Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

masroorm

  • 2 years ago

Does anyone know how to find the Inverse Laplace Transform of 6/(s^3+3s)

  • This Question is Open
  1. athe
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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 \[\frac{2s}{s^2+3}\rightarrow 2\cdot \cos(\sqrt{3}t)\] \[\frac{6}{s^3+3s} \rightarrow 2\cdot H(t) +2\cdot \cos(\sqrt{3}t)\]

  2. honey26
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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)).

  3. Not the answer you are looking for?
    Search for more explanations.

    Search OpenStudy
    • Attachments:

Ask your own question

Ask a Question
Find 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
  • 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.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.