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anonymous
 3 years ago
\[ \ddot{\phi}=k^2 \phi \]
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 \[Acos(kt) \]?
anonymous
 3 years ago
\[ \ddot{\phi}=k^2 \phi \] 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 \[Acos(kt) \]?

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klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.0[Math Processing Error]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I understand it on the level of 'plug it in and it works', but not intuitively.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0While 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.

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

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