First Order ODE/x sine (y over x) y' = y sine (y over x) + x

Theorem
The first order ordinary differential equation:


 * $(1): \quad x \sin \dfrac y x \dfrac {\mathrm d y} {\mathrm d x} = y \sin \dfrac y x + x$

is a homogeneous differential equation with solution:


 * $\cos \dfrac y x + \ln c x = 0$

Proof
Let:
 * $M \left({x, y}\right) = y \sin \dfrac y x + x$
 * $N \left({x, y}\right) = x \sin \dfrac y x$

Put $t x, t y$ for $x, y$:

Thus both $M$ and $N$ are homogeneous functions of degree $1$.

Thus, by definition, $(1)$ is a homogeneous differential equation.

By Solution to Homogeneous Differential Equation, its solution is:
 * $\displaystyle \ln x = \int \frac {\mathrm d z} {f \left({1, z}\right) - z} + C$

where:
 * $f \left({x, y}\right) = \dfrac {y \sin \dfrac y x + x} {x \sin \dfrac y x}$

Thus: