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Math Help - Complex roots

  1. #1
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    Complex roots

    when solving a second order homogeneous equation and if the queficients are complex numbers I know that the general solution has the form: e^ht*(C1cos(ut)+C2sen(ut)) my question is: Does it include the imaginary part of the solution or its just the real part of the solution? I ask this because when getting the two solutions e^((h+iu)t) and e^((h-iu)t) I cant get to the general solution formula by means of linear combinations and eulers formula can someone help me?
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  2. #2
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    the visual efect of "not including the imaginary part" rises from the concept of "linear combination". The idea here is setting functions linearly independet

    agree?? if don't, ask me again ^^
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  3. #3
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    If the coefficients of the differential equation are real, then the coefficients of the characteristic equation are real so that if h- iu is a solution, so is h+ iu. The "general solution" corresponding to e^{(h- iu)t}= e^{ht}(cos(ut)+ i sin(ut)) is C_1e^{(h- iu)t}+ C_2e^{h+ iu}t= C_1(e^{ht}(cos(ut)+ i sin(ut))+ C_2(e^{ht}(cos(-ut)+ i sin(-ut)).

    Because cos(-ut)= cos(ut) and sin(-ut)= -sin(ut) those can be combined:
    e^{ht}((C_1+ C_2)cos(ut)+ i(C_1- C_2)sin(ut).
    Of course, you could then define D_1= C_1+ C_2 and D_2= i(C_1- C_2) so that the general solution is e^{ht}(D_1cos(ut)+ D_2 sin(ut). That includes both real and imaginary parts.

    If your initial or boundary values are also real then C_1 and C_2 will be such that both C_1+ C_2 and i(C_1- C_2) are real numbers and in that case the solution, e^{ht}(D_1 cos(ut)+ D_2 sin(ut), will be a real valued function with 0 imaginary part.

    If the coefficients of the differential equation are NOT real numbers and/or the initial values are not real numbers, then you solution will involve complex numbers and you are probably better off leaving the solution in the exponential form.
    Last edited by HallsofIvy; November 10th 2010 at 04:48 AM.
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