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henpen Group Title

\[ \ddot{\phi}=-k^2 \phi \] I see why the solution should be \[ \phi =A e^{ikt} \], but not why \[ \phi= Acos(kt)+Bsin(wt) \] Surely the real part of \[A e^{ikt} \] is just \[Acos(kt) \]?

  • one year ago
  • one year ago

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  1. klimenkov Group Title
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    [Math Processing Error]

    • one year ago
  2. henpen Group Title
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    I understand it on the level of 'plug it in and it works', but not intuitively.

    • one year ago
  3. phi Group Title
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    if you allow complex coefficients (i.e. B is pure imaginary) it works, doesn't it?

    • one year ago
  4. henpen Group Title
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    So B=iA always?

    • one year ago
  5. henpen Group Title
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    While by the \[Ae^{ikt} \] logic that would make sense, given that the DE is linear I don't see why that has to be so. Is it because the A in \[Ae^{ikt} \] can be a complex number itself? I think that may be it.

    • one year ago
  6. phi Group Title
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    for this specific case. I think it can get more complicated if you have a phase shift

    • one year ago
  7. phi Group Title
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    Ae^(ikt) is complex except for specific values of t (e.g. t=0)

    • one year ago
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