Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
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 \[Acos(kt) \]?
 2 years ago
 2 years 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 \[Acos(kt) \]?
 2 years ago
 2 years ago

This Question is Closed

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
[Math Processing Error]
 2 years 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.
 2 years 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?
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.0
So B=iA always?
 2 years 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.
 2 years 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
 2 years 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)
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.