# ODEs- annihilator method

• Roni1985
In summary, the conversation discusses using the annihilator method to solve for a differential equation involving 4e-2t*cos(2t). The suggested approach is to use the characteristic equation and its roots to find the appropriate annihilator for the given equation. However, the connection between etsin(3t) and etcos(3t) and the roots of the characteristic equation may not be immediately obvious.

## 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)
?

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.

Mark44 said:
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.