(a) Show that
$$\begin{aligned}
D\left(e^{a x} y\right) &=e^{a x}(D+a) y \\
D^{2}\left(e^{a x} y\right) &=e^{a x}(D+a)^{2} y
\end{aligned}$$
and so on; that is, for any positive integral \(n\)
$$D^{n}\left(e^{a x} y\right)=e^{a x}(D+a)^{n} y$$
Thus show that if \(L(D)\) is any polynomial in the operator \(D,\) then
$$L(D)\left(e^{a x} y\right)=e^{a x} L(D+a) y$$
This is called the exponential shift.
(b) Use (a) to show that
$$\begin{aligned}
(D-1)^{3}\left(e^{x} y\right) &=e^{x} D^{3} y \\
\left(D^{2}+D-6\right)\left(e^{-3 x} y\right) &=e^{-3 x}\left(D^{2}-5 D\right)
y
\end{aligned}$$
(c) Replace \(D\) by \(D-a,\) to obtain
$$e^{a x} P(D) y=P(D-a) e^{a x} y$$
This is called the inverse exponential shift.
(d) Using (c), we can change a differential equation whose right-hand side is
an exponential times a polynomial, to one whose right-hand side is just a
polynomial. For example, consider \(\left(D^{2}-D-6\right) y=10 x e^{3 x} ;\)
multiplying both sides by \(e^{-3 x}\) and using \((\mathrm{c}),\) we get
$$\begin{aligned}
e^{-3 x}\left(D^{2}-D-6\right) y &=\left[(D+3)^{2}-(D+3)-6\right] y e^{-3 x}
\\\
&=\left(D^{2}+5 D\right) y e^{-3 x}=10 x
\end{aligned}$$
Show that a solution of \(\left(D^{2}+5 D\right) u=10 x\) is
\(u=x^{2}-\frac{2}{5} x ;\) then \(y e^{-3 x}=x^{2}-\frac{2}{5} x\) or \(y=e^{3
x}\left(x^{2}-\frac{2}{5} x\right) .\) Use this method to solve Problems 23 to
26
