I have an infinitely differentiable function $f(x)$ and I want to find the explicit form of the function $F(x)$ which satisfies one of the relations: $$ \frac{d^n}{dx^n}F(x) = \frac{n!}{C^n}\frac{d^n}{dx^n}f(x),\;\;\;\;\forall n \in \Bbb N $$ or $$ \frac{d^n}{dx^n}F(x) = (-1)^n\frac{n!}{C^n}\frac{d^n}{dx^n}f(x),\;\;\;\;\forall n \in \Bbb N $$ where $C$ is some positive constant. Is there any specific method to find a function if its n$^{th}$ derivative is known?
Note: Satisfying one of the relations is enough, the aim is the neater $F(x)$
Attempt
In terms of $n^{th}$ derivatives the coefficients can be written as: $$ (-1)^n n! = \frac{1}{x^{n+1}}\frac{d^n}{dx^n}\left( \frac{1}{x}\right) $$ and $$ \frac{1}{C^n}=e^{-x/C}\frac{d^n}{dx^n}\left( e^{x/C}\right). $$ I am not very sure about their usability at this point but if there is a way to merge them with $\frac{d^n}{dx^n}f(x)$, the problem will be solved.
Firstly, let’s assume the existence of such function.
Considering the first condition, by Taylor’s theorem, $$F(x)=\sum^\infty_{k=0}F^{(k)}(a)\frac{(x-a)^k}{k!}=\sum^\infty_{k=0}\frac{f^{(k)}(a)}{C^{k}}(x-a)^k$$
Now investigating the radius of convergence:
$$r=|C|\lim_{k\to\infty}\frac{|f^{k}(a)|}{|f^{(k+1)}(a)|}$$
If $r$ is non-zero for some $a$, the above Taylor series represents $F(x)$ within the radius of convergence around $a$.
In the special case $f^{(k)}(a)=1$ for all $k$, $F(x)$ is a geometric series which has a nice closed form: $$F(x)=\frac{C}{x+C-a}$$ within a circle of radius $C$ about $a$.