Inverse Laplace Transform of (2s+1)/(s^2+16)

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SUMMARY

The inverse Laplace transform of the function F(s) = (2s + 1) / (s^2 + 16) can be determined by separating it into two fractions: (2s)/(s^2 + 16) and 1/(s^2 + 16). The inverse Laplace transform of (2s)/(s^2 + 16) is 2cos(4t), while the inverse Laplace transform of 1/(s^2 + 16) is (1/4)sin(4t). Therefore, the complete inverse Laplace transform is L^{-1}{F(s)} = 2cos(4t) + (1/4)sin(4t).

PREREQUISITES
  • Understanding of Laplace transforms
  • Knowledge of trigonometric functions
  • Familiarity with partial fraction decomposition
  • Basic calculus skills
NEXT STEPS
  • Study the properties of Laplace transforms
  • Learn about partial fraction decomposition techniques
  • Explore the inverse Laplace transform of common functions
  • Practice solving differential equations using Laplace transforms
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Students studying engineering mathematics, particularly those focusing on differential equations and Laplace transforms, as well as educators teaching these concepts.

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Homework Statement



Determine the inverse Laplace transform of the given function

F(s) = (2s+1) / (s^2 + 16 )

Homework Equations



the LaPlace transform of different functions

The Attempt at a Solution



I divide the above equation into 2 fractions one with the 2s in the numerator and the other fraction with the 1 in the numerator. I know what the inverse Laplace transform of the 2s is but not what the (1) / (s^2+16).
 
Physics news on Phys.org
The inverse Laplace transform of 1/(s^2 + a^2) is (1/a)sin(at).
 

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