Official solution
\begin{gathered} \, ydy =(xdy-ydx)\sin (\frac{x}{y}) \\ \frac{dy}{y} =(\frac{xdy-ydx}{{y}^{2}})\sin (\frac{x}{y}) \\ \frac{dy}{y} =\sin (\frac{x}{y})d(-\frac{x}{y}) \\ \Rightarrow \, lny =\cos \frac{x}{y}+C
x(1)=\frac{\pi }{2}\Rightarrow 0=\cos \frac{\pi }{2}+C\Rightarrow C=0 \\ =\cos \frac{x}{y} \\ y=2\Rightarrow \cos \frac{x}{2}=\ln 2 \\ \, \cos x =2{\cos }^{2}\frac{x}{2}-1 \\ =2(\ln 2{)}^{2}-1 \end{gathered}