Give the correct answer and solve the given equation dx + (frac{x}{y} ​− sin y)dy = 0

Give the correct answer and solve the given equation $dx+\left(\frac{x}{y}-\mathrm{sin}y\right)dy=0$
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brawnyN

Multiply this equation by y:
$ydx+\left(x-y\mathrm{sin}y\right)dy=0$
Now we will try to find a function F: $R{R}^{2}\to R$ such that
$\frac{\partial F}{\partial x}=y$
and $\frac{\partial F}{\partial x}=x-y\mathrm{sin}y$ (1)
Start with $\frac{\partial F}{\partial x}=y$
and integrate with respect to x:
$F\left(x,y\right)=\int ydx=xy+C\left(y\right)\left(2\right)$
where C(y) is a constant with respect to x (an integrating constant!). From (2)
$\frac{\partial F}{\partial x}=x+{C}^{\prime }\left(y\right)$
Using (1) now, we get that
$x+{C}^{\prime }\left(y\right)=x-y\mathrm{sin}y⇒{C}^{\prime }\left(y\right)=-y\mathrm{sin}y$
Thus,
$C\left(y\right)=\int -y\mathrm{sin}ydy$
$=-\int y\mathrm{sin}ydy$

$=y\mathrm{cos}y-\int \mathrm{cos}ydy$
$=y\mathrm{cos}y-\sim y+D,$
where D is a constant.
Therefore,
$F\left(x,y\right)=xy+C\left(y\right)=xy+y\mathrm{cos}y-\mathrm{sin}y+D$
Now the solution of the initial differential equation is given by $F\left(x,y\right)=0$,
so $xy+y\mathrm{cos}y-\mathrm{sin}y=-D=C,$
where C is some constant.