please give clear solution with all the necessary work! Inverse Laplace transforms can be found using the linearity property of transforms and a reference table. Find the Inverse Laplace transform of \[ F(s) = \frac{-4s - 2}{s^2 - 2s + 26}. \]Show all the work needed to rewrite the functions. Express the inverse Laplace transform as a function of $ t $. [Hint: you may need to first complete the square in the denominator!]\[ f(t) = \underline{\hspace{3cm}} \]

The linearity property of Inverse Laplace Transform:Where and is the constant.And the Inverse…

Answered: Inverse Laplace transforms can be found…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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please give clear solution with all the necessary work!

Inverse Laplace transforms can be found using the linearity property of transforms and a reference table. Find the Inverse Laplace transform of

\[ F(s) = \frac{-4s - 2}{s^2 - 2s + 26}. \]

Show all the work needed to rewrite the functions. Express the inverse Laplace transform as a function of $ t $. [Hint: you may need to first complete the square in the denominator!]

\[ f(t) = \underline{\hspace{3cm}} \]

Expert Solution

Step 1: Define the linearity property of Inverse Laplace Transform:

The linearity property of Inverse Laplace Transform:

$L to the power of negative 1 end exponent open curly brackets a times f open parentheses s close parentheses plus b times g open parentheses s close parentheses close curly brackets equals a times L to the power of negative 1 end exponent open curly brackets f open parentheses s close parentheses close curly brackets plus b times L to the power of negative 1 end exponent open curly brackets g open parentheses s close parentheses close curly brackets$

Where $a$ and $b$ is the constant.

And the Inverse Laplace transform identities:

1). $L to the power of negative 1 end exponent open curly brackets fraction numerator s minus a over denominator open parentheses s minus a close parentheses squared plus b squared end fraction close curly brackets equals e to the power of a t end exponent cos open parentheses b t close parentheses$

2). $L to the power of negative 1 end exponent open curly brackets fraction numerator 1 over denominator open parentheses s minus a close parentheses squared plus b squared end fraction close curly brackets equals 1 over b e to the power of a t end exponent sin open parentheses b t close parentheses$