Solutions for a 3x3 ODE system with one eigenvalues (m=3). I was thinking about an ODE system which has three equal eigenvalues like this situation:
EDIT: $X'=AX$
$$A=\begin{pmatrix}
    1 && 3 && 2 \\
    0 && 1 && 0\\
    0 && 2 && 1
    \end{pmatrix}
$$The eigenvalues are:
$$(\lambda-1)^{3}=0$$
$$\lambda_{1,2,3}=1$$
It's clear that I have $\lambda=1$ with $\mu=3$.
I know very well the form of the solutions for a 2x2 system with eigenvalues with $\mu=2$:
$$ \begin{pmatrix}
    x(t)\\
    y(t)\\
    \end{pmatrix}=c_1e^{\lambda t}\mathbb u+c_2e^{\lambda t}[t\mathbb u+\mathbb w]
$$
Now I'm asking myself which is a possible generalization for a 3x3 case of this formula. I'm expecting to have a $t^{2}$ term to appear in the formula. 
Moreover, I have already computed the eigenvector plus two generalized eigenvectors (for the case I showed in this example for $\lambda=1$).
Eigenvectors:
$$\mathbb u=(1,0,0)$$
$$\mathbb v=(0,0,\frac{1}{2})$$
$$\mathbb w=(0,\frac{1}{4},-\frac{3}{8})$$
Note that $\mathbb v$ and $\mathbb w$ are generalized eigenvectos.
Thank you very much.
 A: For any single Jordan block, the diagonal part and the nilpotent part can be regarded as commuting square matrices. When matrices $A,B$ commute, we get $e^{A+B} = e^A e^B = e^B e^A.$ 
You want $$e^{(I+N)t} = e^{tI} e^{tN} = e^t I e^{tN} = e^t e^{tN}$$
with 
$$
N = 
\left(
\begin{array}{rrr}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{array}
\right)
$$
and
$$
N^2 = 
\left(
\begin{array}{rrr}
0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}
\right)
$$
and $$ N^3 = N^4 = N^5 = \cdots = 0  $$
So
$$ e^{tN} = I + t N + \frac{t^2 }{2} N^2 =
\left(
\begin{array}{rrr}
1 & t & \frac{t^2}{2} \\
0 & 1 & t \\
0 & 0 & 1
\end{array}
\right)
 $$
$$
\frac{1}{8} 
\left(
\begin{array}{rrr}
4 & 3 & 0 \\
0 & 0 & 1 \\
0 & 2 & 0
\end{array}
\right)
\left(
\begin{array}{rrr}
2 & 0 & -3 \\
0 & 0 & 4 \\
0 & 8 & 0
\end{array}
\right) = I
$$
$$
\frac{1}{8} 
\left(
\begin{array}{rrr}
4 & 3 & 0 \\
0 & 0 & 1 \\
0 & 2 & 0
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & 1& 0 \\
0 & 1 & 1 \\
0 & 0 & 1
\end{array}
\right)
\left(
\begin{array}{rrr}
2 & 0 & -3 \\
0 & 0 & 4 \\
0 & 8 & 0
\end{array}
\right) = 
\left(
\begin{array}{rrr}
1 & 3 & 2 \\
0 & 1 & 0 \\
0 & 2 & 1
\end{array}
\right) = A
$$
$$
\frac{e^t}{8} 
\left(
\begin{array}{rrr}
4 & 3 & 0 \\
0 & 0 & 1 \\
0 & 2 & 0
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & t& \frac{t^2}{2} \\
0 & 1 & t \\
0 & 0 & 1
\end{array}
\right)
\left(
\begin{array}{rrr}
2 & 0 & -3 \\
0 & 0 & 4 \\
0 & 8 & 0
\end{array}
\right) =  e^{tA}
$$
The denominators clear nicely,
$$
e^{At} =
\left(
\begin{array}{ccc}
1 & 2 t^2 + 3 t & 2t \\
0 & 1 & 0 \\
0 & 2t & 1 \\
\end{array}
\right) \; \; e^t  \; \; = \; \;
\left(
\begin{array}{ccc}
e^t & (2 t^2 + 3 t)e^t & 2t e^t \\
0 & e^t & 0 \\
0 & 2t e^t & e^t \\
\end{array}
\right)
$$
