# Homework Help: Annihilator method

1. Nov 28, 2009

### jbord39

1. The problem statement, all variables and given/known data

Solve (D^2 - 6D +25)y = (x^2)(e^-x)

2. Relevant equations

D=dy/dx

3. The attempt at a solution

First I found the roots of the left side of the equation, which 4+i and 4-i.
From this,
y(c) = Ae^(3x)sin(4x) + Be^(3x)cos(4x)

Furthermore, the annihilator for (x^2)e^(-x) is (D+1)^3 <--- (Is this correct?).

From this, y(p) = Ce^(-x) + Dxe^(-x) + E(x^2)e^(-x) = e^(-x)[C + Dx + Ex^2]
and y(p)' = e^(-x)[-C + D(1 - x) + E(2x - x^2)]
and y(p)'' = e^(-x)[C + D(x - 2) + E(x^2 - 4x + 2)]

Now plugging this into
(D^2 - 6D +25)y = (x^2)e^(-x)

Yields:
32C + 8D(4x-1) + 2E(16x^2 - 8x + 1) = x^2

From here I cannot figure out how to solve for C, D, and E.

Thanks for any help

2. Nov 28, 2009

### Dick

I won't check all of your steps in getting there, but to solve your final equation write it in the form Lx^2+Mx+N=0. The only way both sides can be equal for all x is for L=M=N=0. That's three equations to solve.

3. Nov 28, 2009

### Staff: Mentor

The actual roots are 3 +/- 4i, which you are using below.
Yes. The annihilator for e^(-x) would be D + 1. For xe^(-x), it would be (D + 1)^2, and for x^2e^(-x) it would be (D + 1)^3.
Your yp' looks fine, but I think you have an error in your yp''. Also, I don't see any advantage in writing the part in square brackets as you have done. I think it would be better to group powers of x together rather than multiples of (1 - x) and (2x - x^2). Writing yp' as e^(-x)[D - C + (2E - D)x - Ex^2] might be simpler to work with.
This is where what I am saying comes into play. You want the left side arranged in powers of x. The right side can be thought of as being in powers of x; namely 0 + 0x + 1x^2. The equation has to be identically true for any value of x, which means that the constant on the left has to equal the constant on the right (0), the coefficient of the x term on the left has to equal the coefficient of x on the right (again, 0), and the coefficient of the x^2 term on the left has to equal the coefficient of x^2 on the right (1).

That will give you three equations in your unknowns C, D, and E.

4. Nov 28, 2009

### jbord39

Thanks.

Now I have :

(32C-8D+2E) + x(32D - 16E) + x^2(32E - 1) = 0

From here:
E = (1/32)
D = (1/64)
and
32C-(2/16) + (1/16) = 0
32C = 1/16
C = 1/512

These coefficients seem really small... Did I do that wrong?

5. Nov 28, 2009

### Staff: Mentor

It doesn't matter whether they are small are large. All that matters is that these coefficients work. If yp'' - 6yp' + 25yp = x2e-x using the coefficients you found, your work is correct - that's what matters.

Be sure to read my previous post, where I mentioned that your yp'' was incorrect.

Last edited: Nov 28, 2009
6. Nov 28, 2009

### jbord39

if y(p)' = e^(-x)[-C + D(1 - x) + E(2x - x^2)]
then y(p)' = -Ce^(-x) + De^(-x) - Dxe^(-x) + 2Exe^(-x) - E(x^2)e^(-x)
so y(p)'' = Ce^(-x) - De^(-x) - D[-xe^(-x) + e^(-x)] + 2E[-xe^(-x) + e^(-x)] -E[-(x^2)e^(-x) + 2xe^(-x)]

So y(p)'' = e^(-x)[C - D + D(1-x) + 2E(1-x) - E(2x-x^2)]
or y(p)'' = e^(-x)[C - D + D - Dx + 2E - 2Ex - 2Ex + Ex^2]
= e^(-x)[C - Dx +Ex^2 - 4Ex + 2E]

Does this y(p)'' look right?

Thanks

7. Nov 28, 2009

### Count Iblis

I'm not a fan of using messy methods involving undetermined coefficients and solving simultaneous equations for them. You can also get the particular soultion by substituting

y(x) = f(x) exp(-x)

Then

D^n y = exp(-x) (D-1)^n f

This then leads to:

[(D-1)^2 - 6(D-1) +25]f = x^2 -------->

[D^2 - 8 D + 32]f = x^2

Then we write the solution formally as:

f(x) = 1/[D^2 - 8 D + 32] x^2

We formally expand the expression involving the differential operator as follows:

1/[D^2 - 8 D + 32] =

1/32 1/[1 - D/4 + D^2/32] =

1/32 [1 + D/4 - D^2/32 + D^2/16 +...] =

1/32 [1 + D/4 + D^2/32 +...]

Applying this to x^2 gives:

f(x) = x^2/32 + x/64 + 1/512

8. Nov 28, 2009

### Staff: Mentor

These look fine. My values now agree with yours and those of Count Iblis.

9. Nov 28, 2009

### jbord39

Thanks so much for the help