac{1}{2}