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General Question about Undetermined Coefficients

  1. Jan 29, 2014 #1
    Hello, I had a question about the method of undetermined coefficients for solving ODE's. I understand it is only useful for certain non-homogeneous functions, and those dictate specific guesses, but what if I had a sum or product of two valid functions, is the guess simply a sum or product of their guesses?

    EX:

    Guess for 3e^-2t = Ae^-2t

    Then would the guess for 3e^-2t(sin(t)) be (Ae^-2t)(Asin(t)+Bsin(t))?


    If this is not valid or too complicated, would variation of parameters be a better method for these types of equations?

    Thanks
     
  2. jcsd
  3. Jan 29, 2014 #2

    mfb

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    2016 Award

    Staff: Mentor

    Do you mean 3e^(-2t) = Ae^(-2t) (that is ##3e^{-2t}=Ae^{-2t}##)? That is equivalent to A=3, and nothing else. There is nothing to guess, or to solve.
     
  4. Jan 29, 2014 #3
    That example is just the non homogeneous term, so extend the example to something like

    y''+9y'+4y=3e^(-2t)(sint)
     
  5. Feb 14, 2014 #4

    lurflurf

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    I presume you are talking about the standard undetermined coefficients, it is possible to extend the method.
    The idea is we want to solve
    Q(D)y=f
    where
    R(D)f=0
    so we solve
    R(D)Q(D)y=0
    then separate
    y=y1+y2
    where Q(D)y1=0
    then determine a particular y2 by
    Q(D)y2=f

    So the functions f for which this method will work will be those that are solutions of some homogenous linear constant coefficient equation. A typical such function is
    $$C \, x^n \, e^{A \, x}\cos(B \, x+\phi)$$
    clearly linear combinations of such function are again such functions as are products of such functions and thus products of linear combinations of such functions.

    I am sure there is some more slick way of showing it.
     
  6. Feb 15, 2014 #5

    AlephZero

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    Homework Helper

    Often, you make a guess and then think "what are some other terms that make this guess work".

    For your example ##y''+9y'+4y=3e^{-2t}\sin t##, since you have ##y## on the left hand side, you need
    at least ##y = A e^{-2t}\sin t##. But if you differentiate that, you get ##y' = -2A e^{-2t}\sin t + A e^{-2t}\cos t## so you will have to equate the coefficients of ##e^{-2t}\cos t## as well.

    So, a better guess is ##y = A e^{-2t}\sin t + B e^{-2t}\cos t##. Now when you substitute into the equation and equate the coefficients of ##e^{-2t}\sin t## and ##e^{-2t}\cos t##, you get two equations for A and B. If you can solve for A and B, you are done.

    But if the equation was ##y'' + 4y' + 5 = 3e^{-2t}\sin t##, your guess ##y = A e^{-2t}\sin t + B e^{-2t}\cos t## is actually the general solution for ##y'' + 4y' + 5 = 0##, so it won't give you a particular solution. If you try it, you can't solve the equations for A and B.

    If that happens, you have to start again with something like ##y = Ax e^{-2t}\sin t##, and then see what other terms you need.
     
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