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Simplifying a solution such that the imaginary number goes away

  1. Sep 25, 2007 #1
    1. The problem statement, all variables and given/known data

    Given a general solution to a system of differential equations:

    Y(t)= C1(1;2i)e^(2it)+C2(1;-2i)e^(-2it)

    side note: i is sqrt of -1, and the (1;2i) is a 2 by 1 matrix. The idea is to simplify the solution such that the imaginary components go away.

    2. Relevant equations

    Euler's formula.


    3. The attempt at a solution
    I have subbed the e^(2it) and e^(-2it) with their appropriate sin and cos counterparts. This is what I have have:

    Y(t)= C1(1;2i)[cos(2t)+isin(2t)]+C2(1;-2i)[cos(2t)-isin(2t)]

    and so from here, I lumped the cosines and sines together:

    Y(t)=[C1(1;2i)+C2(1;-2i)]cos(2t)+i[C1(1;2i)+C2(1;-2i)]sin(2t)


    I know I need to introduce two new free parameters to replace C1 and C2 but I'm not sure if I do that now or is there another way? I'm especially stumped as how to rid of the i from the (1;2i) and (1;-2i). Any suggestions are welcome. thanks.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 25, 2007 #2

    Dick

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    This is really just Y=C1*z+C2*conjugate(z) where z is a complex matrix. The condition for Y to be real is Y=conjugate(Y). Putting it all together, don't you just get C1=conjugate(C2)?
     
  4. Sep 25, 2007 #3
    I'm not sure I totally understand. So you're saying I set C1=the conjugate of C2? Or should I do set C1=C2*conjugate(z)?
     
  5. Sep 25, 2007 #4

    Dick

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    C1 and C2 are constants, right? I'm just saying if C1=(a+bi) then C2=(a-bi). a and b also constants. That will cancel all of the imaginary parts. You need to check this yourself. Shouldn't this:

    Y(t)=[C1(1;2i)+C2(1;-2i)]cos(2t)+i[C1(1;2i)+C2(1;-2i)]sin(2t)

    be this:

    Y(t)=[C1(1;2i)+C2(1;-2i)]cos(2t)+i[C1(1;2i)-C2(1;-2i)]sin(2t)
     
  6. Sep 26, 2007 #5
    Yes, you are right. I think I understand it now. So I should set C1=(a+bi) and C2=(a-bi) and the complex matrices will be resolved. Thank you so much!
     
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