Annihilator operator

by Woopydalan
Tags: annihilator, operator
 P: 704 Hello, I am having trouble when solving non-homogeneous DE's how to find the annihilator to find my particular solution. For example, if you have a DE that equals 24x^2cos(x), how do I find something that will annihilate this? It seems to me no matter how many derivatives you take, you couldn't get this thing to be 0. The annihilator is needed to find the particular solution, right? When using the method of undetermined coefficients, I will have issues finding my particular solution.
Mentor
P: 19,758
 Quote by Woopydalan Hello, I am having trouble when solving non-homogeneous DE's how to find the annihilator to find my particular solution. For example, if you have a DE that equals 24x^2cos(x), how do I find something that will annihilate this? It seems to me no matter how many derivatives you take, you couldn't get this thing to be 0. The annihilator is needed to find the particular solution, right? When using the method of undetermined coefficients, I will have issues finding my particular solution.
The annihilator you choose is tied to the roots of the characteristic equation, and whether these roots are repeated.

If the function on the right side of your DE is sin(x), the annihilator is D2 + 1. The idea is that if y = sin(x), then (D2 + 1)y = 0.

This particular operator also annihilates any constant multiple of sin(x) as well as cos(x) or a constant multiple of cos(x). In fact, the D2 + 1 operator annihilates any linear combination of sin(x) and cos(x).

Let's take a look at the DE I'm considering here: a(D2 + 1)y = 0, or y'' + y = 0. The characteristic equation is r2 + 1 = 0, so r = ħi.

Two solutions of the DE are y1 = eix and y2 = eix. The usual practice is to not work with these exponential function, but to take certain linear combinations of them and use y1 = cos(x) and y2 = sin(x).

Now, let's suppose that we're trying to find solutions of y(4) + 2y'' + y = 0. If we write this in terms of operators, we get (D4 + 2D2 + 1)y = 0, or (D2 + 1)2y = 0.

From either form it doesn't take long to get the characteristic equation, which is (r2 + 1)2 = 0. The solutions are the same as before; namely, r = ħi, but these time each value is repeated. As before, two basic solutions are sin(x) and cos(x), but since the DE is of order 4, we need two more solutions to make a basic set of solutions that would span all possible solutions.

The answer is to tack on a factor of x to sin(x) and to cos(x). The basic solution set is now {sin(x), cos(x), xsin(x), xcos(x)}. I leave it to you to verify that for each of these functions, y(4) + y'' + y = 0. IOW, each of them is a solution to the DE.

Now, can you think of an operator that would annihilate x2sin(x) or x2cos(x)?
 P: 704 I understood how to do those homogeneous DE's, but I can't find the particular solution to a non-homogeneous DE. To find the particular solution, I need the annihilator, right? And for yours, is the annihilator D^3(D^2+1)?
HW Helper
P: 2,080

Annihilator operator

If you want to annihilate (x^n)f(x) and p(D) annihilates f(x) then what will [p(D)]^n annihilate?
Mentor
P: 19,758
 Quote by Woopydalan I understood how to do those homogeneous DE's, but I can't find the particular solution to a non-homogeneous DE. To find the particular solution, I need the annihilator, right? And for yours, is the annihilator D^3(D^2+1)?
No. That's not the pattern I was attempting to show. Lurflurf's post provides a clue.
 Mentor P: 19,758 The think about annihilators is that for certain kinds of forcing functions (the functions that make the DE nonhomogeneous), the application of the right annihilator turns the nonhomogeneous DE into a homogeneous DE of higher order. Let's look at a different example. Suppose the forcing function on the right side is e2x. For this function, the annihilator would be D - 2, with the idea being that (D - 2)e2x = 0. If the nonhomogeneous DE was y' + y = e2x, or (D + 1)y = e2x, we could use the annihilator we found to rewrite the original DE as (D + 1)(D + 2)y = (D + 2)e2x = 0. IOW, instead of solving the nonhomogeneous equation y' + y = e2x, we're now solving the homogeneous equation y'' + 3y' + 2y = 0. With a little work, we see that the characteristic equation is (r + 1)(r + 2) = 0, and that the roots are r = -1 and r = -2. Where things get a little tricky is if there are repeated roots of the homogeneous version of the original DE. Here's an example. y' + 2y = e2x, or (D + 2)y = e2x. This first order nonhomogeneous equation can be converted to a 2nd order homogeneous equation by applying the D + 2 operator: (D + 2)(D + 2)y = (D + 2)e2x = 0. This can be written as (D + 2)2y = 0. The characteristic equation for this DE is (r + 2)2 = 0, so we have a repeated root r = -2. We can get only one basic solution function out of this -- y = e-2x -- but we need another. As it turns out, y = xe2x works, which I leave to you to verify. (IOW if y = xe2x, then y'' + 4y' + 4y = 0.) If the DE had been (D + 2)3y = 0, two of the solutions would be e2x and xe2x. Can you make the leap and guess what the third basic solution would be?
 P: 704 yes I know x^2e^2x. The Problem is when I have a more complicated right side of the equation. something with x and cos(x) or e and cos(x) to find the annihilator for that. Annihilating e is fine
Mentor
P: 19,758
 Quote by Woopydalan yes I know x^2e^2x.
 Quote by Woopydalan The Problem is when I have a more complicated right side of the equation. something with x and cos(x) or e and cos(x) to find the annihilator for that. Annihilating e is fine
Can you be more specific?

As I already showed, the annihilator for xcos(x) is (D2 + 1)2. The same operator annihilates xsin(x).

If the right side is exsin(x) or excos(x) the annihilator is D2 - 2D + 2. It would take a fair amount of explanation to show you why this is true, but you can confirm this fact for yourself by showing that if y = exsin(x), then y'' - 2y' + 2y = 0.

In general, the annihilator of eaxsin(bx) or eaxcos(bx) characteristic equation is (D - (a + bi))(D - (a - bi)). Multiplied out, this is D2 - 2aD + a2 + b2.
 P: 704 Is it the case to solve non-homogeneous DE's (for which there is an annihilator) using the method of undetermined coefficients, the method to solve is 1. solve for homogeneous DE 2. Use annihilator to find particular solution 3. Solve particular solution 4. Add together with homogeneous to get the general solution
Mentor
P: 19,758
 Quote by Woopydalan Is it the case to solve non-homogeneous DE's (for which there is an annihilator) using the method of undetermined coefficients, the method to solve is 1. solve for homogeneous DE 2. Use annihilator to find particular solution 3. Solve particular solution 4. Add together with homogeneous to get the general solution
 P: 704 Yes, I was asking if this method is correct. I just didn't end my sentence with a ''?'' How do you annihilate 24x + 4cos(x)?? Is it just (D^2 + 1)? That was the question I was stuck on. Every example had things multiplied, but never did the DE equal to a sum.
Mentor
P: 19,758
 Quote by Woopydalan Yes, I was asking if this method is correct. I just didn't end my sentence with a ''?''
That's one of the biggest clues that a sentence is a question.
 Quote by Woopydalan How do you annihilate 24x + 4cos(x)?? Is it just (D^2 + 1)?
No. The D2 operator annihilates x or any constant multiple of x. The D2 + 1 operator annihilates cos(x) or any constant multiple of it (also sin(x) or a constant multiple).

To annihilate 24x + 4cos(x), take the product of the two operators: D2(D2 + 1).
 Quote by Woopydalan That was the question I was stuck on. Every example had things multiplied, but never did the DE equal to a sum.
 P: 704 Ok thanks Mark, I got the method down. Now I will attempt to solve it

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