How to solve $x^{\prime\prime\prime} + 2x^{\prime\prime} - x = e^{-t}\cos(2t)$ using the operator method? I did
 $(D^3 + 2D - 1)x = e^{-t}\cos(2t) \Rightarrow x = \frac{1}{D^3 + 2D - 1}e^{-t}\cos(2t) \Rightarrow x= \frac{e^{-t}}{(D-1)^3 + 2(D-1) - 1}\cos(2t) \Rightarrow$
$x =\frac{e^{-t}}{D^3-3D^2+3D-1+2D-2-1}\cos(2t) = \frac{e^{-t}}{D^3-3D^2+5D-4}\cos(2t)$, but I'm having trouble with $D^3-3D^2+5D-4$. 
Am I doing something wrong?

$$(D^3 + 2D^2 - 1)x = e^{-t}\cos(2t) 
$$
$$ \implies x_p = \frac{1}{D^3 + 2D^2 - 1}e^{-t}\cos(2t) $$
You forget the power D for the operator. You have $D^3$ and $D^2$.
$$\implies x_p =e^{-t} \frac{1}{(D-1)^2(D+1)  - 1}\cos(2t)$$
$$\implies x_p =e^{-t} \frac{1}{(-2D-3)(D+1)  - 1}\cos(2t)$$
$$ x_p =e^{-t} \frac{1}{(-5D+4)}\cos(2t)$$
$$ x_p =e^{-t} \frac{(5D+4)}{116}\cos(2t)$$
Finally:
$$ \boxed { x_p =e^{-t} \frac{(-5 \sin(2t)+2 \cos(2t))}{58}}$$
