Write the system in matrix form $u^{\prime }\left(t\right)=$ $Au\left(t\right)$, where:

$u\left(t\right)=$ $\left(\begin{array}{c}x\left(t\right)\\ y\left(t\right)\end{array}\right)$ and $A=$ $\left(\begin{array}{cc}5.5& -8.5\\ -8.5& 5.5\end{array}\right)$

Find the eigenvalues.

The characteristic equation is:

$\det \left(A-\lambda I\right)=$ $0$

That is:

$\det \left(\begin{array}{cc}5.5-\lambda & -8.5\\ -8.5& 5.5-\lambda \end{array}\right)=$ $(5.5-$ $\lambda \left)\right(5.5-$ $\lambda )-$ $\left(-8.5\right)\left(-8.5\right)=$ $0$

${\lambda }^2-$ $11\lambda -$ $42=$ $0$

Solve for $\lambda$:

$\left(\lambda +3\right)\left(\lambda -14\right)=$ $0$

Thus, the eigenvalues are:

${\lambda }_1=-3$ and ${\lambda }_2=14$

Find the eigenvectors.

For each eigenvalue ${\lambda }_i$, solve:

$\left(A-{\lambda }_{i}I\right)v_i=$ $0$

The eigenvectors are:

$v_1=$ $\alpha \left(\begin{array}{cc}1& 1\end{array}\right)$ and $v_2=$ $\alpha \left(\begin{array}{cc}1& -1\end{array}\right)$

where $\alpha$ can be any real number except 0.

Let $\alpha =1$. Now we can build the solution for this question:

$u\left(t\right)=$ $C_1{e}^{{\lambda }_{1}t}{v}_1+$ $C_2{e}^{{\lambda }_{2}t}{v}_2$

That is:

$x\left(t\right)=$ $e^{-3t}{c}_1+$ $e^{14t}{c}_2$

$y\left(t\right)=$ $e^{-3t}{c}_1-$ $e^{14t}{c}_2$