Answer
$$\begin{cases}
x_1'=x_2\\
x_2'=-0.5x_2-2x_1+3\sin t
\end{cases}$$
Work Step by Step
The given equation is
$u''+0.5u'+2u=3\sin t$
Let $x_1=u$ and $x_2=u'$.
We can write $x_1'=x_2$ and $x_1''=u''=-0.5x_2-2x_1+3\sin t$.
Then the system of first-order equations is:
$$\begin{cases}
x_1'=x_2\\
x_2'=-0.5x_2-2x_1+3\sin t
\end{cases}$$
