Matrix form of differential equation, non-diagonalizable 
In this above matrix, normally I just diagonalize to find the eigenvalues/eigenvectors, but how would this work in the case where the resulting matrix is upper triangular and has 0's on the diagonal? Thanks!
 A: This matrix is in Jordan normal form. If you denote it$A$, the solution is 
$$\vec x(t)=\exp(At)\vec x(0),$$
so you have to compute the exponential of this matrix. Now  $a$ is the sum $D+N$, where $D$ is the diagonal matrix $ \lambda I$ and the nilpotent matrix $N=\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}$. Furthermore $D$ and $N$ commute, so 
$$\exp(D+N)t=\exp Dt\,\exp Nt=\mathrm e^{\lambda t} I\,\exp Nt. $$
Now remember $\exp Nt=I+Nt+\dfrac{N^2t^2}{2}+\dfrac{N^3t^3}{3!}+\dotsm$, and observe that
$$N^2=\smash[b]{\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix}}, \qquad N^3=0,$$
so that
$$\exp  Nt=I+Nt+\dfrac{N^2t^2}{2}=
\smash{\begin{bmatrix}1&t&\frac{t^2}2\\0&1&t\\0&0&1\end{bmatrix}},$$
and finally
$$\exp At=\mathrm e^{\lambda t}I\begin{bmatrix}1&t&\frac{t^2}2\\0&1&t\\0&0&1\end{bmatrix}=\begin{bmatrix}\mathrm e^{\lambda t}&te^{\lambda t}&\frac{t^2}2e^{\lambda t}\\0&e^{\lambda t}&te^{\lambda t}\\0&0&e^{\lambda t}\end{bmatrix}.$$
A: It might be helpful to write your matrix at $\lambda I + N$, where $N$ is nilpotent (more precisely, $N^3 = 0$. Then, since these matrices commute, you can exponentiate each matrix separately and then take the product. The first matrix easily exponentiates to
$$\begin{pmatrix}
e^{\lambda t} & 0 & 0 \\
0 & e^{\lambda t} & 0 \\
0 & 0 & e^{\lambda t}\\
\end{pmatrix}$$
The second matrix is also easy, only the first $3$ terms of the exponential series survive. One computes the exponential as
$$\begin{pmatrix}
1 & t & t^2/2 \\
0 & 1 & t \\
0 & 0 & 1 \\
\end{pmatrix}$$
Then the product of these matrices is the solution to your system.
