Chapter 8.6, Problem 31P

(a) Show that $D\left(e^{ax}y\right)=e^{ax}(D+a)y$, $D^2\left(e^{ax}y\right)=e^{ax}{(D+a)}^{2}y$, and so on; that is, for any positive integral $n,$ $D^n\left(e^{ax}y\right)=e^{ax}{(D+a)}^{n}y$.

Thus show that if $L(D)$ is any polynomial in the operator D, then $L(D)\left(e^{ax}y\right)=e^{ax}L(D+a)y$.

This is called the exponential shift.

(b) Use (a) to show that ${(D-1)}^3\left(e^{x}y\right)=e^x{D}^{3}y$, $\left(D^2+D-6\right)\left(e^{-3x}y\right)=e^{-3x}\left(D^2-5D\right)y$.

(c) Replace $D$ by $D-a,$ to obtain $e^{ax}P(D)y=P(D-a)e^{ax}y$.

This is called the inverse exponential shift.

(d) Using (c), we can change a diff5erential equation whose right-hand side is an exponential times a polynomial, to one whose right-hand side is just a polynomial. For example, consider $\left(D^2-D-6\right)y=10x{e}^{3x}$; multiplying both sides by $e^{-3x}$ and using (c), we get $=\left(D^2+5D\right)y{e}^{-3x}=10x$.

Show that a solution of $\left(D^2+5D\right)u=10x$ is $u=x^2-\frac{2}{5}x;$ then $y{e}^{-3x}=x^2-\frac{2}{5}x$ or Use this method to solve Problems 23 to 26.