Let \(D\) stand for \(d / d x,\) that is, \(D y=d y / d x ;\) then $$D^{2} y=D(D
y)=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d^{2} y}{d x^{2}}, \quad
D^{3} y=\frac{d^{3} y}{d x^{3}}, \text { etc. }$$ \(D\) (or an expression
involving \(D\) ) is called a differential operator. Two operators are equal if
they give the same results when they operate on \(y\). For example, $$D(D+x)
y=\frac{d}{d x}\left(\frac{d y}{d x}+x y\right)=\frac{d^{2} y}{d x^{2}}+x
\frac{d y}{d x}+y=\left(D^{2}+x D+1\right) y$$ so we say that $$D(D+x)=D^{2}+x
D+1$$ In a similar way show that:
(a) \(\quad(D-a)(D-b)=(D-b)(D-a)=D^{2}-(b+a) D+a b\) for constant \(a\) and \(b\).
(b) \(\quad D^{3}+1=(D+1)\left(D^{2}-D+1\right)\)
(c) \(\quad D x=x D+1 .\) (Note that \(D\) and \(x\) do not commute, that is, \(D x
\neq x D .\) )
(d) \(\quad(D-x)(D+x)=D^{2}-x^{2}+1,\) but \((D+x)(D-x)=D^{2}-x^{2}-1\)
Comment: The operator equations in (c) and (d) are useful in quantum
mechanics; see Chapter 12, Section 22.