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Undetermined coefficients with annihilator approach (DE)

  1. Mar 16, 2013 #1
    There isn't a specific problem that's making me stuck, but I was hoping if someone could point me in the right direction here. I've looked up the topic online, but most of what I could find was through another approach or very unclear. The book I'm using also does not use the method my professor uses.

    In my ODE class I've been using the annihilator approach to solve homogenous DEs. We moved on to solving nonhomogenous ODEs a while ago and covered undetermined coefficients and variation of parameters, but I had to miss the lecture on undetermined coefficients and there's something I missed which is messing me up.

    I understand that if I have a simple ODE such as
    [tex] y'' + 16y = e^{3x} [/tex]

    I can see why the particular solution would take the form

    [tex] y_p = Ae^{3x} [/tex]

    You'd then differentiate as needed, plug in, and solve for the coefficients. This way of thinking has been recurring throughout the course so it doesn't feel foreign. However, in the later exercises, I'm having problems.

    Consider the ODE

    [tex] y^{(3)} - y = e^x + 7 [/tex]

    If I just look at the RHS and think of what'll annihilate it, let D be the differential operator, then D(D-1) looks like it'll do the job. This means that the particular solution would look like the form:

    [tex] y_p = c_1 + c_2e^x [/tex]

    But if I were to differentiate that three times and plug it in to the ODE, it wouldn't give me a nice result.

    I noticed that the problems I've been struggling with have something in common though.
    The LHS of the ODE is annihilated by (D-1)(D2 + D + 1). So (D-1) is a common annihilator of both sides of the equation and I suspect this has to do with what's throwing me off.

    Could someone please shed some light on what's the algorithm for these types of situations?
     
    Last edited: Mar 16, 2013
  2. jcsd
  3. Mar 16, 2013 #2

    vela

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    If you apply the annihilator, you'll have D(D-1)2(D2+D+1)y = 0. What should be the form of the solution to this homogeneous equation? Compare it to yh you get from the original DE. The terms left over make up yp.
     
  4. Mar 16, 2013 #3
    Thanks for the reply, I hadn't thought of it that way. You're kind of shifting the nonhomogeneous equation to a homogeneous one.

    The solution formed by applying the annihilator would take the form:

    [tex] y = c_1 + c_2e^x + c_3xe^x + e^{-x/2}(c_4cos(\sqrt{3/4}x) + c_5sin(\sqrt{3/4}x)) [/tex]

    Since

    [tex] y_h = a_1e^x +e^{-x/2}(a_2cos(\sqrt{3/4}x) + a_3sin(\sqrt{3/4}x)) [/tex]

    I'm guessing

    [tex] y_p = Axe^{x/2} + B [/tex]

    So in order to find the particular solution, it would simply just be a manner of differentiating the last equation above and plugging in to find the coefficients?
     
    Last edited: Mar 16, 2013
  5. Mar 16, 2013 #4

    vela

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    I'm not sure what's going on with the exponents there. Where did the ##a_2## term and the A term come from?
     
  6. Mar 16, 2013 #5
    Oh, I just switched the letters up because I didn't feel like keeping track of which constant is which.

    Sorry about that, I thought of adding the explanation when I made the choice to do that after I was finished typing the equations, but I forgot to actually go ahead and type that out when I got to finishing...

    That is, B would be equal to c_1, a_1 would be equal to c_2...
    Should've just gone ahead and used the same names...
     
  7. Mar 16, 2013 #6

    vela

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    The exponent in yp should be x, not x/2, but otherwise it looks fine now.

    Do you see how you would generally modify your guess for the particular solution when the forcing function has terms that satisfy the homogeneous equation?
     
  8. Mar 16, 2013 #7
    Yep, that exponent's a typo.

    Just write out the general solution for the DE annihilated by both annihilators (the RHS and LHS) and then filter out yp from the terms that already appeared in yh and then just go through the mindless computation and equate coefficients.

    Pretty simple procedure then, thanks a lot for taking the trouble to help me out. :smile:
     
  9. Mar 16, 2013 #8

    vela

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    You can streamline that procedure a bit. You just need to multiply what you might naively guess for yp by appropriate powers of x. For example, if you had (D-1)2y=ex, you might initially guess that yp=Cex. But since yh=Aex+Bxex, you need to multiply by x2 to make the particular solution linearly independent of yh, so you'd actually use yp=Cx2ex.
     
  10. Mar 16, 2013 #9
    Eh, the thought process of converting the nonhomogeneous eqn to a homogeneous eqn and then choosing what you want to solve seems like the most intuitive thing for me and I hate memorizing things, so I'll just stick to what you told me a couple of posts ago. What you're saying now might seem more obvious after I work out a couple of problems, but for now I'm happy the other way.
     
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