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Imaginary Numbers in a general homogenous solution for a differential equation

  1. May 11, 2010 #1

    TG3

    User Avatar

    The problem statement, all variables and given/known data
    Find the general solution for:
    y''+2y'+5y=3sin2t


    The attempt at a solution

    y''+2y'+5y=3sin2t

    First step is to find the general solution to the homogenous equation, so skipping 2 steps (letting y=e^rt and dividing)
    R^2+2r+5
    (-2+/- sqroot(4-4*5))/2
    =-1 +/- 2i

    How do I put this complex number into my equation? Were it not for the 2i I'd say c1E^-T +c2Te^-T. Were it not for the -1 I'd say Acos(T) +Bsin(T).

    But how do I combine them?
     
  2. jcsd
  3. May 11, 2010 #2

    Mark44

    Staff: Mentor

    Two linearly independent solutions are y1 = e(-1 + 2i)t and y2 = e(-1 - 2i)t. These can be written as e-tei2t and e-te-i2t. Using the fact that eix = cosx + isinx, and skipping a few steps myself, a different pair of linearly independent solutions is e-tcos(2t) and e-tsin(2t).
     
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