Let $\phi({D})$ be a differential operator defined by $$D^n+a_{n-1}D^{n-1}+a_{n-2}D^{n-2}+...+a_0=\sum_{j=0}^{n}a_jD^j$$
with constant coefficients $a_j$ (setting $a_n=1$) and the derivative operator $D=\frac{d}{dx}$. Let $f(x)$ be an arbitrarily often differentiable function. Show that:
$\phi(D)[e^{\lambda x}f(x)]=e^{\lambda x} \phi(D+\lambda)f(x)$
$\phi(D)[x f(x)]=x \phi(D)f(x)+\phi'(D)f(x)$
It's relatively easy to show that this is true for the first couple of terms but I don' know how to show it for all $n$. Is there some elegant quick way to show that 1. and 2. are true?
In both cases first verify the claim for $\phi(D)=D,$ then by induction for $\phi(D)=D^n$ (this is the most interesting step), then by linearity for a general polynomial $\phi.$
Or if you are as lazy as I am, your first step could even be the constant $\phi(D)=1.$