The particular solution of differential equation \(\mathrm{e}^{\frac{d y}{d x}}=(x+1), y(0)=3\) is

\(
\begin{aligned}
& \mathrm{e}^{\frac{\mathrm{d}}{\mathrm{d} x}}=(x+1) \\
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\log (x+1)
\end{aligned}
\)
Integrating on both sides, we get
\(\int \mathrm{d} y =\int \log (x+1) \mathrm{d} x+\mathrm{c} \)
\( \Rightarrow y =x \log (x+1)-\int \frac{x}{x+1} \mathrm{~d} x+\mathrm{c} \)
\( =x \log (x+1)-\int \frac{x+1-1}{x+1} \mathrm{~d} x+\mathrm{c} \)
\( =x \log (x+1)-\int\left(1-\frac{1}{x+1}\right) \mathrm{d} x+\mathrm{c} \)
\( y =x \log (x+1)-x+\log (x+1)+\mathrm{c}.\)
Since \(y(0)=3\), i.e., \(y=3\) when \(x=0\)
\(
\begin{aligned}
& 3=0+\mathrm{c} \Rightarrow \mathrm{c}=3 \\
& y=x \log (x+1)+\log (x+1)-x+3 \\
& y=(x+1) \log (x+1)-x+3
\end{aligned}
\)