If $F$ is a field of characteristic $0$ and $t$ is an indeterminate, then the ring $\text{End}_F(F[t])$ contains the operators$$x: f(t) \mapsto tf(t), \quad y: f(t) \mapsto {d\over{dt}}f(t).$$Let $A$ be the subalgebra of $\text{End}_F(F[t])$ generated by $x$ and $y$.
Question. Why is the $F$-algebra of differential operators with polynomial coefficients in one variable precisely $A$? How do I see that the $F$-algebra of differential operators with polynomial coefficients in one variable is a subalgebra of $\text{End}_F(F[t])$, and how do I see that it is generated by $x$ and $y$?
Any differential operator with polynomial coefficients can be written as a product of $\partial$ and $X$: it suffices you consider operators of the form $p(X)\partial^j$, and further, operators of the form $X^i\partial^j$, which are manifestly in the algebra generated by $X$ and $\partial$.
The fact that the set of differential operators with polynomial coefficients is an algebra stems from the fact that $\partial X$ can be written in terms of $X\partial$ by virtue of the product rule of the derivative: $\partial(Xp) = p+X\partial p$. Thus you can always write any composition of operators in terms of the monomials $X^i\partial^j$.