Having trouble using eigenvectors to solve differential equations The question asked to solve $$\frac{dx}{dy} = \begin{pmatrix}
        5 & 4 \\
        -1 & 1\\
        \end{pmatrix}x$$ ,where $$ x = \begin{pmatrix} x_1 \\ 
x_2 \\ \end{pmatrix}$$
I went ahead an found the determinant of matrix $$ |A - I\lambda| = \lambda^2 - 4\lambda + 9$$
And found $\lambda = 3$
Then the $\alpha$ matrices was found to be 
$$ \begin{pmatrix}
5 & 4\\
-1 & 1\\
\end{pmatrix} \alpha = 3\alpha$$ where $$\alpha = \begin{pmatrix} 
\alpha_1 \\
\alpha_2 \\
\end{pmatrix}$$
Ultimately $-\alpha_1 = 2\alpha_2$ so I write $$\alpha = \begin{pmatrix} 
-1 \\
2 \\
\end{pmatrix}$$
Then because $\lambda$ is a repeated root I know the solution is supposed to look something like this:
$$x = c_1\alpha e^{\lambda t} + c_2( \alpha t + \beta) e^{\lambda t}t$$
And then this is where it gets tricky for me. I know we find the $\beta$ matrix by figuring this out:
$$(A - I\lambda)\beta = \alpha$$
Now when I multiply all that out I get
$$2\beta_1 + 4\beta_2 = -1$$
$$\beta_1 + 2\beta_2 = -2$$
This is the system of equations I can't seem to solve to get a suitable $\beta$. One option I have is to make $$\beta = \begin{pmatrix} 
0\\
-1\\
\end{pmatrix}$$
But this doesn't work for the second system of equations. Help please.

The characteristic polynomial is:
$$|A - \lambda I| = 0 \rightarrow \lambda^2-6 \lambda+9 = 0 \rightarrow \lambda_{1,2} = 3$$
Substituting in the first eigenvalue to find the first eigenvector:
$$[A - \lambda I]v_1 = 0 \rightarrow \begin{pmatrix}
        2 & 4 \\-1 & -2\\\end{pmatrix}v_1 = 0$$
After RREF, for the first eigenvector, I would have chosen:
$$a + 2b = 0 \rightarrow b = 1 \rightarrow a = -2$$
So, the first eigenvector is $v_1 = (-2, 1)$.
Since we have a repeated eigenvalue, we need a generalized eigenvector and you did the right approach, we have:
$$[A - \lambda I]v_2 = v_1$$
$$\begin{pmatrix}
        2 & 4 \\-1 & -2 \end{pmatrix}v_2 = \begin{pmatrix}-2\\1 \end{pmatrix}$$
The RREF is:
$$\begin{pmatrix}1 & 2 \\ 0 & 0 \end{pmatrix}v_2 = \begin{pmatrix}-1\\0 \end{pmatrix} $$
This yields:
$$a + 2b = -1 \rightarrow b = 0 \rightarrow a = -1$$
From this, the second eigenvector is $v_2 = (-1, 0)$.
