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Solve the first equation for $x_2$: $x_2 = (-x'_1 + x_1 + 10)/9$. So $x_2' = (-x''_1 + x_1')/9 = x_1 + x_2 = (-x'_1 + 10 x_1 + 10)/9$. Thus $x_1'' - 2 x'_1 + 10 x_1 = -10$. The characteristic polynomial $r^2 - 2 r + 10$ has roots $(2 \pm \sqrt{4^2-40})/2 = 1 \pm 3 i$, and a particular solution is $x_1 = -1$, so the general solution is $x_1 = -1 + c_1 {\rm e}^t \cos(3 t) + c_2 {\rm e}^t \sin(3 t)$. Then $x_2 = (-x'_1 + x_1 + 10)/9 = 1 - (c_2/3) {\rm e}^t \cos(3 t) + (c_1/3) {\rm e}^t \sin(3 t)$.