Answer
$x_1=\sin 2t\\
x_2=\cos 2t$
Work Step by Step
Taking the Laplace transform of both sides of the given differential equation and imposing the initial conditions yields:
$\frac{dx_1}{dt}=2x_2\\
\rightarrow sL(x_1)=2(s_2)\\
sL(x_1)-2(x_2)=0$ (1)
and $\frac{dx_2}{dt}=-2x_1\\
sL(x_2)+2(x_1)=1$ (2)
Substitute $sL(x_1)=2(s_2)$ to (2):
$s[\frac{s}{2}L(x_1)]+2L(x_1)=1$
Hence, $L(x_1)=\frac{2}{s^2+4} \rightarrow x_1=\sin 2t\\
\rightarrow x_2=\cos 2t$
