Answer
$$\begin{cases}
x_1'=x_2\\
x_2'=\frac{-tx_2-(t^2-0.25)x_1}{t^2}
\end{cases}$$
Work Step by Step
The given equation is
$t^2u''+tu'+(t^2-0.25)u=0$
Let $x_1=u$ and $x_2=u'$.
We can write $x_1'=x_2$ and $x_1''=u''=\frac{-tx_2-(t^2-0.25)x_1}{t^2}$.
Then the system of first-order equations is:
$$\begin{cases}
x_1'=x_2\\
x_2'=\frac{-tx_2-(t^2-0.25)x_1}{t^2}
\end{cases}$$
