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Second Order Homogeneous Eq's, Auxiliary Eq for complex roots - Help!

  1. Aug 8, 2009 #1
    Hi

    I'v got a maths exam on Tuesday for my 2nd year of chemial engineering.
    Been going through a past paper and have been going over 2nd order homogenous DE's

    Im at the stage of calculating the roots (wether repeated or 2 distinct roots) I take the easy path like so:

    E.g m2 + 4m + 4 = 0

    (m + 2)(m + 2) = 0 so m=-2 (twice)

    Thats easy but that doesn't work for complex roots.

    My lechtrer does it this way which I dont really understand (for complex numbers that is, fine for everything else)

    So something like:

    m2 - 6m + 10 = 0

    He then does:


    m = ( 6 ± SQRT[36-40] / 2 ) = 3 ± i


    I dont understand this! A general trend i see here is that the 6m = the 6 before the square root, the 36 is the 6 squared and the 40 is always the
    last number * 4 (so 10*4 = 40) This holds true for every problem but I don't have a clue how to get that answer?

    I don't understand really what ' i ' is but if i got that answer I can work throuh my textbook and work out the rest, I just need to know how to get 3 ± i in the first place.

    Thanks in advance
     
  2. jcsd
  3. Aug 8, 2009 #2

    rock.freak667

    User Avatar
    Homework Helper

    You should post this in the correct forum. But I will answer you anyway.

    m2 - 6m + 10 = 0

    your teacher used the quadratic equation formula

    [tex]x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]

    and then used the fact that i=√-1 for the minus sign in the the square root when it is computed

    generally for complex roots λ±μi, your general solution (or homogeneous solution) is given by

    eλx(Asin(μx)+Bcos(μx)) where A & B are constants.
     
  4. Aug 8, 2009 #3
    Fantastic, thanks very much!
     
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