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Is It Possible to Use the Method of Undetermined Coefficients Here?

  1. Jul 13, 2014 #1
    I'm doing a practice problem I found online, and I get a solution, but I think it should have a sine term in it. I looked up the solution, and most sites say to use variation of parameters, but is it possible to use the method of undetermined coefficients?

    The problem is as follows: y'' + 4y = 8cos2t

    r^2 + 4 = 0 ----> r^2 = -4 -----> r = +/- 2i

    yh = c1*sin2t + c2*cos2t

    Particular solution has form At*sin2t + Bt*cos2t (because you add a "t" to both terms so they don't clash with the cos2t or sin2t terms)

    First derivative is -2At*sin2t + A*cos2t + 2Bt*cos2t + B*sin2t

    Second derivative is -4At*cos2t - 2A*sin2t - 2A*sin2t - 4Bt*sin2t + 2B*cos2t + 2B*cos2t

    So then you add that term to the particular solution and get:

    -4At*cos2t - 2A*sin2t - 2A*sin2t - 4Bt*sin2t + 2B*cos2t + 2B*cos2t + 4At*sin2t + 4Bt*cos2t

    Canceling out terms leaves you with -2A*sin2t - 2A*sin2t + 2B*cos2t + 2B*cos2t = 8*cos2t.

    Now the problem is that 8*cos2t is really 8*cos2t + 0*sin2t. Solving for B gives you B = 2, but that means A = 0. But the final solution should have a sine term in it. (And I looked up the solutions, and they do have a sine term)

    I thought you could use the Method of Undetermined Coefficients when you were dealing with a sine/cosine function, a regular polynomial, or an exponential function. (I think the fourth one was a function of e^t. I forget off the top of my head.)
     
  2. jcsd
  3. Jul 14, 2014 #2

    pasmith

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    Homework Helper

    You are inconsistent in your definitions of [itex]A[/itex] and [itex]B[/itex]. You start by defining
    [tex]
    y(t) = At\sin(2t) + Bt\cos(2t)
    [/tex] but in computing the derivatives you somehow switched to [tex]
    y(t) = At\cos(2t) + Bt\sin(2t).[/tex]
    Thus when you substituted these into the ODE you obtained
    where you failed to notice that the terms I've bolded don't actually cancel.

    But continuing on, you find that [itex]B = 2[/itex]. Because you flipped the definitions of [itex]A[/itex] and [itex]B[/itex] halfway through, this is actually telling you that [itex]y(t) = 2t\sin(2t)[/itex], which is correct.
     
    Last edited: Jul 14, 2014
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