Inverse $D$ operators question

Question

Find $$\dfrac 1 {D^2+6D+9}e^{-3x}$$

So I am new to the topic of $2$nd order linear ordinary differential equations and on $d$-operators. I attempted the question below and am I not supposed to just replace $D$ with $-3$? That would give me $\frac{1}{0}$, which makes me wrong. Can anyone help me?

Answer 1

You have to solve for $y(x)$ once continuously differentiable and piecewise twice differentiable with $y(x)=0$ for $x<0$ and for $x\ge 0$ $$ (D+3)^2y(x)=e^{-3x}\iff (e^{3x}y(x))''=1 \\~\\ \implies e^{3x}y(x)=\frac{x^2}2u(x) $$ where $u(x)$ is the unit jump from $0$ to $1$ at $x=0$.

Answer 2

$$y_p=\dfrac 1 {(D^2+6D+9)}e^{-3x}$$ Write the operator as: $$y_p=\dfrac 1 {(D+3)^2}e^{-3x}$$ Apply the rule $$y_p=\dfrac 1 {f(D)}e^{ax}C(x)=e^{ax}\dfrac 1 {f(D+a)}C(x)$$ Here $C(x)=1$ So that you have; $$y_p=e^{-3x}\dfrac 1 {D^2}1$$ $$y_p=e^{-3x}\dfrac 1 D (x+c_1)=e^{-3x}\left(\dfrac {x^2}{2}+c_1x+c_2\right)$$