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Method of undetermined coefficients

  1. Nov 14, 2006 #1
    y" + 0.5y' + y = 1-cos(t); y(0) = y'(0) = 0

    I used method of undetermined coefficients to get particular solution:

    Y(t) = -2sin(t) + 1

    To get homogeneous solution, I solved characteristic equation to get complex roots:

    r_1,2 = -1/4 +- i*sqrt(15)/4

    so homogeneous solution is:

    y = c1*exp(-1/4*t)cos(sqrt(15)/4*t) + c2*exp(-1/4*t)sin(sqrt(15)/4*t)

    But when I use the initial conditions, I get c1 = c2 = 0.
    Is this right, or did I make a mistake?

    I simplified the original problem with an easier forcing function to try first, so it is possible the ODE doesn't
    make sense?
     
  2. jcsd
  3. Nov 15, 2006 #2

    HallsofIvy

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    Yes, you made a mistake! You didn't add the particular solution to the solution to the associated homogeneous equation!
    The general solution to the entire equation is y(t)= c1*exp(-1/4*t)cos(sqrt(15)/4*t) + c2*exp(-1/4*t)sin(sqrt(15)/4*t)-2sin(t) + 1

    Put x= 0 into that and its derivative.
    y(0)= c1+ 1= 0 so c1= -1.

    You do the derivative.
     
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