How to find the general solution using annihilators method?

Question

the problem is:

y''' - 6y'' + 11y' - 6y = e^t, y = y(t), y'=dy/dt

I don't know how to use annihilators method. Can you show how to do the problem?

Answer 1

First express the LHS in terms of differential operators,

(D^3 - 6D^2 + 11D - 6)y = e^t
Now try to find a differential operator you could apply to the RHS to make it zero. Since the derivative of e^t is e^t, we know that (e^t)' - (e^t) = 0. So (D-1)e^t = 0. This is why they call it the annihilator method, since (D-1) annihilates e^t.

(D-1)(D^3 - 6D^2 + 11D - 6)y = 0
Now factor and solve for D.
(D-1)(D-1)(D-2)(D-3)y = 0
D = 1 (repeated root) or D = 2 or D = 3
So the general solution will be:
y = Ae^t + Bte^t + Ce^(2t) + De^(3t)
Now plug it into the original equation to try to solve for the coefficients.
We know the Ae^t + Ce^(2t) + De^(3t) will give us zero when we plug it into y''' - 6y'' + 11y' - 6y, so we have the following:
(Bte^t)"' - 6(Bte^t)'' + 11(Bte^t)' - 6(Bte^t) = e^t
(3Be^t + Bte^t) - 6(2Be^t + Bte^t) + 11(Be^t + Bte^t) - 6Bte^t = e^t
3Be^t - 12Be^t + 11Be^t = e^t
2Be^t = e^t
B = 1/2

Final general solution:
y = Ae^t + Ce^(2t) + De^(3t) + (1/2)te^t

Note that this method is pretty much the same as the method of undetermined coefficients. It's really only useful if you're unsure of what form to guess for undetermined coefficients, the annihilator method can tell you.

Answer 2

Method Of Annihilators

Answer 3

It's a term I haven't hear of either. But if you propose the y = Ae^t, you just keep generating Ae^t for all the derivitives, and since the coefficients of the derivitives sum to 0, these terms "annilihate" each other.