Isolate $x_2$ in first equation we get $x_2 = \frac{4}{3}x'_1 - \frac{5}{3}x_1$
Differentiate both sides with respect to t we get $x'_2 = \frac{4}{3}x''_1 - \frac{5}{3}x'_1$
Sub into second equation , we get $x''_1 - \frac{5}{2}x'_1 + x_1 = 0$
Characteristic equation is $r^2 - \frac{5}{2} r + 1 = 0$,
hence $r_1 = \frac{1}{2}, r_2 = 2$
General solution for $x_1$ is $x_1 = c_1e^\frac{t}{2}+ c_2 e^{2t}$
Plug into $x_2 = \frac{4}{3}x'_1 - \frac{5}{3}x_1$, we get $x_2 = -c_1 e^\frac{t}{2}+ c_2 e^{2t}$,
So, $x_1 = c_1e^\frac{t}{2}+ c_2e^{2t}$, $x_2 = -c_1 e^\frac{t}{2} + c_2 e^{2t}$
Plug in $x_1(0) = -2, x_2(0) = 1$,
$c_1 + c_2 = -2, -c_1 + c_2 = 1$
hence $c_1 = -\frac{3}{2}, c_2 = -\frac{1}{2}$
Therefore, $x_1 = -\frac{3}{2} e^\frac{t}{2}-\frac{1}{2} e^{2t}$,
$x_2 = \frac{3}{2} e^\frac{t}{2} -\frac{1}{2} e^{2t}$
