Solution to differential equation $f^{(n)}-(n+1)f^{(n-1)}-(n+1)nf^{(n-2)}-\dotsc-(n+1)!f=g$ Let $n$ be a given positive integer and $g$ be a continuous function. We are looking for a function $f \in C^n(\mathbb{R})$ such that
$$f^{(n)}-(n+1)f^{(n-1)}-(n+1)nf^{(n-2)}-\dotsc-(n+1)!f=g.$$
It is of course a linear equation of order $n$ but if I try to solve its characteristic equation it gets complicated even for small $n$.
Is there a way to find some operator $L$ (possibly quite "complicated") such that $f = L(g)$?
 A: First, we study the associated homogeneous ODE :
$$F^{(n)}-(n+1)F^{(n-1)}-(n+1)nF^{(n-2)}-\dotsc-(n+1)!F=0$$
$$F^{(n)}-\sum_{k=1}^n \frac{(n+1)!}{k!}F^{(k-1)}=0$$
The solution is :
$$F=\sum_{k=1}^n c_k \: e^{r_k x}$$
where $c_k$ are arbitrary constants and $r_k$ are the roots of the next polynomial equation :
$$r^n-\sum_{k=1}^n \frac{(n+1)!}{k!}r^{k-1}=0$$
The roots (realand/orcomplex) can be computed by numerical methods for any $n$ and the formulas above give the answer to the question. But, if we are looking for analytic solutions, the answer becomes more limited :
It is known that there is no elementary closed form for the roots of the general polynomial equation of degree $n>4$.
Hense, there is no elementary closed form for the solution of the ODE if $n>4$
So, an answer to the question can be given only in the cases $1\leq n \leq 4$
Case $n=1$ :
$F'-2F=0 \quad \to \quad F=c_1 e^{2x}\quad$ and the solution of $f'-2f=g \quad$ is :
$$f=c_1 e^{2x}+F_p(x)$$
where $F_p(x)$ is a particular solution which can be found thanks to the method of "variation of parameter"(for example).
$$F_p(x)=e^{2x}\int e^{-2x}g(x)dx$$
Case $n=2$ :
$F''-3F'-6F=0$ 
The roots of the polynomial equation  $r^2-3r-6=0$ are :
$r_1=\frac{1}{2}\left(3-\sqrt{33}\right) \quad , \quad r_2=\frac{1}{2}\left(3+\sqrt{33}\right)$
The solution of $f''-3f'-6f=g \quad$ is :
$$f=c_1 e^{r_1 x}+c_2 e^{r_2 x}+F_p(x)$$
where $F_p(x)$ is a particular solution in which the function $g(x)$ is involved. $F_p(x)$ can be found thanks to the method of "variation of parameters".
The cases $n=3$ and $n=4$ are more arduous to solve, but that can be done thanks to the same method.
