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Math Help - Complex cofficients in differencial equation

  1. #1
    Senior Member tukeywilliams's Avatar
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    Complex cofficients in differencial equation

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  2. #2
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    Post in correct subforum. This has absolutely nothing to do with algebra!
    Attached Thumbnails Attached Thumbnails Complex cofficients in differencial equation-picture16.gif  
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  3. #3
    Senior Member tukeywilliams's Avatar
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    Why did you set c1 = c2 = 1/2 and c1 = 1/2i and c2 = -1/2i? How did you get that to equal e^(\alpha t) \cos t and e^(\alpha t) \sin t?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by tukeywilliams View Post
    NOTE: I have corrected the 9:30AM version of this!

    You need to split y into its real and complex parts:

    I'm going to call alpha = a and beta = b and I^2 = -1. (Don't confuse these with the coefficients of the differential equation!)

    y = k1*exp{(a + Ib)t} + k2*exp{(a - Ib)t}

    y = k1*exp{at}*exp{Ibt} + k2*exp{at}*exp{-Ibt}

    y = exp{at}*[k1*exp{Ibt} + k2*exp{-Ibt}]

    y = exp{at}*[k1*cos(bt) + I*k1*sin(bt) + k2*cos(bt) - I*k2*sin(bt)]


    Define k1 = c + Id and k2 = e + If where c, d, e, f are real.

    Then
    y = exp{at}*[c*cos(bt) + I*d*cos(bt) + I*c*sin(bt) - d*sin(bt) + e*cos(bt) + I*f*cos(bt) - I*e*sin(bt) + f*sin(bt)]

    y = exp{at}*[c*cos(bt) - d*sin(bt) + e*cos(bt) + f*sin(bt)] + I*exp{at}*[d*cos(bt) + c*sin(bt) + f*cos(bt) - e*sin(bt)]

    The complex part of y is
    exp{at}*[d*cos(bt) + c*sin(bt) + f*cos(bt) - e*sin(bt)]

    = exp{at}*[(d + f)*cos(bt) + (c - e)*sin(bt)]

    This must be 0. To be true for all b and all t we must have that
    d + f = 0
    c - e = 0

    Thus f = -d
    Thus e = c.

    So k2 = e + If = c - Id, which is the complex conjugate of k1 = c + Id.

    -Dan
    Last edited by topsquark; March 27th 2007 at 10:10 AM.
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  5. #5
    Senior Member tukeywilliams's Avatar
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    Quote Originally Posted by topsquark View Post
    NOTE: I have corrected the 9:30AM version of this!

    You need to split y into its real and complex parts:

    I'm going to call alpha = a and beta = b and I^2 = -1. (Don't confuse these with the coefficients of the differential equation!)

    y = k1*exp{a + Ib} + k2*exp{a - Ib}

    y = k1*exp{a}*exp{Ib} + k2*exp{a}*exp{-Ib}

    y = exp{a}*[k1*exp{Ib} + k2*exp{-Ib}]

    y = exp{a}*[k1*cos(b) + I*k1*sin(b) + k2*cos(b) - I*k2*sin(b)]


    Define k1 = c + Id and k2 = e + If where c, d, e, f are real.

    Then
    y = exp{a}*[c*cos(b) + I*d*cos(b) + I*c*sin(b) - d*sin(b) + e*cos(b) + I*f*cos(b) - I*e*sin(b) + f*sin(b)]

    y = exp{a}*[c*cos(b) - d*sin(b) + e*cos(b) + f*sin(b)] + I*exp{a}*[d*cos(b) + c*sin(b) + f*cos(b) - e*sin(b)]

    The complex part of y is
    exp{a}*[d*cos(b) + c*sin(b) + f*cos(b) - e*sin(b)]

    = exp{a}*[(d + f)*cos(b) + (c - e)*sin(b)]

    This must be 0. To be true for all b we must have that
    d + f = 0
    c - e = 0

    Thus f = -d
    Thus e = c.

    So k2 = e + If = c - Id, which is the complex conjugate of k1 = c + Id.

    -Dan
    Shouldn't there be a t in y = k1*exp{a + Ib} + k2*exp{a - Ib}?
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by tukeywilliams View Post
    Shouldn't there be a t in y = k1*exp{a + Ib} + k2*exp{a - Ib}?
    Yup. It doesn't actually change the result though. I went back and fixed (again!) my original answer.

    -Dan
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