Solve the differential equation
$\;$ $\left[x \cos \left(\dfrac{y}{x}\right) + y \sin \left(\dfrac{y}{x}\right)\right] y = \left[y \sin \left(\dfrac{y}{x}\right) - x \cos \left(\dfrac{y}{x}\right)\right] x \; \dfrac{dy}{dx}$
The given differential equation is
$\left[x \cos \left(\dfrac{y}{x}\right) + y \sin \left(\dfrac{y}{x}\right)\right] y = \left[y \sin \left(\dfrac{y}{x}\right) - x \cos \left(\dfrac{y}{x}\right)\right] x \; \dfrac{dy}{dx}$
i.e. $\dfrac{dy}{dx} = \left[\dfrac{x \cos \left(\dfrac{y}{x}\right) + y \sin \left(\dfrac{y}{x}\right)}{y \sin \left(\dfrac{y}{x}\right) - x \cos \left(\dfrac{y}{x}\right)}\right] \times \dfrac{y}{x}$
i.e. $\dfrac{dy}{dx} = \left[\dfrac{\cos \left(\dfrac{y}{x}\right) + \dfrac{y}{x} \sin \left(\dfrac{y}{x}\right)}{\dfrac{y}{x} \sin \left(\dfrac{y}{x}\right) - \cos \left(\dfrac{y}{x}\right)}\right] \times \dfrac{y}{x}$ $\;\;\; \cdots \; (1)$
Let $\dfrac{y}{x} = v$ $\;\;\; \cdots \; (2a)$
$\implies$ $y = v \; x$ $\;\;\; \cdots \; (2b)$
Differentiating equation $(2b)$ w.r.t x gives
$\dfrac{dy}{dx} = v + x \; \dfrac{dv}{dx}$ $\;\;\; \cdots \; (2c)$
$\therefore$ $\;$ In view of equations $(2a)$ and $(2c)$, equation $(1)$ becomes
$v + x \; \dfrac{dv}{dx} = \left[\dfrac{\cos \left(v\right) + v \; \sin \left(v\right)}{v \; \sin \left(v\right) - \cos \left(v\right)}\right] \times v$
i.e. $x \; \dfrac{dv}{dx} = \left[\dfrac{\cos \left(v\right) + v \; \sin \left(v\right)}{v \; \sin \left(v\right) - \cos \left(v\right)} - 1\right] \times v$
i.e. $x \; \dfrac{dv}{dx} = \left[\dfrac{\cos \left(v\right) + v \; \sin \left(v\right) - v \; \sin \left(v\right) + \cos \left(v\right)}{v \; \sin \left(v\right) - \cos \left(v\right)}\right] \times v$
i.e. $x \; \dfrac{dv}{dx} = \dfrac{2 \; v \; \cos \left(v\right)}{v \; \sin \left(v\right) - \cos \left(v\right)}$
i.e. $\left[\dfrac{v \; \sin \left(v\right) - \cos \left(v\right)}{v \; \cos \left(v\right)}\right] \; dv = 2 \; \dfrac{dx}{x}$
i.e. $\left[\tan \left(v\right) - \dfrac{1}{v}\right] \; dv = 2 \; \dfrac{dx}{x}$ $\;\;\; \cdots \; (3)$
Integrating equation $(3)$ gives
$\displaystyle \int \left[\tan \left(v\right) - \dfrac{1}{v}\right] \; dv = 2 \int \dfrac{dx}{x}$
i.e. $\log \left|\sec \left(v\right)\right| - \log \left|v\right| = 2 \; \log \left|x\right| + \log \left|c\right|$ $\;\;\; \cdots \; (4)$
Substituting the value of $v$ from equation $(2a)$ in equation $(4)$ gives
$\log \left|\sec \left(\dfrac{y}{x}\right)\right| - \log \left|\dfrac{y}{x}\right| - 2 \log \left|x\right| = \log \left|c\right|$
i.e. $\log \left|\dfrac{\sec \left(\dfrac{y}{x}\right)}{\dfrac{y}{x} \times x^2}\right| = \log \left|c\right|$
i.e. $\dfrac{\sec \left(\dfrac{y}{x}\right)}{x \; y} = c$
i.e. $\sec \left(\dfrac{y}{x}\right) = c \; x \; y$ $\;\;\; \cdots \; (5)$
Equation $(5)$ is the general solution of the given differential equation.