You have an eigenvalue $\lambda$ and its eigenvector $v_1$. So one of your solutions will be
$$ x(t) = e^{\lambda t} v_1$$
As you've noticed however, since you only have two eigenvalues (each with one eigenvector), you only have two solutions total, and you need four to form a fundamental solution set. For each eigenvalue $\lambda$, you will calculate what's called a generalized eigenvector $v_2$, which is the solution to
$$ (A - \lambda I)v_2 = v_1, \quad \text{where } (A - \lambda I)v_1 = 0;$$
in other words, $v_1$ was the first eigenvector. Then this contributes a new solution
$$ x(t) = e^{\lambda t} v_2 + te^{\lambda t} v_1 $$
Now you have two linearly independent solutions corresponding to one eigenvalue. Now repeat the process for the second eigenvalue to get all four elements of your fundamental solution set.
Note: What I'm calling $v_2$ is NOT the same as what you're calling $v_2$ in your question.
You have done most of the work already. $A's$ eigenvalues are:
$$\lambda_{1, 2} = \dfrac{-5\pm\sqrt{33}}{2}$$
We now have to find the eigenvectors and solve $[A - \lambda_i I]v_i = 0$. So for $\lambda_{1,2}$, we have:
$$[A - \lambda_{1,2} I]v_{1,2} = [A - \dfrac{1}{2} (-5 \pm \sqrt{33}) I]v_{1,2} = \begin{pmatrix}
- \dfrac{1}{2} (-5 \pm \sqrt{33}) & 1 \\
2 & -5 - \dfrac{1}{2}(-5 \pm \sqrt{33})
\end{pmatrix}v_{1, 2} = \begin{pmatrix}
0 \\
0
\end{pmatrix}$$
When we do the RREF of this matrix, we have:
$$\begin{pmatrix}
1 & \dfrac{1}{4} (-5 \mp\sqrt{33}) \\
0 & 0 \end{pmatrix}v_{1, 2}= \begin{pmatrix}
0 \\
0
\end{pmatrix}$$
This gives us:
$$v_{1,2} = \begin{pmatrix}
\dfrac{1}{4}(5 \pm \sqrt{33})\\1
\end{pmatrix}$$
We can now write:
$$X(t) = \begin{pmatrix}
x(t)\\y(t)
\end{pmatrix} = c_1 e^{\lambda_1 t} v_1 + c_2 e^{\lambda_2 t} v_2$$
You have two initial conditions to solve for $c_1$ and $c_2$ and should get:
$$X(t) = \begin{pmatrix}
x(t)\\y(t)
\end{pmatrix} =\begin{pmatrix}
\dfrac{2}{\sqrt{33}}~e^{-1/2 (5+\sqrt{33}) t} (e^{\sqrt{33} t}-1) \\ \dfrac{1}{33} e^{-1/2 (5+\sqrt{33}) t} ((33-5 \sqrt{33}) e^{\sqrt{33} t}+33+5 \sqrt{33})
\end{pmatrix}$$
Update For $\lambda_1$, our RREF is:
$$\begin{pmatrix}
1 & \dfrac{1}{4} (-5 -\sqrt{33}) \\
0 & 0 \end{pmatrix}v_1= \begin{pmatrix}
0 \\
0
\end{pmatrix}$$
Let the vector $v_1 = \begin{pmatrix} a \\ b \end{pmatrix}$, so this gives us:
$$a + \dfrac{1}{4} (-5 -\sqrt{33})b = 0 \implies a = \dfrac{1}{4} (5 +\sqrt{33})b$$
We are free to choose whatever $b$ we'd like, so choose $b = 1$ and conclude.
This is how we arrived at:
$$v_1 = \begin{pmatrix}
\dfrac{1}{4}(5 + \sqrt{33})\\1
\end{pmatrix}$$
I should also mention, note how this choice also satisfies the second equation.
Best Answer
You solved the homogeneous equation $\dot{y} = By$ to get your solution $y$. Now you need to get a particular solution $y_p$, and the actual solution will be $y + y_p$, then you can plug in the initial condition.