Proof of G(D)(e^{rx}Y(x))=e^{rx}G(D+r)Y(x)

[3.1415926535]

As you might have already found out G(D) is a differential operator. I can prove this property for G(D)=D^n , but I would like to know the general proof.
I would like an answer with the same notation as I use in this following proof, Thank you in advance

$$G(D)(e^{rx}Y(x))=e^{rx}G(D+r)Y(x)$$

Proof of
$$D^{n}(e^{rx}Y(x)))=e^{rx}(D+r)^{n}Y$$

$$D^{n}(e^{rx}Y(x)))=(e^{rx}Y)^{(n)}=(e^{rx}Y)^{n}Y+\binom{n}{1}(e^{rx})^{(n-1)}{Y}'+...e^{rx}Y^{(n)}=e^{rx}(r^nY+(\binom{n}{1})r^{n-1}Y'+...Y^{(n)})=e^{rx}(r^{n}+(\binom{n}{1})r^{n-1}D+...+D^{n})Y=e^{rx}(D+r)^{n}Y$$

 

[fresh_42]

It looks a bit like the representations of Lie groups and their Lie algebras commuting with the exponential function. If so, then I would look in that direction.