# D.E. Annihilator method question

1. Mar 18, 2013

### Jeff12341234

When I solve for A,B,C, and D, A and D disappear. Is that ok? Are the terms that contained A and D simply omitted from the solution equation or have I made some other mistake somewhere?

2. Mar 18, 2013

### Staff: Mentor

The complementary function, yc = c1 + c2e3x + c2xe3x. The particular solution can't have either of the last two functions, above. It has to have x2e3x in it. The particular solution also can't have a pure constant in it, as you have with Ae0x, which is really just A. It has to have x and x2.

3. Mar 18, 2013

### Jeff12341234

So where did I go wrong?
Which specific line or step is incorrect and what should it be?

4. Mar 18, 2013

### Staff: Mentor

The first line that starts off "y = ..." You have too many terms in it. The particular solution should be yp = Ax + Bx2 + Cx2e3x.

What you need to do is to apply enough annihilators to get to a homogeneous equation. That would be D3(D - 3)3y = 0.

The roots of the char. equation are {0, 0, 0, 3, 3, 3}. These would correspond to 1, x, x2, e3x, xe3x, and x2e3x.

The characteristic equation for the homogeneous version of the original DE has roots {0, 3, 3}, corresponding to 1, e3x, and xe3x. The particular solution can't have any of these three functions. It will have the others that are listed in the paragraph above this one.

5. Mar 18, 2013

### Jeff12341234

yc, when expanded out, has a highest power of 3 so it has 3 roots; {0, 3, 3}. Right?
yp is to the 4th power when expanded out so doesn't it have to have 4 roots? {0, 0, 0, 3}?
Is that logic correct so far?

6. Mar 18, 2013

### Staff: Mentor

No, because of a mistake you made early on. The right side of your original DE is 8 + 7x + e3x. The annihilator for it is D2(D - 3), not D2(D - 3)D = D3(D - 3) as you show. Note that D2(8 + 7x) = D(D(8 + 7x) = D(0 + 7) = 0. You have an extra factor of D.

The original equation, as a homogeneous DE is y''' - 6y'' + 9y' = 0, or (D3 - 6D2 + 9D)y = 0. This can be written as D(D - 3)2y = 0.

The rootso of the char. equation are r = 0, and r = 3 (multiplicity 2). These roots correspond to, as I've already said, the functions {1, e3x, xe3x}.

In converting the origininal nonhomogeneous equation of order 3 to a homogeneous equation of order 6, you're appending two factors of D and one factor of D - 3. There are a lot of repeated roots there: r = 0 is of multiplicity 3 and r = 3 is of multiplicity 3 as well. The set of solutions are {1, e3x, xe3x, x, x2, x2e3x}.

The first three in the set above (in blue) are a basis for yc, which represents a basis for the solution of y''' - 6y'' + 9y' = 0. The second group of three (in green) are a basis for yp, a particular solution of the nonhomogenous equation y''' - 6y'' + 9y' = 8 + 7x + e3x. A function that is part of yc CANNOT ALSO BE IN yp, and that is what you are doing. You have 1 (= e0x) in both yc and yp and you have e3x in both, as well.

7. Mar 18, 2013

### Jeff12341234

I understand what you're saying. I need to figure out why the initial mistake was made though. From what I understand:
7x = D^2
e^(3x) = (D-3)
8 = 8x^0 = D
When put together I get D^2*(D-3)*D. Which one of those is wrong?

8. Mar 19, 2013

### Staff: Mentor

You don't need (and shouldn't have) separate factors of D and D2. D2 annihilates A + Bx for any constants A and B.

D2(A + Bx) = D(D(Ax + B)) = D(A) = 0.