If \(y=y(x)\) is the solution of the differential equation \(\left(\frac{5+\mathrm{e}^x}{2+y}\right) \frac{\mathrm{d} y}{\mathrm{~d} x}+\mathrm{e}^x=0\) satisfying \(y(0)=1\), then a value of \(y(\log 13)\) is

\(\begin{aligned}
& \left(\frac{5+\mathrm{e}^x}{2+y}\right) \frac{\mathrm{d} y}{\mathrm{~d} x}+\mathrm{e}^x=0 \\
& \Rightarrow \frac{\mathrm{~d} y}{2+y}=\frac{-\mathrm{e}^x}{5+\mathrm{e}^x} \mathrm{~d} x
\end{aligned}\)
Integrating on both sides, we get
\(\begin{aligned}
& \log |2+y|=-\log \left|5+\mathrm{e}^x\right|+\log |\mathrm{c}| \\
& \Rightarrow \log |2+y|=\log \left|\frac{\mathrm{c}}{5+\mathrm{e}^x}\right|
\end{aligned}\)
Since \(y(0)=1\) i.e., \(y=1\) when \(x=0\)
\(\begin{aligned} & \therefore \quad \log 3=\log \left|\frac{\mathrm{c}}{6}\right| \\ & \quad \Rightarrow 3=\frac{\mathrm{c}}{6} \\ & \Rightarrow \quad \mathrm{c}=18 \\ & \therefore \quad \log |2+y|=\log \left|\frac{18}{5+\mathrm{e}^x}\right| \quad \ldots[\text { From (i) }] \\ & \Rightarrow 2+y=\frac{18}{5+\mathrm{e}^x} \\ & \Rightarrow y=\frac{18}{5+\mathrm{e}^x}-2\end{aligned}\)
\(\begin{aligned} \Rightarrow y(\log 13) & =\frac{18}{5+\mathrm{e}^{\log 13}}-2 \\ & =\frac{18}{5+13}-2 \\ & =-1\end{aligned}\)