Chapter 8.6, Problem 30P

(a) Show that $(D-a)e^{cx}=(c-a)e^{cx}$; $\left(D^2+5D-3\right)e^{cx}=\left(c^2+5c-3\right)e^{cx}$; $L(D)e^{cx}=L(c)e^{cx},$ where $L(D)$ is any polynomial in $D$; $(D-c)x{e}^{cx}=e^{cx}$; ${(D-c)}^2{x}^2{e}^{cx}=2e^{cx}$.

(b) Define the expression $y=[1/L(D)]u(x)$ to mean a solution of the differential equation $L(D)y=u.$

Using part (a), show that $\frac{1}{D-a}{e}^{cx}=\frac{e^{cx}}{c-a},c\ne a$; $\frac{1}{D^2+5D-3}{e}^{cx}=\frac{e^{cx}}{c^2+5c-3}$; $\frac{1}{L(D)}{e}^{cx}=\frac{e^{cx}}{L(c)},L(c)\ne 0$; $\frac{1}{D-c}{e}^{cx}=x{e}^{cx}$; $\frac{1}{{(D-c)}^2}{e}^{cx}=\frac{1}{2}{x}^2{e}^{cx}$.

(c) The expressions $1/L(D)$ in (b) are called inverse operators. They can be used to find particular solutions of differential equations. As an example consider Problem 3. We write $\left(D^2+D-2\right)y=e^{2x}$, $y=\frac{1}{D^2+D-2}{e}^{2x}=\frac{e^{2z}}{2^2+2-2}=\frac{e^{2x}}{4}$.

Using inverse operators, find particular solutions of Problems 4 to 20. Be careful to use parts 4 or 5 of (b) if c is a root of the auxiliary equation. For example, $\frac{1}{(D-a)(D-c)}{e}^{cx}=\frac{1}{D-c}\frac{1}{D-a}{e}^{cx}=\frac{1}{D-c}\frac{e^{cx}}{c-a}=\frac{x{e}^{cx}}{c-a}$.,