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Finding the solution to a DE with complex roots

  1. Jan 26, 2010 #1
    I can't figure out how to get the answer in the back of the book.

    y''+6y'+13y=0

    So far, I have.....

    ert (r2 + 6r + 13) = 0
    r2 + 6r + 13 = 0
    r2 + 6r + ___ = -13 + ___
    r2 + 6r + 9 = -13 + 9
    (r+3)2 = -4
    r = -3 +/- 2i

    Then we have a crazy-looking thing after assuming y = C1e(r1)(t) + C2e(r2)(t)

    y = e-3t (C1e(2t)i + C2e(-2t)i

    And using Euler's formula and simplifying a little bit

    y = e-3t ( C1 ( cos(2t) + i*sin(2t) ) + C2( cos(-2t) i*sin(-2t) ) )

    The answer in the back of the book, however, is something that doesn't even involve imaginary numbers.

    What am I missing?
     
  2. jcsd
  3. Jan 26, 2010 #2

    Pengwuino

    User Avatar
    Gold Member

    What was the answer in the back of the book that you are finding conflict with? Looks right to me assuming you meant the [tex]e^{-3t}[/tex] to multiply the whole thing in yoru last line.
     
  4. Jan 26, 2010 #3
    y=C1e-3tcos(2t) + C2e-3tsin(2t)

    There's probably some trig identity eluding me :confused:
     
  5. Jan 26, 2010 #4
    I'm pretty sure the answer from the book is just a rearranged form of your equation
    y = e-3t ( C1 ( cos(2t) + i*sin(2t) ) + C2( cos(2t) - i*sin(2t) ) )

    Just multiply out everything, then you can factor the sine and cosine terms, so it's of the form e-3tcos(2t)+e-3tsin(2t). The rest is just renaming constants, the book solution just absorbed the i into their C2.
     
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