$$
\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}=y+3 \\
& \Rightarrow \frac{\mathrm{d} y}{y+3}=\mathrm{d} x
\end{aligned}
$$
Integrating on both sides, we get
$$
\begin{array}{ll}
& \int \frac{\mathrm{d} y}{y+3}=\int \mathrm{d} x+\mathrm{c} \\
\Rightarrow & \log (y+3)=x+\mathrm{c} \\
& y=2 \text { when } x=0 \\
\therefore \quad & \log (2+3)=0+\mathrm{c} \Rightarrow \mathrm{c}=\log 5 \\
\therefore \quad & \log (y+3)=x+\log 5 \\
& \Rightarrow y+3=5 \mathrm{e}^x \\
\Rightarrow y=5 \mathrm{e}^x-3 & \\
\therefore \quad & y(\log 2)=5 \mathrm{e}^{\log 2}-3=10-3 \\
& =7
\end{array}
$$