Let $y^{\mathrm{\prime }}+\sin \frac{x+y}{2}=\sin \frac{x-y}{2}$

Statement-1:  A solution satisfying $y\left(0\right)=\pi$ is a periodic function with period $4\pi$

Statement-1: $y$ can be explicitly represented in terms of x

$y^{\mathrm{\prime }}=\sin \frac{x-y}{2}-\sin \frac{x+y}{2}=-2\sin \frac{y}{2}\cos \frac{x}{2}$

$\mathrm{cosec} y/2dy=-2\cos (x/2)dx$

$\begin{array}{l}\Rightarrow \mathrm{cosec} y/2dy=-2\cos (x/2)dx\\ \Rightarrow 2\log |\tan (y/4\left)\right|=-4\sin (x/2)+\text{ Const }\\ \Rightarrow \log |\tan (y/4\left)\right|=-2\sin (x/2)+\text{ Const }\\ \Rightarrow |\tan (y/4\left)\right|=\text{ Const }{e}^{-2\sin (x/2)}\\ \text{ Since }y\left(0\right)=\pi \text{ so Const }=1\text{. Thus }|\tan (y/4\left)\right|=e^{-2\sin x/2}\end{array}$

so $y=4{\tan }^{-1} \left(\pm e^{-2\sin x/2}\right)$ which is periodic with period 

$4\pi$