TITLE: "Natural" proof of $P\left(\frac{d}{dx}\right)\bigl(e^{xy}Q(x)\bigr)=Q\left(\frac{d}{dy}\right)\bigl(e^{xy}P(y)\bigr)$.
QUESTION [9 upvotes]: In the context of linear differential equations, I've stumbled upon the following identity for an arbitrary pair of polynomials $P$ and $Q$ with real or complex coefficients:
$$
  P\left(\frac{d}{dx}\right)\bigl(e^{xy}Q(x)\bigr)
  =\sum_{n=0}^\infty\frac{P^{(n)}(y)e^{xy}Q^{(n)}(x)}{n!}
  = Q\left(\frac{d}{dy}\right)\bigl(e^{xy}P(y)\bigr).
$$
This can be more or less easily checked by using Taylor expansions of $P\bigl(\frac{d}{dx}\bigr)$ at $y$ and of $Q\bigl(\frac{d}{dy}\bigr)$ at $x$:
$$
  P\left(\frac{d}{dx}\right)
  =\sum_{n=0}^\infty\frac{P^{(n)}(y)}{n!}\left(\frac{d}{dx} - y\right)^n,
  \quad
  Q\left(\frac{d}{dy}\right)
  =\sum_{n=0}^\infty\frac{Q^{(n)}(x)}{n!}\left(\frac{d}{dy} - x\right)^n.
$$
Is there any easy way to "see" that
$P\bigl(\frac{d}{dx}\bigr)\bigl(e^{xy}Q(x)) = Q\bigl(\frac{d}{dy}\bigr)\bigl(e^{xy}P(y)\bigr)$
without "getting hands dirty"?
Is this identity a part of some general theory?
It makes me think of Fourier analysis, but I do not know much about it.

REPLY [3 votes]: Here I post a proof based on the comment by KCd since he has not done it himself (yet):
$$
P\left(\frac{\partial}{\partial x}\right)\bigl(e^{xy}Q(x)\bigr)
= P\left(\frac{\partial}{\partial x}\right)Q\left(\frac{\partial}{\partial y}\right)e^{xy}
= Q\left(\frac{\partial}{\partial y}\right)P\left(\frac{\partial}{\partial x}\right)e^{xy}
= Q\left(\frac{\partial}{\partial y}\right)\bigl(e^{xy}P(y)\bigr).
$$