# ODEs- annihilator method

## Homework Statement

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

## 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)
?

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Mark44
Mentor
If you wanted to annihilate, for example, etsin(3t) or etcos(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 etsin(3t) and etcos(3t) on the one hand, and 1 +/- 3i on the other, is not coincidental.

If you wanted to annihilate, for example, etsin(3t) or etcos(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 etsin(3t) and etcos(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 =etcos(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 etcos(3t)

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

<=>(D2 - 2D + 10)2y=0

Thank you.