\[ y^{\prime }x-y-2\sin \left ( 3\frac {y}{x}\right ) =0 \]
The first step is to see if we can write the above as \begin {equation} y^{\prime }=\frac {y}{x}+g\left ( x\right ) f\left ( b\frac {y}{x}\right ) ^{\frac {n}{m}} \tag {1} \end {equation} Hence \begin {align} y^{\prime }x-y-2\sin \left ( 3\frac {y}{x}\right ) & =0\nonumber \\ y^{\prime } & =\frac {y}{x}-\frac {2}{x}\sin \left ( 3\frac {y}{x}\right ) \tag {2} \end {align}
Comparing (2) to (1) shows that\begin {align*} n & =1\\ m & =1\\ g\left ( x\right ) & =-\frac {2}{x}\\ b & =3\\ f\left ( b\frac {y}{x}\right ) & =\sin \left ( 3\frac {y}{x}\right ) \end {align*}
Hence the solution is \begin {equation} y=ux \tag {A} \end {equation} Where \(u\) is the solution to \begin {equation} u^{\prime }=\frac {1}{x}g\left ( x\right ) f\left ( u\right ) \tag {3} \end {equation} Therefore \(f\left ( u\right ) =\sin \left ( 3u\right ) \) and \(\left ( 3\right ) \) becomes\[ u^{\prime }=-\frac {2}{x^{2}}\sin \left ( 3u\right ) \] This is separable. \begin {align*} \frac {1}{\sin \left ( 3u\right ) }du & =-\frac {2}{x^{2}}dx\\ \int \frac {1}{\sin \left ( 3u\right ) }du & =-2\int \frac {1}{x^{2}}dx\\ \frac {1}{3}\left ( \ln \sin \left ( \frac {3u}{2}\right ) -\ln \cos \left ( \frac {3u}{2}\right ) \right ) & =\frac {2}{x}+c_{1}\\ \ln \sin \left ( \frac {3u}{2}\right ) -\ln \cos \left ( \frac {3u}{2}\right ) & =-\frac {6}{x}+c_{2}\\ \ln \frac {\sin \left ( \frac {3u}{2}\right ) }{\cos \left ( \frac {3u}{2}\right ) } & =-\frac {6}{x}+c_{2}\\ \ln \tan \left ( \frac {3u}{2}\right ) & =-\frac {6}{x}+c_{2}\\ \tan \left ( \frac {3u}{2}\right ) & =c_{3}e^{-\frac {6}{x}}\\ \frac {3u}{2} & =\arctan \left ( c_{3}e^{-\frac {6}{x}}\right ) \\ u & =\frac {2}{3}\arctan \left ( c_{3}e^{-\frac {6}{x}}\right ) \end {align*}
And (A) becomes\[ y=\frac {2}{3}x\arctan \left ( c_{3}e^{-\frac {6}{x}}\right ) \]