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Complex-valued solutions to real-valued solutions

  1. Dec 3, 2005 #1
    I'm going crazy trying to figure out how to do this. Here are two eigen value solutions: r=-1+-2i to a problem. And the two eigen vectors are (-2i, 1) and (2i, 1). How do I covert this to a real-valued solution? The book used (-2i, 1) to come up with the solution x = c * e^(-t)( -2sin(2t), cos(2t) ) + d * e^(-t)( 2cos(2t), sin(2t) )
    c and d are constants.
    I know that real value of r=1+-2i should be e^(-t)sin(2t) + e^(-t)cos(2t) but I don't how they came up with their answer using this fact.
     
  2. jcsd
  3. Dec 3, 2005 #2

    HallsofIvy

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    [tex]e^{ix}= cos(x)+i sin(x)[/tex]. Does that help?
     
    Last edited: Dec 6, 2005
  4. Dec 5, 2005 #3

    jim mcnamara

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    [tex]e^{ix}= cos(x)+i sin(x)[/tex]
     
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