When solving first order systems of differential equations(i.e $x'=Ax$). If $A$ here is diagonalizable, the solutions are always like $\sum e^{\lambda_i x}$ without any terms like $e^{\lambda x}x$.
When the system is actually solving some high-order linear ode (e.g. $x''+2x'+x=0$), $A$ here become the matrix of the recurrence relation.(e.g. $A=\begin{pmatrix} 0& 1\\ -1&-2 \end{pmatrix}$) It seems that when the eigenvalues of characteristic polynomial are not distinct, the equation does have solutions like $e^{\lambda x}x$, is that mean $A$ here is not diagonalizable?