Explanation

(b)
Given, $\frac{dy}{dx} = \frac{y + x\ \tan\frac{y}{x}}{x}$
Put $y = vx$
$\Rightarrow$ $\frac{dy}{dx} = v + x\ \frac{dv}{dx}$
$\therefore$ $v + x\ \ \frac{dv}{dx} = \frac{vx + x\ \tan\left( \frac{vx}{x} \right)}{x}$
$\Rightarrow$ $x\ \frac{dv}{dx} = v + \tan{v - v}$
$\Rightarrow$ $\int_{}^{}{\cot v}\ dv = \int_{}^{}\frac{dx}{x}$
$\Rightarrow$ $\log{\sin v} = \log x + \log c\ \ \Rightarrow \sin\frac{y}{x} = xc$