anonymous
  • anonymous
\[ \ddot{\phi}=-k^2 \phi \] 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 \[Acos(kt) \]?
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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klimenkov
  • klimenkov
[Math Processing Error]
anonymous
  • anonymous
I understand it on the level of 'plug it in and it works', but not intuitively.
phi
  • phi
if you allow complex coefficients (i.e. B is pure imaginary) it works, doesn't it?

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anonymous
  • anonymous
So B=iA always?
anonymous
  • anonymous
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.
phi
  • phi
for this specific case. I think it can get more complicated if you have a phase shift
phi
  • phi
Ae^(ikt) is complex except for specific values of t (e.g. t=0)

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