The general solution of the differential equation \(\mathrm{e}^{y-x} \frac{\mathrm{~d} y}{\mathrm{~d} x}=y\left(\frac{\sin x+\cos x}{1+y \log y}\right)\). is

\(\begin{array}{ll}
& \mathrm{e}^{y-x} \frac{\mathrm{~d} y}{\mathrm{~d} x}=y\left(\frac{\sin x+\cos x}{1+y \log y}\right) \\
\therefore & \frac{\mathrm{e}^y}{\mathrm{e}^x} \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{y}{(1+y \log y)}(\sin x+\cos x) \\
\therefore & \mathrm{e}^y \frac{(1+y \log y)}{y} \mathrm{~d} y=\mathrm{e}^x(\sin x+\cos x) \mathrm{d} x \\
\therefore \quad & \mathrm{e}^y\left(\log y+\frac{1}{y}\right) \mathrm{d} y=\mathrm{e}^x(\sin x+\cos x) \mathrm{d} x
\end{array}\)
Integrating both sides, we get
\(\begin{aligned}
& \mathrm{e}^y \log y=\mathrm{e}^x \sin x+\mathrm{c} \\
& \quad \ldots\left[\because \int \mathrm{e}^x\left[\mathrm{f}(x)+\mathrm{f}^{\prime}(x)\right] \mathrm{d} x=\mathrm{e}^x \mathrm{f}(x)+\mathrm{c}\right]
\end{aligned}\)