First Order ODE/-1 over y sine x over y dx + x over y^2 sine x over y dy

Theorem
The first order ordinary differential equation:


 * $(1): \quad -\dfrac 1 y \sin \dfrac x y + \left({\dfrac x {y^2} \sin \dfrac x y}\right) \dfrac {\mathrm d y} {\mathrm d x} = 0$

is an exact differential equation with solution:


 * $\dfrac x y = C$

This can also be presented as:
 * $\dfrac {\mathrm d y} {\mathrm d x} = \dfrac {\dfrac 1 y \sin \dfrac x y} {\dfrac x {y^2} \sin \dfrac x y}$

Proof
Let:
 * $M \left({x, y}\right) = -\dfrac 1 y \sin \dfrac x y$
 * $N \left({x, y}\right) = \dfrac x {y^2} \sin \dfrac x y$

Then:

Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be exact.

By Solution to Exact Differential Equation, the solution to $(1)$ is:
 * $f \left({x, y}\right) = C$

where:

Hence:

and:

Set $z = \dfrac x y$:

Thus by Integration by Substitution:

Thus:
 * $f \left({x, y}\right) = - \cos \dfrac x y$

and by Solution to Exact Differential Equation, the solution to $(1)$, after simplification, is: