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Chapter 4: Problem 13

In Problems, find either \(F(s)\) or \(f(t)\), as indicated. $$ \mathscr{L}^{-1}\left\\{\frac{1}{s^{2}-6 s+10}\right\\} $$

Short Answer

Expert verified
The inverse Laplace transform is \( f(t) = e^{3t} \cos(t) \).

Step by step solution

01

Recognize the Expression

The given expression is in the form of a Laplace inverse transform: \[ \mathscr{L}^{-1}\left\{\frac{1}{s^{2} - 6s + 10}\right\} \] which suggests finding a time-domain function \(f(t)\).
02

Complete the Square

Start by rewriting the denominator \(s^2 - 6s + 10\) in a more familiar form by completing the square. First, take the coefficient of \(s\), which is \(-6\), divide by \(2\), and square it to get \(9\). \[ (s^2 - 6s + 10) = (s - 3)^2 + 1 \] Now, the expression becomes\[ \frac{1}{(s-3)^2 + 1} \].
03

Identify the Inverse Laplace Transform Formula

The expression \(\frac{1}{(s-a)^2 + b^2}\) can be matched with the inverse Laplace transform formula for the exponential times cosine function: \[ \mathscr{L}^{-1}\left\{\frac{b}{(s-a)^2 + b^2}\right\} = e^{at}\cos(bt) \]Since \(b=1\), we need to adjust the expression by considering: \[ \mathscr{L}^{-1}\left\{\frac{1}{(s-3)^2 + 1^2}\right\} = e^{3t}\frac{1}{1}\cos(t) = e^{3t}\cos(t) \].
04

Write the Final Time-Domain Function

After identifying the corresponding time-domain function from the Laplace inverse, it becomes clear that the inverse Laplace transform is:\[ f(t) = e^{3t} \cos(t) \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Completing the Square
In the realm of mathematics, completing the square is a critical technique frequently employed to simplify quadratic expressions. When dealing with quadratic expressions like \( s^2 - 6s + 10 \), completing the square helps us reshape it into a more easily recognizable form. This technique becomes tremendously useful when solving for inverse Laplace transforms.
To complete the square for \( s^2 - 6s + 10 \), one starts by taking the coefficient of \( s \), which is \(-6\). Then, divide it by 2, resulting in \(-3\), and square the result to obtain \(9\).
This allows us to rewrite the original expression as \((s - 3)^2 + 1\).
  • It's about creating a perfect square trinomial, allowing easier manipulation in later steps.
  • Particularly useful for recognizing patterns or forms that align with known Laplace transform pairs.
Time-domain Function
The time-domain function is essentially the outcome of the inverse Laplace transform. It represents a function expressed in terms of \( t \), or time.
In our exercise, the time-domain function \( f(t) \) is derived through the inverse Laplace transform of \( \frac{1}{(s-3)^2 + 1} \).
Through this process, we unravel how functions behave over time, transitioning from the frequency domain to the time domain.
  • This transition helps us better analyze systems, especially in control systems and signal processing.
  • Providing insights into how a system evolves with time.
Exponential Function
Exponential functions are prevalent in natural phenomena and numerous mathematical applications. They are characterized by the formula \( e^{at} \), where \( a \) is a constant dictating the rate of exponential growth or decay.
In this exercise, the time-domain function \( f(t) = e^{3t} \cos(t) \), incorporates an exponential function demonstrating growth. Here, the exponential term \( e^{3t} \) suggests that the function increases exponentially as time progresses.
  • Exponential functions are crucial in modeling growth processes, radioactive decay, and interest calculations.
  • The combination with trigonometric functions, as seen here, often occurs in oscillating systems subjected to exponential damping or forcing.
Laplace Transform Formula
The Laplace transform formula is a cornerstone in transforming functions from the time domain to the frequency domain and vice versa. It enables simplification of complex differential equations. In this exercise, the expression \( \frac{1}{(s-a)^2 + b^2} \) is tied to the inverse Laplace formula for \( e^{at}\cos(bt) \).
By recognizing this, we leverage standard Laplace transform pairs to deduce the time-domain function. This transformation is critical in engineering and physics for its role in solving linear time-invariant systems.
  • The formula helps decode the relationships between different mathematical representations of the same physical realities.
  • Key for tackling and analyzing problems in electrical circuits, mechanical systems, and beyond.

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Most popular questions from this chapter

Use the Laplace transform to solve the given integral equation or integrodifferential equation. $$ y^{\prime}(t)=1-\sin t-\int_{0}^{t} y(\tau) d \tau, \quad y(0)=0 $$

Use the Laplace transform to solve the given system of differential equations. $$ \begin{aligned} &\frac{d^{2} x}{d t^{2}}+3 \frac{d y}{d t}+3 y=0 \\ &\frac{d^{2} x}{d t^{2}}+3 y=t e^{-t} \\ &x(0)=0, x^{\prime}(0)=2, \\ &y(0)=0 \end{aligned} $$

In Problems, find either \(F(s)\) or \(f(t)\), as indicated. $$ \mathscr{L}\left\\{e^{2-t} q(t-2)\right\\} $$

Use the Laplace transform to solve the given integral equation or integrodifferential equation. $$ \frac{d y}{d t}+6 y(t)+9 \int_{0}^{t} y(\tau) d \tau=1, \quad y(0)=0 $$

Use the Laplace transform to solve the given initial-value problem. Use the table of Laplace transforms in Appendix III as needed. $$ \begin{gathered} y^{\prime \prime}+y=f(t), \quad y(0)=1, y^{\prime}(0)=0, \text { where } \\ f(t)=\left\\{\begin{array}{lr} 1, & 0 \leq t<\pi / 2 \\ \sin t, & t \geq \pi / 2 \end{array}\right. \end{gathered} $$
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