The general solution of the differential equation. $\frac{\mathrm{dy}}{\mathrm{d} x}+\sin \left(\frac{x+\mathrm{y}}{2}\right)=\sin \left(\frac{x-\mathrm{y}}{2}\right)$ is

The differential equation is $\frac{\mathrm{d}y}{\mathrm{d}x} + \sin\left(\frac{x+y}{2}\right) = \sin\left(\frac{x-y}{2}\right)$.

Rearranging terms gives $\frac{\mathrm{d}y}{\mathrm{d}x} = \sin\left(\frac{x-y}{2}\right) - \sin\left(\frac{x+y}{2}\right)$.

Applying the sum-to-product identity $\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$ with $A = \frac{x-y}{2}$ and $B = \frac{x+y}{2}$ yields $\frac{\mathrm{d}y}{\mathrm{d}x} = -2 \cos\left(\frac{x}{2}\right) \sin\left(\frac{y}{2}\right)$.

Separating variables results in $\frac{\mathrm{d}y}{\sin\left(\frac{y}{2}\right)} = -2 \cos\left(\frac{x}{2}\right) \mathrm{d}x$.

Integrating both sides:

$\int \csc\left(\frac{y}{2}\right) \mathrm{d}y = -2 \int \cos\left(\frac{x}{2}\right) \mathrm{d}x$

The left side integrates to $2 \log\left|\tan\left(\frac{y}{4}\right)\right|$ and the right side to $-4 \sin\left(\frac{x}{2}\right)$, giving $2 \log\left|\tan\left(\frac{y}{4}\right)\right| = -4 \sin\left(\frac{x}{2}\right) + C'$.

Simplifying yields $\log\left|\tan\left(\frac{y}{4}\right)\right| = c - 2 \sin\left(\frac{x}{2}\right)$ where $c$ is an integration constant.

This matches option B.