# Find the solution (xsiny/x)dy=(ysiny/x -x)dx

Find the solution $\left(x\frac{\mathrm{sin}y}{x}\right)dy=\left(y\frac{\mathrm{sin}y}{x}-x\right)dx$
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The DE is x. $\mathrm{sin}\left(\frac{y}{x}\right)dy=\left(y.\mathrm{sin}\left(\frac{y}{x}\right)-x\right)dx$, giving
$\frac{dy}{dx}=\frac{y}{x}-\frac{1}{\mathrm{sin}\left(\frac{y}{x}\right)}$
let $v=\frac{y}{x}$, then $y=vx\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{dy}{dx}=v+x.d\frac{v}{dx}$
Substituting for v and $\frac{dy}{dx}$ into the DE,
$v+x.d\frac{v}{dx}=v-\frac{1}{\mathrm{sin}\left(v\right)}$
$x.d\frac{v}{dx}=-\frac{1}{\mathrm{sin}\left(v\right)}$
$-\mathrm{sin}\left(v\right)dv=\left(\frac{1}{x}\right)dx$
integrating both sides,
$\mathrm{cos}\left(v\right)=\mathrm{ln}\left(x\right)+K$
$\mathrm{cos}\left(\frac{y}{x}\right)=\mathrm{ln}\left(x\right)+K$
$y=x.a\mathrm{cos}\left(\mathrm{ln}\left(x\right)+K\right)$