## 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 $Acos(kt)$?

1. klimenkov

[Math Processing Error]

2. anonymous

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

3. phi

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

4. anonymous

So B=iA always?

5. 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.

6. phi

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

7. phi

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