
#1
Nov1612, 12:08 AM

P: 746

Hello,
I am having trouble when solving nonhomogeneous 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. 



#2
Nov1612, 12:44 AM

Mentor
P: 21,059

If the function on the right side of your DE is sin(x), the annihilator is D^{2} + 1. The idea is that if y = sin(x), then (D^{2} + 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 D^{2} + 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(D^{2} + 1)y = 0, or y'' + y = 0. The characteristic equation is r^{2} + 1 = 0, so r = ħi. Two solutions of the DE are y_{1} = e^{ix} and y_{2} = e^{ix}. The usual practice is to not work with these exponential function, but to take certain linear combinations of them and use y_{1} = cos(x) and y_{2} = 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 (D^{4} + 2D^{2} + 1)y = 0, or (D^{2} + 1)^{2}y = 0. From either form it doesn't take long to get the characteristic equation, which is (r^{2} + 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 x^{2}sin(x) or x^{2}cos(x)? 



#3
Nov1612, 01:12 AM

P: 746

I understood how to do those homogeneous DE's, but I can't find the particular solution to a nonhomogeneous DE. To find the particular solution, I need the annihilator, right?
And for yours, is the annihilator D^3(D^2+1)? 



#4
Nov1612, 06:39 AM

HW Helper
P: 2,166

Annihilator operator
If you want to annihilate (x^n)f(x) and p(D) annihilates f(x) then what will [p(D)]^n annihilate?




#5
Nov1612, 09:49 AM

Mentor
P: 21,059





#6
Nov1612, 10:07 AM

Mentor
P: 21,059

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 e^{2x}. For this function, the annihilator would be D  2, with the idea being that (D  2)e^{2x} = 0. If the nonhomogeneous DE was y' + y = e^{2x}, or (D + 1)y = e^{2x}, we could use the annihilator we found to rewrite the original DE as (D + 1)(D + 2)y = (D + 2)e^{2x} = 0. IOW, instead of solving the nonhomogeneous equation y' + y = e^{2x}, 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 = e^{2x}, or (D + 2)y = e^{2x}. 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)e^{2x} = 0. This can be written as (D + 2)^{2}y = 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 = xe^{2x} works, which I leave to you to verify. (IOW if y = xe^{2x}, then y'' + 4y' + 4y = 0.) If the DE had been (D + 2)^{3}y = 0, two of the solutions would be e^{2x} and xe^{2x}. Can you make the leap and guess what the third basic solution would be? 



#7
Nov1612, 11:20 AM

P: 746

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 



#8
Nov1612, 12:41 PM

Mentor
P: 21,059

As I already showed, the annihilator for xcos(x) is (D^{2} + 1)^{2}. The same operator annihilates xsin(x). If the right side is e^{x}sin(x) or e^{x}cos(x) the annihilator is D^{2}  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 = e^{x}sin(x), then y''  2y' + 2y = 0. In general, the annihilator of e^{ax}sin(bx) or e^{ax}cos(bx) characteristic equation is (D  (a + bi))(D  (a  bi)). Multiplied out, this is D^{2}  2aD + a^{2} + b^{2}. 



#9
Nov1612, 12:50 PM

P: 746

Is it the case to solve nonhomogeneous 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 



#10
Nov1612, 04:20 PM

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P: 21,059





#11
Nov1612, 08:55 PM

P: 746

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. 



#12
Nov1612, 10:55 PM

Mentor
P: 21,059

To annihilate 24x + 4cos(x), take the product of the two operators: D^{2}(D^{2} + 1). 



#13
Nov1712, 01:01 AM

P: 746

Ok thanks Mark, I got the method down. Now I will attempt to solve it



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