Show that the differential equation $x \cos \left(\frac{y}{x}\right) \frac{d y}{d x}=y \cos \left(\frac{y}{x}\right)+x$ is homogeneous and solve it.

(N/A) The given differential equation can be written as
$\frac{d y}{d x}=\frac{y \cos \left(\frac{y}{x}\right)+x}{x \cos \left(\frac{y}{x}\right)}$ ............$(1)$
It is a differential equation of the form $\frac{d y}{d x}=F(x, y)$.
Here $F(x, y) = \frac{y \cos \left(\frac{y}{x}\right) + x}{x \cos \left(\frac{y}{x}\right)}$.
Replacing $x$ by $\lambda x$ and $y$ by $\lambda y$,we get
$F(\lambda x, \lambda y) = \frac{\lambda y \cos \left(\frac{\lambda y}{\lambda x}\right) + \lambda x}{\lambda x \cos \left(\frac{\lambda y}{\lambda x}\right)} = \frac{\lambda [y \cos (y/x) + x]}{\lambda [x \cos (y/x)]} = \lambda^0 F(x, y)$.
Since $F(x, y)$ is a homogeneous function of degree zero,the given differential equation is homogeneous.
To solve it,substitute $y = vx$,which implies $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into equation $(1)$:
$v + x \frac{dv}{dx} = \frac{vx \cos v + x}{x \cos v} = \frac{v \cos v + 1}{\cos v}$.
$x \frac{dv}{dx} = \frac{v \cos v + 1}{\cos v} - v = \frac{v \cos v + 1 - v \cos v}{\cos v} = \frac{1}{\cos v}$.
Separating variables: $\cos v \, dv = \frac{dx}{x}$.
Integrating both sides: $\int \cos v \, dv = \int \frac{1}{x} \, dx$.
$\sin v = \log |x| + C$.
Substituting $v = y/x$,the general solution is $\sin \left(\frac{y}{x}\right) = \log |x| + C$.