Setting $x = \xi t^r$ results in the algebraic equations
$$\begin{pmatrix}1-r & -1\\\ 5 & -3-r\end{pmatrix} \begin{pmatrix}\xi_1 \\ \xi_2 \end{pmatrix} = \begin{pmatrix}0 \\ 0 \end{pmatrix}$$
The characteristic equation is $r^2+2r+2=0$, with root $r = -1+i, r = -1-i$. Substituting reduces $r = -1-i$ the system of equations to $(2+i)\xi_{1}-\xi_{2}=0$,The eigenvectors are $\xi^{(1)} = (1,2+i)^T$, $\xi^{(2)} = (1,2-i)^T$.Hence one of the complex-valued solutions is given by $$x^{(1)} = \begin{pmatrix}1 \\\ 2+i \end{pmatrix}e^{-(1+i)t} =\begin{pmatrix}1 \\\ 2+i \end{pmatrix}e^{-t}(\cos t - i \sin t) $$
Therefore the general solution is $$x = c_1 e^{-t}\begin{pmatrix}\cos t \\\ 2\cos t + \sin t \end{pmatrix} + c_2 e^{-t}\begin{pmatrix}\sin t \\\ -\cos t + 2\sin t \end{pmatrix}$$.
Attached is the graph.
