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

The equation of the curve passing through $\left(2, \frac{9}{2}\right)$ and having the slope $\left(1-\frac{1}{x^2}\right)$ at $(x, y)$ is

If $y=y(x)$ and $\left(\frac{2+\sin x}{y+1}\right) \frac{\mathrm{d} y}{\mathrm{~d} x}=-\cos x, y(0)=1$, then $y\left(\frac{\pi}{2}\right)=$

The degree of the differential equation $\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}+3\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2=x^2 \log \left(\frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}\right)$ is