What is the Inverse Laplace Transform of (s+1)/ (s^2 + 4s + 5) + e^-2s / 3s^4?

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Homework Help Overview

The discussion revolves around finding the inverse Laplace transform of the expression (s+1)/(s^2 + 4s + 5) + e^-2s / 3s^4, which involves concepts from Laplace transforms and their properties.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to complete the square for the denominator of the first term and expresses uncertainty about finding the inverse Laplace transform from that point. They also question the approach for the second term involving e^-2s / 3s^4.
  • Some participants suggest referring to a table of inverse Laplace transforms, noting specific forms that may be relevant to the original poster's expression.
  • There are questions regarding the understanding of LaTeX code used in the discussion, indicating a need for clarification on notation.

Discussion Status

The discussion is ongoing, with participants exploring different methods to approach the problem. Some guidance has been offered regarding the use of tables for inverse Laplace transforms, but there is no explicit consensus on the best method to proceed.

Contextual Notes

Participants express confusion over specific notation and the use of LaTeX, which may affect their understanding of the problem. There is an indication that the original poster may be constrained by their current resources or familiarity with the topic.

math_04
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Homework Statement



Find the inverse laplace transform of (s+1)/ (s^2 + 4s + 5) + e^-2s / 3s^4

Homework Equations





The Attempt at a Solution



For the first one

(s+1) / (s^2 + 4s + 5), I completed the square for the denominator so

(s+1) / [(s+1)^2 + 1]

Now it gets confusing, how do I find out the inverse laplace of this.

And for e^-2s / 3s^4 , is the inverse laplace 1/9 x 1/ (s+2)^4 = 1/9(s+2)^4
 
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The same way you find the inverse Laplace transform of just about anything: look it up in a table.
If you don't have one in your book, here's one:
http://www.vibrationdata.com/Laplace.htm

You will notice that they give an inverse Laplace transform for [itex]s/(s^2+ \alpha[/itex] as well as the inverse Laplace transform for [itex]F(x-\alpha)[/itex]. Use them together for [itex](s+1)/((s+1)^2+ 2)[/itex].
 
The problem is I don't get the LaTeX Code: s/(s^2+ \\alpha thing. I only get the f(x +a) thing.
 
math_04 said:
The problem is I don't get the LaTeX Code: s/(s^2+ \\alpha thing. I only get the f(x +a) thing.

In the last line of Ivy you should recognize a form that can be found in the table. Ivy rewrote the equation in that form by taking out a square.
 

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