# 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 $f = L(g)$?

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I would start looking at solutions for small values of $k$, maybe there's a pattern that allows to formulate a hypothesis or two. –  TZakrevskiy Dec 15 '13 at 0:41
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 radicals –  Ewan Delanoy Jun 24 at 8:37

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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What do you exactly mean by: It is known that there is no closed form for the roots of the polynomial equation of degree $n>4$? $(x-1)^5$ is a polynomial equation of degree $n>4$ but you can easily find the roots. How are you so sure that it isn't the case here? –  wythagoras Jun 24 at 7:48
Every body understand (and I am sure, you too) what I mean. Of course, this concerns the GENERAL polynomial equations. Also, please, no tedious questions about the solutions involving special functions such as the theta functions in case $n=5$. This was discussed many times elsewhere. –  JJacquelin Jun 24 at 8:01
@wythagoras : I understand your remark and agree with you in the sake of theoretical exactitude. This draw me to modify the wording of my answer on this point. Thanks you for the comment. –  JJacquelin Jun 24 at 8:28
For $n=3$ and $n=4$, we have two complex roots for the characteristic equation. For $n=5$ and $n=6$, four complex roots. –  Claude Leibovici Jun 25 at 8:50
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. –  JJacquelin Jun 25 at 14:11