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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 $f = L(g)$?

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    For $n=5$, the characteristic polynomial in $r^5 - 6r^4 - 30r^3 - 120r^2 - 360r - 720$, and PARI-GP confirms that its Galois group is $S_5$, so it is not solvabel by radicals2015-06-24

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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.

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    Hi Claude ! As usual I agree with you. More generally, I conjecture that for any even $n$ there are two real roots and for any odd $n$ there is one real root, with all other roots complex.2015-06-25