Answer
\[y(x)=x \cos^{-1}\left(\frac{1}{Cx}\right) \]
Work Step by Step
\[\sin\left(\frac{y}{x}\right)(xy'-y)=x\cos\left(\frac{y}{x}\right)\]
\[y'=\frac{x\cot\left(\frac{y}{x}\right)+y}{x}\]
\[y'=\cot\left(\frac{y}{x}\right)+\frac{y}{x}\;\;\;\ldots (1)\]
Substitute $\:y=Vx\;\;\;\ldots (2)$
Differentiate (2) with respect to $x$
\[y'=V+x\frac{dV}{dx} \;\;\;\ldots (3)\]
From (1), (2) and (3)
\[V+x\frac{dV}{dx}=\cot V+V\]
\[x\frac{dV}{dx}=\cot V\]
Separating variables,
\[\tan V dV=\frac{1}{x}dx\]
Integrating,
\[\int\tan V dV=\int\frac{1}{x}dx+\ln C\]
$\ln C$ is constant of integration
\[-\ln |\cos V|=\ln |x|+\ln |C|\]
\[\frac{1}{\cos V}=Cx\]
\[V=\cos^{-1}\left(\frac{1}{Cx}\right)\]
From (2)
\[\frac{y}{x}= \cos^{-1}\left(\frac{1}{Cx}\right) \]
\[y=x \cos^{-1}\left(\frac{1}{Cx}\right) \]
Hence general solution of (1) is \[y(x)=x \cos^{-1}\left(\frac{1}{Cx}\right) \]
