Find an integrating factor and solve the given equation.
$1+(\frac{x}{y}-sin(y))y'=0$
$M=1$
$N=(\frac{x}{y}-sin(y))$
$R=\frac{My-Nx}{M}$
$R=\frac{0-(\frac{1}{y}-0)}{1}$
$R=-\frac{1}{y}$
$\mu=e^{-\int -\frac{1}{y}dy}$
$\mu=e^{lny}$
$\mu=y$
Multiply $\mu$ to both side:
$y+(x-ysin(y))y'=0$
$My=1$
$Nx=1$
It is exact now, so $\mu=y$ is the integrating factor.
Integrate M with respect to x and we get:
$xy+h(y)$
Take derivative on both side with respect to $y$ and we get:
$x+h'(y)$
$h'(y)=-ysiny$
$h(y)=-\int ysiny$
