Can systems of linear ODE's be solved when their matrix is not diagonalizable? I'm working through my ODE homework right now and I've run into a repeated issue of ODE systems not being diagonalizable. I am not aware of any other methods to solve systems and my lecture notes do not have any comments on unsolvable systems. Do I need to approach the problem in a different way or is there simply not solutions to some systems?
One problem I'm working with is $$x' =\pmatrix{1&1&1 \\ 2&1&-1 \\ -3&2&4}x$$
I can't find any generalized eigenvectors to work with and I'm stuck on how to solve it. I ask that you don't solve the problem and just nudge me in the right direction.
 A: The Jordan normal form can be written as $QAQ^{-1}=D+N$ with $D$ diagonal and $N$ nil-potent, $N^{p+1}=0$, and commuting with $D$. Then
$$
Qx(t)=e^{(D+N)t}Qx_0=e^{Dt}(I+Nt+\tfrac12N^2t^2+...+\tfrac1{p!}N^pt^p)Qx_0
$$
This should at least indicate what extra terms in addition to the exponentials to the eigenvalues you will get.
A: As a supplementary hint to Lutz Lehmann's answer, I point out that corresponding to each block of the Jordan normal form you'll get a system of ODE's of the following form
$$
y'=\pmatrix{\lambda&1&0&0&\dots&\dots&0\\
             0&\lambda&1&0&\dots&\dots&0\\
             0&0&\lambda&1&\dots&\dots&0\\
    \vdots&\vdots&&\ddots&\ddots&&\vdots\\
    \vdots&\vdots&&&\ddots&\ddots&\vdots\\
    0&0&\dots&\dots&\dots&\lambda&1\\
    0&0&\dots&\dots&\dots&0&\lambda}y\ ,
$$
where the entries of $\ y\ $ are a subset of the entries of $\ Qx\ $.  You can write this system of ODEs as
\begin{align}
y_n'=&\lambda y_n\\
y_{n-1}'=&\lambda y_{n-1}+y_n\\
&\vdots\\
y_i'=&\lambda y_i+y_{i+1}\\
&\vdots\\
y_1'=&\lambda y_1+y_2\ ,
\end{align}
where $\ n\ $ is the number of rows and columns in the above matrix.  Can you see how you might be able to solve this system by successively finding $\ y_n, y_{n-1}, \dots, y_1\ $?
