What is the Inverse Laplace Transform of (9s+5)/(s2-9) with s>3?

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SUMMARY

The inverse Laplace Transform of the function (9s+5)/(s²-9) for s>3 can be determined using the properties of hyperbolic functions. Specifically, the terms "sinh" and "cosh" refer to the hyperbolic sine and cosine functions, respectively. The inverse Laplace Transform can be derived by recognizing that sinh(at) corresponds to a/(s²-a²) and cosh(at) corresponds to s/(s²-a²). Thus, the solution involves applying these definitions to split the original function appropriately.

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  • Understanding of Laplace Transforms
  • Familiarity with hyperbolic functions (sinh and cosh)
  • Ability to manipulate algebraic expressions
  • Knowledge of inverse functions in calculus
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  • Study the properties of hyperbolic functions, specifically sinh and cosh
  • Learn how to apply the inverse Laplace Transform using tables of transforms
  • Explore examples of inverse Laplace Transforms involving rational functions
  • Practice solving differential equations using Laplace Transforms
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Students in engineering or mathematics, particularly those studying differential equations and Laplace Transforms, will benefit from this discussion.

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Homework Statement


Find the inverse Laplace Transform of (9s+5)/(s2-9) given s>3


Homework Equations


I have a table of Laplace Transforms.
it says "sinh at" goes with a/(s2-a2) and "cosh at" goes with s/(s2-a2)


The Attempt at a Solution


I know how to split it up and find the inverse Laplace Transform but I don't know what it means by "cosh". I assume it is some variation of cosine? Someone said it means degrees but I don't see how that would change the problem.
 
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cosh(x) = 1/2 [exp(x) + exp(-x)]


sinh(x) = 1/2 [exp(x) - exp(-x)]
 
thanks, apparently the professor somewhat explained this but I missed it.
 

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