Solve $y’’ – 4y’ + 5y = 4e^{2x}\sin(x)$ using $\mathcal D$ operator Hello – I am working through the following question and get stuck at step 6. Could someone please advise in simple terms which I can hopefully understand. Thanks
$$y'' – 4y' + 5y = 4e^{2x}\sin(x)$$
Step one – Order equation so that differential operator is in front of the RHS of the equation
$\newcommand{\D}{\mathcal D}$
$$1 = \frac 1 {\D^2 – 4\D + 5}  \cdot 4e^{2x}\sin(x)$$
Step two – move constant and exponential in front of the $\D$ operator
$$1 = 4e^{2x}\cdot \frac 1 {\D^2 – 4\D + 5}\cdot \sin(x)$$
Step three – calculate $a$
Because of $e^{2x}$, $a = 2$, and because of $\sin(x)$, $a = 2 + i$.
Step four – insert $a$ into the $\D$ operator and then calculate to see if it equals zero
\begin{align}
1 &= 4e^{2x}\cdot \frac 1 {(2+i)^2 – 4(2+i) + 5}\cdot \sin(x)
\\
&= 4e^{2x}\cdot \frac 1 {(4+4i+4-8-4i+5)}\cdot\sin(x)
\\
&= 4e^{2x}\cdot \frac 1 {(0)}\cdot \sin(x)
\end{align}
Step 5 – because there is a zero, note $a = 2$ therefore make it $\D+2$
\begin{align}
1 &= 4e^{2x}\cdot \frac 1 {(\D + 2)^2 – 4(\D + 2) + 5}\cdot \sin(x)
\\
&= 4e^{2x}\cdot\frac 1 {\D^2 + 1}\cdot\sin(x)
\end{align}
What do I do for step 6? Please explain in simply terms and assume my calculus knowledge is low.

$e^{ix}=\cos(x)+i\sin(x)$
$\sin(x)= Im(e^{ix})$
$$
{1 \over D^2 + 1} \ \sin x = {1 \over D^2 + 1} \ \ Im(e^{ix})=Im{1 \over D^2 + 1} \ \ e^{ix}=Im{1 \over 2D} \ \ xe^{ix} =x Im{1 \over 2i} \ \ e^{ix}=x Im{1 \over 2i} \ \ (\cos(x)+i\sin(x))=\frac{x}{2} Im\ \ (-i\cos(x)+\sin(x))
= \left[ -{x \over 2} \cos x \right]
$$
