Repeated Eigenvalues Initial Value Problem If someone could help me step by step in solving this initial value problem. There's a lot I'm confused about since the solution is supposed to be expressed in the form $x_1(t) =$ and $x_2(t) =$:
$$
x' = \left[\begin{matrix}
\frac{22}{3} & \frac{1}{3} \\
-\frac{25}{3} & \frac{32}{3} \\
\end{matrix}\right]
$$
$$
x(0) = \left[\begin{matrix}
3 \\
24
\end{matrix}\right]\\
$$
 A: The characteristic polynomial is given by $|A - \lambda I| = 0$, yielding:
$$(\lambda-9)^2 = 0 \rightarrow \lambda_{1,2} = 9$$
Next we need to find two linearly independent eigenvectors. We setup and solve $[A - \lambda I]v_1 = 0$, yielding:
$$\begin{bmatrix}-\dfrac{5}{3} & \dfrac{1}{3}\\ -\dfrac{25}{3} & \dfrac{5}{3}\end{bmatrix}v_1 = 0$$
The RREF gives:
$$\begin{bmatrix}1 & -\dfrac{1}{5}\\ 0 & 0 \end{bmatrix}v_1 = 0$$
This gives us an eigenvector of:
$$v_1 = \left(\dfrac{1}{5},1\right)$$
Unfortunately, this yields a single eigenvector, so to find a second one we try:
$$[A - \lambda I]v_2 = v_1$$
This yields (work the details) a second, generalized eigenvector of:
$$v_2 = \left( -\dfrac{3}{25},0 \right)$$
Now, we can write the solution as:
$$X(t) = e^{\lambda t}\left(c_1 v_1 + c_2(v_2 + tv_1)\right)$$
We substitute the initial condition:
$$X(0) = \begin{bmatrix}3 \\ 24 \end{bmatrix}$$
The final solution is (spoiler):

 $$X(t) = \begin{bmatrix} x(t) \\ y(t) \end{bmatrix} = 3e^{9t}\begin{bmatrix} t + 1 \\ 5t+8 \end{bmatrix}$$

