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Complexification of rsin(at)+rcos(at)

  1. Mar 1, 2009 #1
    Sorry for this dumb questions! its that i am just trying to find multiple ways of solving problems: Basically i have

    y'' + 9y' = 648sin(9 t) + 648cos(9 t)

    i dont know how can you complexify this, would it be p(D)y=648exp(9t)?

    if so, what part do i extract, the imaginary or real? because it has both....

    i can solve this with uknown coefficients but i am trying to look for efficient ways of doing this.
     
  2. jcsd
  3. Mar 1, 2009 #2

    tiny-tim

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    complexify?

    Hi marmot! :smile:

    Do you mean y'' + 9y' = i648sin(9 t) + 648cos(9 t)? :confused:

    (and do you mean 3t?)

    if so, yes, that's 648ei9t
     
  4. Mar 1, 2009 #3
    no, it lacks the imaginary number, its just like that. i am trying to complexify it so that the differential equation becomes relatively straightforward. the problem i encounter is that generally when i want to complexify something, its either sin(x) or cos(x) that is in the right of the equation, not both. so i am wondering how to complexify it and then, whether if i have to extract from the solution the real or imaginary or both parts.

    thanks
     
  5. Mar 1, 2009 #4

    Ben Niehoff

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    You need both [itex]e^{i\omega t}[/itex] and its complex conjugate.
     
  6. Mar 1, 2009 #5
    I would rewrite the right hand side in the form R*cos(9t+a), where R and a are two new constants, because then it is the real part of R*e^i(9t+a).
     
  7. Mar 1, 2009 #6
    thank you!

    however marmoset

    i am trying to do this

    p(D)y=P(D)exp(at)

    where p(D)=D^2+D

    thus

    the solution would be

    y=exp(at)/P(a)

    however, how can i do this when i have two unknown variables, which in your case, are R and a?
     
  8. Mar 1, 2009 #7
    Last edited: Mar 2, 2009
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