A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing

This Question is Closed

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1do you know the identity \(\huge L^{1}[e^{as}F(s)]=f(ta)u(ta)\) ?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1so, you first need laplace inverse of 1/(2s^2+s+2) right ?

gorica
 one year ago
Best ResponseYou've already chosen the best response.0yes, I think I have it.

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1ohh...so you were able to find laplace inverse of 1/(2s^2+s+2) ??

gorica
 one year ago
Best ResponseYou've already chosen the best response.0yes. It's this \[\frac{ 2 }{ \sqrt{15} }e ^{t/4}\sin(\frac{ \sqrt{15} }{ 4 }t)\]

gorica
 one year ago
Best ResponseYou've already chosen the best response.0and this is where I don't know what to do next

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1yes, thats correct, now just replace, t by 't5' and add u(t5) in the end. using that formula....

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1\(\huge \frac{ 2 }{ \sqrt{15} }e ^{(t+5)/4}\sin(\frac{ \sqrt{15} }{ 4 }(t5))u(t5)\)
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
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.