Question

Use Table \(9.1\) to find \(\mathcal{I}^{-1}\\{F(s)\\}\) for each of the functions \(F\) defined \(F(s)=\frac{2}{s^{2}+9}\)

Short answer

The inverse Laplace transform of \(F(s) = \frac{2}{s^2 + 9}\) is \(\mathcal{I}^{-1}\{F(s)\} = \frac{2}{3} \sin(3t)\).

Step-by-step solution

Identify the form of the given function

Compare the given function \(F(s) = \frac{2}{s^2 + 9}\) with the standard form of the Laplace transform \(\frac{1}{s^2 + \omega^2}\) in Table 9.1. We can see that it is indeed similar to that standard form, and we now identify the parameter \(\omega\): \[ \frac{2}{s^2 + 9} = \frac{2}{s^2 + (3)^2}. \] So, \(\omega = 3\).

Find the corresponding inverse Laplace transform function

According to Table 9.1, the inverse Laplace transform of the standard function \(\frac{1}{s^2 + \omega^2}\) is \(f(t) = \frac{1}{\omega} \sin(\omega t)\). Now, we can find the inverse Laplace transform of our given function by using the identified parameter \(\omega\) and the fact that it is multiplied by \(2\): \[ \mathcal{I}^{-1}\left\{\frac{2}{s^2 + 9}\right\} = \mathcal{I}^{-1}\left\{\frac{2}{s^2 + (3)^2}\right\} = 2 \mathcal{I}^{-1}\left\{\frac{1}{s^2 + (3)^2}\right\}. \]

Complete the inverse Laplace transform

Use the corresponding inverse Laplace transform function determined in Step 2 and substitute \(\omega = 3\): \[ 2 \mathcal{I}^{-1}\left\{\frac{1}{s^2 + (3)^2}\right\} = 2 \left(\frac{1}{3} \sin(3t)\right) = \frac{2}{3} \sin(3t). \] So, the inverse Laplace transform of the given function \(F(s) = \frac{2}{s^2 + 9}\) is: \[ \mathcal{I}^{-1}\{F(s)\} = \frac{2}{3} \sin(3t). \]

Key concepts

Inverse Laplace Transform

The inverse Laplace transform is a method used in mathematics to convert functions from the Laplace domain back to the time domain. Imagine you have a function expressed in terms of the complex variable "s", and you wish to find its equivalent in terms of time "t".

• This technique is crucial in solving differential equations, especially those involving initial conditions.
• It is a key tool in engineering and physics for system analysis and control theory.

The fundamental idea is to reverse the process of Laplace transformation. Given a function in the form of \( F(s)=\frac{2}{s^2 + 9} \), you can use transformation tables (like Table 9.1 used here) to find its inverse.
In this specific example, comparing with the standard form \( \frac{1}{s^2 + \omega^2} \), we determined that the parameter \( \omega = 3 \). Then, using the known inverse formula \( f(t) = \frac{1}{\omega} \sin(\omega t) \), we were able to convert \( F(s) \) back to a time function, \( \frac{2}{3} \sin(3t) \).

Sinusoidal Functions

Sinusoidal functions play a pivotal role in both physics and mathematics. These functions describe wave-like phenomena and are often expressed in terms of sine or cosine.

• They are periodic, meaning they repeat at regular intervals, which makes them ideal for modeling cycles like sound waves, alternating current, and even circadian rhythms.
• The amplitude determines the peak value of the wave, while the frequency or angular frequency indicates the rate at which the cycle repeats.

In the context of Laplace transforms, sinusoidal functions emerge when you convert certain rational algebraic expressions in the "s" domain back to the time domain.
For instance, in our exercise, after applying the inverse Laplace transform, we obtained \( \frac{2}{3} \sin(3t) \). This represents a sinusoidal function with an amplitude of \( \frac{2}{3} \) and an angular frequency of \( 3 \).

Differential Equations

Differential equations are equations that involve derivatives of a function. They describe various physical systems and phenomena, such as motion, heat, waves, and more.

• These equations can be either ordinary (ODEs) when they contain one independent variable, or partial (PDEs) when they involve multiple variables.
• Solving these equations provides insights into the behavior of dynamic systems.

Laplace transformations are particularly advantageous when dealing with differential equations. They convert differentiation operations into multiplication in the Laplace domain, making it easier to handle initial conditions and solve the equations.
In our initial exercise, finding the inverse Laplace transform can efficiently solve differential equations by providing a way to revert from the frequency domain back to the time-dependent solution. This process illustrates how integral transforms like Laplace aid in transforming complicated calculus problems into simpler algebraic ones.