Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix} 4-\lambda & 0 & 0 \\ 1 & 4-\lambda & 0\\ 0 & 1 & 4-\lambda\end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0 \end{bmatrix}$
$\begin{bmatrix} 4-\lambda & 0 & 0 \\ 1 & 4-\lambda & 0\\ 0 & 1 & 4-\lambda\end{bmatrix}=0$
$(\lambda-4)^3=0$
$\lambda_1=\lambda_2=\lambda_3=4$
2. Find eigenvectors:
For $\lambda=4$
let $B=A-\lambda_1I$
$B=\begin{bmatrix} 4-\lambda & 0 & 0 \\ 1 & 4-\lambda & 0\\ 0 & 1 & 4-\lambda\end{bmatrix}=\begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0\\ 0 & 1 & 0 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0 \end{bmatrix} $
Let $r,s,t$ be free variables.
$\vec{V}=r(0,0,1)+s(0,1,0)+t(1,0,0)\\
E_1=\{(0,0,1);(0,1,0);(1,0,0)\} \\
\rightarrow dim(E_2)=3$
Hence, $S=\begin{bmatrix} 0 & 0 & 1\\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} $
We have $x=Sy \rightarrow x'=Ax \rightarrow y'=\begin{bmatrix} 4 & 1 & 0 \\ 0 & 4 &1 \\ 0 & 0 & 4 \end{bmatrix}y $
then $y_1=c_1e^{4t}+c_2te^{4t}+\frac{1}{2}c_3t^2e^{4t}\\
y_2=c_2e^{4t}+c_3te^{4t}\\
y_3=c_3e^{4t}$
Hence,
$x(t)=S.y(t)=\begin{bmatrix} 0 & 0 & 1\\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} c_1e^{4t}+c_2te^{4t}+\frac{1}{2}c_3t^2e^{4t}\\ c_2e^{4t}+c_3te^{4t} \\ c_3e^{4t} \end{bmatrix} \\
=\begin{bmatrix} c_3e^{4t}\\ c_2e^{4t}+c_3e^{4t}\\ c_1e^{4t}+c_2te^{4t} +\frac{1}{2}c_3t^2e^{4t}\end{bmatrix} $
