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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
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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klimenkov Group TitleBest ResponseYou've already chosen the best response.0
[Math Processing Error]
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.0
I understand it on the level of 'plug it in and it works', but not intuitively.
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
if you allow complex coefficients (i.e. B is pure imaginary) it works, doesn't it?
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.0
So B=iA always?
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.0
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

phi Group TitleBest ResponseYou've already chosen the best response.1
for this specific case. I think it can get more complicated if you have a phase shift
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
Ae^(ikt) is complex except for specific values of t (e.g. t=0)
 one year ago
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