$$ \ddot{\phi}=-k^2 \phi $$
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 $$Acos(kt) $$?

Question

klimenkov · Answer

[Math Processing Error]

anonymous · Answer

I understand it on the level of 'plug it in and it works', but not intuitively.

phi · Answer

if you allow complex coefficients  (i.e. B is pure imaginary)  it works, doesn't it?

anonymous · Answer

So B=iA always?

anonymous · Answer

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 · Answer

for this specific case.  I think it can get more complicated if you have a phase shift

phi · Answer

Ae^(ikt) is complex except for specific values of t (e.g. t=0)