Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
masroorm
Group Title
Does anyone know how to find the Inverse Laplace Transform of 6/(s^3+3s)
 one year ago
 one year ago
masroorm Group Title
Does anyone know how to find the Inverse Laplace Transform of 6/(s^3+3s)
 one year ago
 one year ago

This Question is Open

athe Group TitleBest ResponseYou've already chosen the best response.1
It is combination of simple Laplace transformation:\[\frac{6}{s^3+3s}=\frac{6}{s(s^2+3)}\] Use Method of undetermined coefficients: \[\frac{6}{s(s^2+3)} \rightarrow \frac{2}{s}\frac{2s}{s^2+3}\] Use well know transformation for this functions: \[\frac{2}{s} \rightarrow 2\cdot H(t)\], H(t)  Heaviside step function \[\frac{2s}{s^2+3}\rightarrow 2\cdot \cos(\sqrt{3}t)\] \[\frac{6}{s^3+3s} \rightarrow 2\cdot H(t) +2\cdot \cos(\sqrt{3}t)\]
 one year ago

honey26 Group TitleBest ResponseYou've already chosen the best response.0
we know that L[sin 3^(1/2) t]= (3^(1/2))/(s^2+3) and L[integral f(k) dk (in limits 0 to t)]=L[f(t)]/s. Here we can consider f(t)=sin 3^(1/2) t,then (3^(1/2))/s(s^2 +3) =integral f(k) dk in limits 0 to t .the integral value is 1(cos3^(1/2) t)/(3^(1/2)). therefore, L[6/((s^3)+3s)] = 2(1(cos3^(1/2) t)/(3^(1/2)).
 one year 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.