Solving the first equation for B , we obtain $x_2 = x_1'/2+x_1/4$. Substitution into the second equation results in $$x_1''/2+x_1'/4 = -2x_1-(x_1'/2+x_1'/4)/2$$.
Rearranging the terms, the single differential equation for $x_1$ is $$x_1''+x_1'+\frac{17}{4}x_1=0$$
The general soln is $$x_1(t) = e^{-t/2}(c_1cos2t+c_2sin2t)$$. With $x_2$ given in terms of $x_1$, it has
$$x_2(t) = e^{-t/2}(-c_1cos2t+c_2sin2t)$$.
Imposing the specified initial conditions, we obtain $c_1 = -2, c_2 = 2$. Hence,
$$x_1(t) = e^{-t/2}(-2cos2t+2sin2t)$$.
$$x_2(t) = e^{-t/2}(2cos2t+2sin2t)$$.
Attached is the graph.
