Show that the differential equation $x \frac{dy}{dx} - y + \sin \left(\frac{y}{x}\right) = 0$ is a homogeneous equation and find its solution.

(N/A) $x \frac{dy}{dx} - y + \sin \left(\frac{y}{x}\right) = 0$
$\Rightarrow x \frac{dy}{dx} = y - x \sin \left(\frac{y}{x}\right)$
$\Rightarrow \frac{dy}{dx} = \frac{y - x \sin \left(\frac{y}{x}\right)}{x} \quad \dots (1)$
Let $F(x, y) = \frac{y - x \sin \left(\frac{y}{x}\right)}{x}$.
$\therefore F(\lambda x, \lambda y) = \frac{\lambda y - \lambda x \sin \left(\frac{\lambda y}{\lambda x}\right)}{\lambda x} = \frac{y - x \sin \left(\frac{y}{x}\right)}{x} = \lambda^0 F(x, y)$.
Therefore,the given differential equation is a homogeneous equation.
To solve it,we substitute $y = vx$,which implies $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into equation $(1)$:
$v + x \frac{dv}{dx} = \frac{vx - x \sin v}{x} = v - \sin v$
$\Rightarrow x \frac{dv}{dx} = -\sin v$
$\Rightarrow -\csc v \, dv = \frac{dx}{x}$
Integrating both sides:
$\int -\csc v \, dv = \int \frac{dx}{x}$
$\Rightarrow \ln |\csc v + \cot v| = \ln |x| + C$
$\Rightarrow \ln \left| \frac{1 + \cos v}{\sin v} \right| = \ln |x| + C$
Using $1 + \cos v = 2 \cos^2 \left(\frac{v}{2}\right)$ and $\sin v = 2 \sin \left(\frac{v}{2}\right) \cos \left(\frac{v}{2}\right)$:
$\ln \left| \cot \left(\frac{v}{2}\right) \right| = \ln |x| + C$
$\Rightarrow \cot \left(\frac{y}{2x}\right) = Cx$
Thus,the required solution is $\cot \left(\frac{y}{2x}\right) = Cx$.