The low road to showing that $y^l e^{\lambda y}$ is a solution of a higher order linear ODE via the characteristic polynomial of the companion matrix Can someone give me hint on how to show that the fact that $\alpha$ is a root of order $l+1$ of the polynomial $p(x)=x^n+a_1x^{n-1}+\cdots+a_n$ implies that $t\mapsto y^l e^{\lambda y}$ is a solution of the ODE $x^{(n)}+a_1 x^{(n-1)}+\cdots + a_n x=0$ ?
I tried doing a direct calculation, but didn't succeed (altough I would prefer this direct route, since I don't yet see all the connection between the different theorems of the theory for linear systems of ODEs, so it would be easier for me to understand)...then I tried using some theory about ODEs but didn't get far: The ODE is equivalent to a system of linear ODEs of order 1 and the characteristic polynomial of the matrix of that system is precisely the polynomial above, but there I kind of stop dead, since calculating the Jordan normal form is just horrible, so I thought there must be some easier, less calculation-heavy way. Could you show me the way ?
 A: Define a series of variables: $x_1 = x,x_2=\dot x, x_3= \ddot x, \ldots$. Then, rephrase the given ODE as 
$$
x^{(n)}= -a_1 x^{(n-1)}+\cdots - a_n x
$$ 
This allows us to write the differential equation in a compact matrix equation form 
$$
\dot x(t) := A x(t) =  \pmatrix{0 &1 &0 &\ldots &0\\0 &0 &1 &\ldots& 0\\ \vdots &&\ddots &\ddots &0\\0 &&&0&1\\-a_n &-a_{n-1} &\ldots &-a_2 &-a_1}x(t)
$$
Then, the solution of this matrix differential equation is given by 
$$
x(t) = e^{At}x_0 = V^{-1}e^{Jt}Vx_0
$$
The determinant of $(x I-A)$ is $p(x)=x^n+a_1x^{n-1}+\cdots+a_n$ and can be shown by expanding along the last column. Then, you don't need to calculate much. The remaining step is assuming an arbitrary sized Jordan block in $J$ and then computing e^{Jt}. There you will see that when $\alpha$ is an eigenvalue of $A$ associated with a Jordan block of size at least two then, the result follows. Look at the state-transition matrix in this link for the general formula. 
A: The most succinct way I know how to do this is to think about differential operators. Letting $D$ denote the differentiation operator, we want to compute the nullspace of $p(D)$ where $p$ is some polynomial. The first basic observation is that if $p(z) = \prod (z - \lambda_i)^{m_i}$, then
$$p(D) = \prod (D - \lambda_i)^{m_i}$$
and each of the operators in this product commute with each other; in particular, if $(D - \lambda_i)^{m_i} f = 0$ for some function $f$, then $p(D) f = 0$ as well. So we are reduced to the problem of solving
$$(D - \lambda_i)^{m_i} f = 0.$$
Since we know that $(D - \lambda_i) f = 0$ has solution $f = C e^{\lambda_i x}$, it might occur to us to write $f = e^{\lambda_i x} g$ where $g$ is some function. 
Proposition: $(D - \lambda_i)^{m_i} e^{\lambda_i x} g = e^{\lambda_i x} D^{m_i} g$.
Proof. It suffices to compute that $(D - \lambda_i) e^{\lambda_i x} g = Dg$. 
It follows that $(D - \lambda_i)^{m_i} e^{\lambda_i x} g = 0$ if and only if $D^{m_i} g = 0$, hence if and only if $g$ is a polynomial of degree at most $m_i - 1$.
