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Homework Help: D.E. Annihilator method Question

  1. Apr 1, 2013 #1
    I lost the A variable when I simplified this. Did I go wrong somewhere?

  2. jcsd
  3. Apr 1, 2013 #2


    Staff: Mentor

    You're still not getting this - you're making the same mistakes you made in a previous post a week or so ago.

    The roots of the homogeneous equation 2y''' - 4y'' = 0 are r = 0, r = 0, and r = 2.

    This means that the solutions of this equation look like y = c1 + c2t + c3e2x.

    Now for the nonhomogeneous equation 2y''' - 4y'' = e2x + 2 + e-2x, we would ordinarily seek a particular solution of the form yp = Ae2x + B + Ce-2x.

    BUT, and this is important, and is what you're not understanding, e2x and 1 are already solutions of the homogeneous equation, so could not possibly be particular solutions of the nonhomogeneous equation.

    The root r = 0 is already repeated in the homogeneous equation, which led to solutions of the form c1 + c2t. Since there is another repetition of r = 0 in the nonhomogeneous equation, we need a solution for the nonhomogeneous equation of what form?

    The root r = 2 is already used in the homogeneous equation, which led to the solution c3e2x. So in the nonhomogeneous equation we need a solution of what form?

    Note that e-2x is NOT a solution of the homogeneous equation, so it will appear unchanged in the particular solution of the nonhomogeneous equation.
  4. Apr 1, 2013 #3
    I ended up realizing my mistake I think. Yes, it was the same error I made before. I got:
    y = c1+c2x+c3e2x+Axe2x+Be-2x

    which resulted in a final answer of:
    y = c1+c2x+c3e2x+[itex]\frac{1}{8}[/itex]xe2x-[itex]\frac{1}{32}[/itex]e-2x

    Is that right?
  5. Apr 1, 2013 #4


    Staff: Mentor

    No, not quite. The general solution of your nonhomogeneous equation will be composed of six functions, which corresponds to the number of roots (six) of the characteristic equation of the sixth-order homogeneous equation.

    To expand on that a bit, your original 3rd order nonhomogeneous DE could be written as 2(D3 - 2D2)y = e2x + 2 + e-2x, or 2D2(D - 3)y = e2x + 2 + e-2x.

    The homogeneous equation is 2D2(D - 3)y = 0, and its characteristic equation is 2r2(r - 2) = 0, with roots r = 0, r = 0, and r = 2.
    An independent set of solutions is {e0 = 1, xe0 = x, e2x}

    Using annihilators, we see that the operator that annihilates the right side of the original DE is (D - 2)(D)(D + 2). This means that the 3rd-order nonhomogeneous equation can be transformed to a 6th-order homogeneous equation.

    2(D - 2)(D)(D + 2)D2(D - 2)y = (D - 2)(D)(D + 2)(e2x + 2 + e-2x) = 0.

    The general solution of this 6th-order DE will consist of six linearly independent functions, if which three will be the complementary solution of the 3rd-order homogeneous equation, and the other three will come from the right side of the nonhomogeneous equation, and that we'll use for a particular solution.

    So, yc = c1 + c2x + c3e2x.
    And, yp = A____ + B_____ + Ce-2x

    Remember, the root r = 0 was of multiplicity three, and we have accounted for two of them in yc, so what do we need in yp?

    Also, the root r = 2 is of multiplicity two, and one of them is in yc, so what do we need in yp?

  6. Apr 1, 2013 #5
    Hold on, I divided the original DE by two. Is that not valid?
  7. Apr 1, 2013 #6


    Staff: Mentor

    Not unless you also divide the right side by 2. If it were a homogeneous equation, you could do that, but it's a nonhomogeneous equation.
  8. Apr 1, 2013 #7
    Oh, I should've had a one on the right side. Simple math mistake. 2/2 does not equal 0! I did need to divide the entire thing by two didn't I?
    Last edited: Apr 1, 2013
  9. Apr 1, 2013 #8


    Staff: Mentor

    Yes, it is. Speaking of images, please take some time and shrink your images so that they fit the window. The max. size should be no larger than 900 x 600. When they're wider than that, you have to scroll to see the entire image.
  10. Apr 1, 2013 #9


    Staff: Mentor

  11. Apr 1, 2013 #10
    Is this right?
  12. Apr 1, 2013 #11


    Staff: Mentor

    You can check it yourself, you know. You can do it in two parts.
    1. Verify that D2(D - 2)[c1 + c2x +c3e2x] = 0.
    2. Verify that D2(D - 2)[(1/8)xe2x - (1/4)x2 - (1/32)e-2x] = (1/2)[e2x + 2 + e-2x].
  13. Apr 2, 2013 #12
    Well step one doesn't equal zero.. What did I do wrong?

    if y = c1 + c2x +c3e2x
    y"(y'-2) = 4c3e2x(2c3e2x+c2-2)
  14. Apr 2, 2013 #13


    Staff: Mentor

    Step 1 does result in 0. You are confusing multiplication with operator composition. The operator D2(D - 2) does not mean y'' * (y' - 2). If you expand D2(D - 2), you get D3 - 2D2.

    So D2(D - 2)y = (D3 - 2D2)y = y''' - 2y''. What do you get when you calculate that difference?
  15. Apr 2, 2013 #14
    Wait a minute, You're getting D2(D - 2), but I got D(D-2)(D+2). Yours is correct but I don't see how you go it.

    [itex]\frac{1}{2}[/itex]e2x = (D-2)

    [itex]\frac{1}{2}[/itex]e-2x = (D+2)

    1 = D

    D3-2D2 is for *Yc*
    and D(D-2)(D+2) is Yp right?
    Last edited: Apr 2, 2013
  16. Apr 2, 2013 #15


    Staff: Mentor

    Look at the original DE, which was 2y''' - 4y'' = e2x + 2 + e-2x.

    The homogeneous version of this equation is 2y''' - 4y'' = 0, or equivalently, y''' - 2y'' = 0. Using operators, this is (D3 - 2D2)y = 0, or D2(D - 2)y = 0.

    y = c1 + c2x + c3e2x is the general solution of the homogeneous 3rd order equation. This means that D2(D - 2)[c1 + c2x + c3e2x] has to equal 0.
  17. Apr 2, 2013 #16
    That's what I already have. It's just a one step process. You replace y''' with D3 and y'' with D2 resulting in D3-2D2. What would be the point of even mentioning, or arranging it as D2(D-2)?? I thought the test was for Yp; the D(D-2)(D+2) part.. I'm kind of confused now..

    If I use Yc, as you seem to suggest, and make sure everything is expanded out, then it passes both tests. However some examples from my notes don't pass the test
    y"-2y'+y turns into D2-2D+1, you get c1ex+c2xex. That test doesn't pass for step one.
    Last edited: Apr 2, 2013
  18. Apr 2, 2013 #17


    Staff: Mentor

    Almost. y''' = D3y, and y'' = D2y, so y''' - 2y'' = (D3 - 2D2)y. Note that, for example, Dy does not mean D * y, any more than ##\frac{dy}{dx}## means ## \frac{d}{dx} * y##.
    Because of the close relationship between the operator factorization and the characteristic equation.
    Using annihilators, there are three diff. equations you need to keep track of:
    1. The original nonhomogeneous equation. In this case, 2y''' - 4y'' = (e2x + e-2x)2.
    In operator form, this is 2D2(D - 2)y = (ex + e-x)2.
    2. The homogeneous form of the original equation. In this case, 2y''' - 4y'' = 0. In operator form, it can be reduced to D2(D - 2)y = 0.
    3. The higher order homogenous diff. equation. In this case, in its operator form, it is
    2D3(D - 2)2(D + 2)y = 0. I don't want to do the extra work of writing this in the form 2y(6) + ... = 0.

    The general solution of the equation in #3 is made up of six functions, three of which are the solution to #1 (the complementary function), and three of which are the solution to #2 (the particular solution.

    From what you've already done, yc = c1 + c2x + c3e2x. Notice that the operator form of #1 has D2 and D - 2. The characteristic equation for #1 has roots r = 0 (twice) and r = 2. The three functions that were chosen tie closely to the roots of the characteristic equation and the operators: {e0x = 1, xe0x, e2x}.

    As a check, it has to be the case that 2D2(D - 2)[c1 + c2x + c3e2x] = 0. This is the first part of the check I said you should do. If you don't get zero here, it means you didn't pick the right functions for the complementary solutions

    The particular solution, yp, is made up of the other three functions, with yp = Ax2 + Bxe2x + Ce-2x. Since this is supposed to be a solution of #2, the nonhomogeneous equation, it has to be the case that 2D2(D - 2)[Ax2 + Bxe2x + Ce-2x] = e2x + 2 + e2x. This is the second part of the check. If it doesn't end up as it should, it means you didn't pick the right particular solution.
    What does this have to do with the problem in this thread?
  19. Apr 2, 2013 #18
    Right. You basically re-explained what you already explained earlier. The point i was trying to make was that the test you said to use marks problems incorrect that were example problems given by the instructor. An example of this was:
    y"-2y'+y turns into D2-2D+1.
    You then get yc=c1ex+c2xex.
    That's all the info you need to try test one on this problem and it fails it.
  20. Apr 2, 2013 #19


    Staff: Mentor

    In operator notation, the DE above is (D2 - 2D + 1)y = x3+4x, or (D - 1)2y = x3+4x .

    The homogeneous equation is y'' - 2y' + y = 0, or (D - 1)2y = 0.
    The characteristic equation is r2 - 2r + 1 = 0, with roots r = 1 (multiplicity 2).
    The general solution of the homogeneous equation is yc = c1ex + c2xex.

    The nonhomogeneous equation is y'' - 2y' + y = x3+4x.

    The annihilator of the right side is D4, so applying this operator to both sides, we get:
    D4(D - 1)2y = D4(x3+4x) = 0.
    This is a 6th order homogeneous diff. equation, whose characteristic equation is r4(r - 1)2 = 0.

    The general solution would be y = c1ex + c2xex + A + Bx + Cx2 + Dx3. This problem is easier than the one of this thread, since there are no roots that are repeated between the homogeneous equation and the one that is produced when the order of the equation is bumped up by the application of annihilators.

    The first two terms are yc, the general solution of the homogeneous 2nd order equation. The last four terms are yp, the particular solution of the 2nd order nonhomogeneous equation. You would use the method of undetermined coefficients to solve for A, B, C, and D.

    It should be the case that (D - 1)2(c1ex + c2xex) = 0.

    It should also be the case that (D - 1)2(A + Bx + Cx2 + Dx3) = x3 + 4x.

    So how do you see this failing?
  21. Apr 2, 2013 #20
    If you set f(x)=c1ex + c2xex
    and then do ([itex]\frac{d}{dx}[/itex][f(x)]-1)2 into a calc like mine then you don't get zero. You get ((c2x+c1+c2)ex-1)2. Even if you expand (D - 1)2 out and then do it, you still don't get zero.
    When using the exact same technique on the original problem that this thread was based on, you do get 0. So what's going on here?
  22. Apr 2, 2013 #21


    Staff: Mentor

    Let me set you straight since you are obviously confused.

    (D - 1)2y means (D2 - 2D + 1)y means y'' - 2y' + y.

    Let y = c1ex + c2xex.
    Calculate y' and y''.
    What do you get for y'' - 2y' + y? I get 0.
  23. Apr 2, 2013 #22


    Staff: Mentor

    This thing -- ([itex]\frac{d}{dx}[/itex][f(x)]-1)2 -- has nothing to do with the operator (D - 1)2.
  24. Apr 2, 2013 #23
    Aaaaah, that's the real calculation. No need for that silly D notation except for finding zeros. Yes, when you cut straight to what actually needs to be calculated, I also get zero. (ex (c1+c2 (x+2)))-2(ex (c1+c2x+c2))+(c1ex + c2xex)=0
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