How can I use the annihilator method to solve for 4e-2t*cos(2t)?

[Roni1985]

Homework Statement

How can I annihilate the following ?
4e^-2t*cos(2t)

Homework Equations

The Attempt at a Solution

I know that if I want to annihilate e^-t
it would be (D-1) and to annihilate cos(2t) it would be (D²+2²)

but what happens if they are multiplied ?
how do I annihilate this ?
I tried something and I'm not sure it's correct but I got
(D²+4)/(D-2)
?

 

[Mark44]

If you wanted to annihilate, for example, e^tsin(3t) or e^tcos(3t), the annihilator would be D² - 2D + 10.

The characteristic equation for the differential equation y'' - 2y' + 10y = 0, or (D² - 2D + 10)y = 0, is r² - 2r + 10 = 0. The roots of this latter equation are r = 1 +/- 3i. The connection between e^tsin(3t) and e^tcos(3t) on the one hand, and 1 +/- 3i on the other, is not coincidental.

 

[Roni1985]

If you wanted to annihilate, for example, e^tsin(3t) or e^tcos(3t), the annihilator would be D² - 2D + 10.

The characteristic equation for the differential equation y'' - 2y' + 10y = 0, or (D² - 2D + 10)y = 0, is r² - 2r + 10 = 0. The roots of this latter equation are r = 1 +/- 3i. The connection between e^tsin(3t) and e^tcos(3t) on the one hand, and 1 +/- 3i on the other, is not coincidental.

I am having a little problem following your explanation.

Say this is your differential equation:
y'' - 2y' + 10y =e^tcos(3t)

or for the left part only
(D² - 2D + 10)y
and what you are basically saying is that the left part of the equation => (D² - 2D + 10) annihilates e^tcos(3t)

so it should look this way :
(D² - 2D + 10)*(D² - 2D + 10)y =(D² - 2D + 10)*e^tcos(3t)

<=>(D² - 2D + 10)²y=0

Thank you.