The general solution of \(\frac{\mathrm{d} y}{\mathrm{~d} x}+\sin \left(\frac{x+y}{2}\right)=\sin \left(\frac{x-y}{2}\right)\) is

\(\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}+\sin \left(\frac{x+y}{2}\right)=\sin \left(\frac{x-y}{2}\right) \\
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\sin \left(\frac{x-y}{2}\right)-\sin \left(\frac{x+y}{2}\right) \\
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=-2 \sin \left(\frac{y}{2}\right) \cdot \cos \left(\frac{x}{2}\right)
\end{aligned}\)
Integrating on both sides, we get
\(\begin{aligned}
& \int \operatorname{cosec}\left(\frac{y}{2}\right) \mathrm{d} y=-\int 2 \cos \left(\frac{x}{2}\right) \mathrm{d} x+\mathrm{c}_1 \\
& \Rightarrow \frac{\log \tan \left(\frac{y}{4}\right)}{\frac{1}{2}}=-\frac{2 \sin \left(\frac{x}{2}\right)}{\frac{1}{2}}+\mathrm{c}_1 \\
& \Rightarrow \log \tan \left(\frac{y}{4}\right)=C-2 \sin \left(\frac{x}{2}\right), \text { where } C=\frac{1}{2} c_1
\end{aligned}\)