Question

Using a partial fraction expansion, find \(\mathcal{L}^{-1}\\{F(s)\\} .\) In Exercise 40 , compare your answer with (6) in Table 5.1. \(F(s)=\frac{12}{(s-3)(s+1)}\)

Short answer

Question: Find the inverse Laplace transform of the function \(F(s) = \frac{12}{(s-3)(s+1)}\). Answer: The inverse Laplace transform of the function \(F(s) = \frac{12}{(s-3)(s+1)}\) is \(3e^{3t} - 3e^{-t}\).

Step-by-step solution

Express F(s) as a Partial Fraction Expansion

First, we need to express the given function as a partial fraction expansion. We write the given function \(F(s)=\frac{12}{(s-3)(s+1)}\) in the form: \[F(s) = \frac{A}{s-3} + \frac{B}{s+1}\] Now, we will find the values of A and B.

Find the Numerator

Multiply both sides of the equation by the denominator \((s-3)(s+1)\) to eliminate the denominator: \[12 = A(s+1) + B(s-3)\]

Solve for A and B

To solve for A, let s = 3. The equation becomes: \[12 = A(4)\] So, \(A = 3\). To solve for B, let s = -1. The equation becomes: \[12 = B(-4)\] So, \(B = -3\). Now, we have found the values of A and B: \[F(s) = \frac{3}{s-3} - \frac{3}{s+1}\]

Apply Inverse Laplace Transform

Now, we will apply the inverse Laplace transform to each term of F(s): \[\mathcal{L}^{-1}\\{F(s)\\} = \mathcal{L}^{-1}\\{\frac{3}{s-3} - \frac{3}{s+1}\\}\] Using the linearity property of inverse Laplace transforms, we can write: \[\mathcal{L}^{-1}\\{F(s)\\} = 3\cdot \mathcal{L}^{-1}\\{\frac{1}{s-3}\\} - 3\cdot \mathcal{L}^{-1}\\{\frac{1}{s+1}\\}\]

Use Table 5.1 to Find Inverse Laplace Transforms

Using Table 5.1 (entry 6), we can find the inverse Laplace transforms for both terms: \[\mathcal{L}^{-1}\\{\frac{1}{s-3}\\} = e^{3t}\] \[\mathcal{L}^{-1}\\{\frac{1}{s+1}\\} = e^{-t}\] Now, substitute these results back into the equation: \[\mathcal{L}^{-1}\\{F(s)\\} = 3e^{3t} - 3e^{-t}\] So, the inverse Laplace transform of the given function is: \[\mathcal{L}^{-1}\\{F(s)\\} = 3e^{3t} - 3e^{-t}\]

Key concepts

Partial Fraction Expansion

Partial Fraction Expansion is a method used to break down complex rational expressions into simpler fractions, which are easier to work with. When facing a function like \( F(s) = \frac{12}{(s-3)(s+1)} \), it's challenging to directly apply a Laplace transform. Instead, we express it as a sum of simpler fractions using partial fraction expansion. This approach allows us to solve for constants \( A \) and \( B \), such that:

• \( F(s) = \frac{A}{s-3} + \frac{B}{s+1} \)

By solving equations derived from multiplying out the denominators, we can find the values of \( A \) and \( B \). In our example, these values are 3 and -3, respectively, resulting in:

• \( F(s) = \frac{3}{s-3} - \frac{3}{s+1} \)

Expanding in this way simplifies the function and makes subsequent inverse transformations feasible.

Linearity Property

The Linearity Property of Laplace transforms states that the transform of a sum is the sum of the transforms. Similarly, this property holds for inverse Laplace transforms. Essentially, it allows us to handle each part of the expanded function separately. For instance, if we have:

• \( F(s) = \frac{3}{s-3} - \frac{3}{s+1} \)

Using the linearity property, the inverse Laplace transform can be calculated as follows:

• \( \mathcal{L}^{-1}\{ F(s) \} = 3 \cdot \mathcal{L}^{-1}\{ \frac{1}{s-3} \} - 3 \cdot \mathcal{L}^{-1}\{ \frac{1}{s+1} \} \)

This property simplifies computations significantly by allowing for the expression of complex inverse transforms in terms of simpler ones.

Laplace Transform Table

A Laplace Transform Table is an essential tool in solving inverse transformations efficiently. It lists common Laplace transforms and their corresponding inverse pairs. These tables save computation time when applying inverse transforms to functions that match listed forms. In our problem, the table directly provides:

• \( \mathcal{L}^{-1}\{ \frac{1}{s-3} \} = e^{3t} \)
• \( \mathcal{L}^{-1}\{ \frac{1}{s+1} \} = e^{-t} \)

Using such a table allows for easy substitution once the partial fraction expansion is completed, dramatically reducing complexity. Without the table, deriving these transforms manually would require more advanced calculus.

Exponential Function

Exponential Functions often appear as results of inverse Laplace transformations due to their properties that align with the behaviors of growth and decay processes in systems. In the context of our example:

• The function \( \mathcal{L}^{-1}\{ \frac{1}{s-3} \} \) results in \( e^{3t} \), indicating a growth function since the rate \( 3 \) is positive.
• Conversely, \( \mathcal{L}^{-1}\{ \frac{1}{s+1} \} \) results in \( e^{-t} \), which is a decay function due to the negative rate \( -1 \).

These exponential functions help model real-world phenomena such as population growth, radioactive decay, and charging and discharging capacitors in electric circuits. Understanding exponential functions is crucial for interpreting the inverse Laplace transformations in applied sciences.