Can systems of linear ODE's be solved when their matrix is not diagonalizable?

Question

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.

Answer 1

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.

Answer 2

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\ $?