Find the eigenvalues of $A$ (the matrix): $(1-\lambda)(3-\lambda) +1=\lambda^2-4\lambda + 4=(\lambda -2)^2=0$. Unfortunately, if you haven't seen how to solve systems this way the repeated eigenvalue case is not a good one to start from.
An eigenvector associated to this eigenvalue is $\left(\begin{matrix} 1 \\ -1 \end{matrix}\right)$. This tells us one solution is $Y_1=\left(\begin{matrix} 1 \\ -1 \end{matrix}\right)e^{2t}$.
We have to get another linearly independent solution, and to do this we need to find a generalized eigenvector, or an $\eta$ such that $(A-2I)\eta = \left(\begin{matrix} 1 \\ -1 \end{matrix}\right)$. Such a vector is $\eta=\left(\begin{matrix} 1 \\ -2 \end{matrix}\right)$.
This gives us a second solution which is linearly independent of the form $Y_2= \left(\begin{matrix} 1 \\ -1 \end{matrix}\right)te^{2t}+\left(\begin{matrix} 1 \\ -2 \end{matrix}\right)e^{2t}$
The general solution is of the form $Y(t)=C_1 Y_1 + C_2 Y_2$. If you plug in $Y(0)=\left(\begin{matrix} 0 \\ 2 \end{matrix}\right)$ you can solve for $C_1$ and $C_2$ using standard linear algebra.