Understanding the Inverse Laplace Transform of 1/s^3 - 8

When working with the inverse Laplace transform of a function such as F(s) 1 / (s^3 - 8), it's crucial to break down the problem using algebraic and calculus techniques. In this article, we will explore how to solve the inverse Laplace transform of this particular function in a step-by-step manner, explaining each part of the process to ensure clarity and understanding.

Factoring the Denominator

The first essential step is to factor the denominator of the given function. The expression s^3 - 8 can be factored using the difference of cubes formula: s^3 - 8 (s - 2)(s^2 2s 4). This simplifies our original function to:

F(s) 1 / ((s - 2)(s^2 2s 4))

Partial Fraction Decomposition

Next, we decompose the function into simpler fractions using partial fraction decomposition. We express F(s) as follows:

F(s) A / (s - 2) (Bs C) / (s^2 2s 4)

where A, B, C are constants that we need to determine. Multiplying both sides by the denominator (s - 2)(s^2 2s 4) gives:

1 A(s^2 2s 4) (Bs C)(s - 2)

Expanding the right-hand side and combining like terms:

1 As^2 2As 4A Bs^2 - 2Bs Cs - 2C

1 (A B)s^2 (2A - 2B C)s (4A - 2C)

Solving the System of Equations

Setting up a system of equations by equating coefficients, we have:

A B 0

2A - 2B C 0

4A - 2C 1

From the first equation, we find:

B -A

Substituting B -A into the second equation:

2A - 2(-A) C 0 implies 4A C 0 implies C -4A

Substituting C -4A into the third equation:

4A - 2(-4A) 1 implies 12A 1 implies A 1/12

Thus, we get:

B -A -1/12 and C -4A -4/12 -1/3

Reconstructing the Function

Substituting these values, we can write:

F(s) 1/12(s - 2) (-1/12s - 1/3) / (s^2 2s 4)

Inverse Laplace Transform

To find the inverse Laplace transform, we handle each term separately:

Term 1: 1/12(s - 2)

The inverse Laplace transform of 1/12(s - 2) is:

L^{-1}{1/12(s - 2)} 1/12 e^{2t}

Term 2: (-1/12s - 1/3) / (s^2 2s 4)

First, we rewrite s^2 2s 4 as ((s 1)^2 3). The inverse Laplace transform of:

L{1 / ((s 1)^2 3)} 1/sqrt{3} e^{-t} sin(sqrt{3}t)

and for:

L{(s 1) / ((s 1)^2 3)} e^{-t} cos(sqrt{3}t)

Thus:

L^{-1}{(-1/12s - 1/3) / (s^2 2s 4)} -1/12 e^{-t} cos(sqrt{3}t) - 1/3sqrt{3} e^{-t} sin(sqrt{3}t)

Final Inverse Laplace Transform

Combining these results, the complete inverse Laplace transform is:

L^{-1}{1 / (s^3 - 8)} 1/12 e^{2t} - 1/12 e^{-t} cos(sqrt{3}t) - 1/3sqrt{3} e^{-t} sin(sqrt{3}t)