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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![**Understanding Laplace Transforms in Differential Equations**
In the study of differential equations, the Laplace transform is often employed to transform differential equations into algebraic equations, which are simpler to solve. Here we consider the Laplace transforms of derivatives of a function \( x(t) \).
### Laplace Transform of Derivatives
**Given:**
1. **Second Derivative:**
\[
\mathcal{L}\{x''\} = s^2 \cdot x(s) - x(0) \cdot s - x'(0)
\]
This formula represents the Laplace transform of the second derivative of \( x(t) \). Here, \( x'' \) denotes the second derivative of \( x \) with respect to \( t \). \( s \) is a complex frequency parameter, \( x(s) \) is the Laplace transform of the function \( x(t) \), \( x(0) \) is the initial value of \( x \) at \( t = 0 \), and \( x'(0) \) is the initial value of the first derivative of \( x \).
2. **First Derivative:**
\[
\mathcal{L}\{x'\} = s \cdot x(s) - x(0)
\]
This represents the Laplace transform of the first derivative of \( x(t) \). Similar to the second derivative formula, \( s \) is the complex frequency parameter, \( x(s) \) is the Laplace transform of \( x(t) \), and \( x(0) \) is the initial value of \( x \).
3. **Original Function:**
\[
\mathcal{L}\{x\} = x(s)
\]
This simple expression states that the Laplace transform of the function \( x(t) \) is \( x(s) \).
By utilizing these transformations, solving differential equations can be converted into solving algebraic equations, which are generally easier to manage. The initial conditions \( x(0) \) and \( x'(0) \) play a crucial role in these transformed equations.
For more detailed explanations and examples, visit our Differential Equations section on our educational website.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F336090ec-d2ca-4d15-a2ae-85a71ff5580d%2F2539e405-646f-4ef7-89d6-c94e94980ffa%2Fa9wdljk_processed.png&w=3840&q=75)
Transcribed Image Text:**Understanding Laplace Transforms in Differential Equations**
In the study of differential equations, the Laplace transform is often employed to transform differential equations into algebraic equations, which are simpler to solve. Here we consider the Laplace transforms of derivatives of a function \( x(t) \).
### Laplace Transform of Derivatives
**Given:**
1. **Second Derivative:**
\[
\mathcal{L}\{x''\} = s^2 \cdot x(s) - x(0) \cdot s - x'(0)
\]
This formula represents the Laplace transform of the second derivative of \( x(t) \). Here, \( x'' \) denotes the second derivative of \( x \) with respect to \( t \). \( s \) is a complex frequency parameter, \( x(s) \) is the Laplace transform of the function \( x(t) \), \( x(0) \) is the initial value of \( x \) at \( t = 0 \), and \( x'(0) \) is the initial value of the first derivative of \( x \).
2. **First Derivative:**
\[
\mathcal{L}\{x'\} = s \cdot x(s) - x(0)
\]
This represents the Laplace transform of the first derivative of \( x(t) \). Similar to the second derivative formula, \( s \) is the complex frequency parameter, \( x(s) \) is the Laplace transform of \( x(t) \), and \( x(0) \) is the initial value of \( x \).
3. **Original Function:**
\[
\mathcal{L}\{x\} = x(s)
\]
This simple expression states that the Laplace transform of the function \( x(t) \) is \( x(s) \).
By utilizing these transformations, solving differential equations can be converted into solving algebraic equations, which are generally easier to manage. The initial conditions \( x(0) \) and \( x'(0) \) play a crucial role in these transformed equations.
For more detailed explanations and examples, visit our Differential Equations section on our educational website.
![**Title: Solving Initial Value Problems (IVP) using Laplace Transforms**
**Objective:**
Learn how to solve initial value problems involving differential equations using Laplace Transforms.
**Problem Statement:**
**Solve the given IVP using Laplace Transforms.**
*Given Differential Equation:*
\[ x'' + 9x = \cos(3t) \]
*Initial Conditions:*
\[ x(0) = 0, \quad x'(0) = 2 \]
**Task:**
*Solve for \( x(s) \) and STOP! Express \( x(s) \) as a single fraction.*
**Step-by-Step Solution:**
1. **Apply the Laplace Transform to both sides of the differential equation.**
- Recall that:
\[ \mathcal{L} \{ x''(t) \} = s^2 X(s) - sx(0) - x'(0) \]
\[ \mathcal{L} \{ x'(t) \} = s X(s) - x(0) \]
\[ \mathcal{L} \{ x(t) \} = X(s) \]
\[ \mathcal{L} \{ \cos(3t) \} = \frac{s}{s^2 + 9} \]
2. **Substitute the Laplace Transforms into the differential equation:**
Given:
\[ x'' + 9x = \cos(3t) \]
After applying Laplace Transforms, we get:
\[ s^2 X(s) - sx(0) - x'(0) + 9X(s) = \frac{s}{s^2 + 9} \]
3. **Substitute the initial conditions \( x(0) = 0 \) and \( x'(0) = 2 \):**
\[ s^2 X(s) - 0 - 2 + 9X(s) = \frac{s}{s^2 + 9} \]
Simplifies to:
\[ (s^2 + 9)X(s) - 2 = \frac{s}{s^2 + 9} \]
4. **Solve for \( X(s) \):**
Move \(-2\) to the other side:
\[ (s](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F336090ec-d2ca-4d15-a2ae-85a71ff5580d%2F2539e405-646f-4ef7-89d6-c94e94980ffa%2Friz7pq_processed.png&w=3840&q=75)
Transcribed Image Text:**Title: Solving Initial Value Problems (IVP) using Laplace Transforms**
**Objective:**
Learn how to solve initial value problems involving differential equations using Laplace Transforms.
**Problem Statement:**
**Solve the given IVP using Laplace Transforms.**
*Given Differential Equation:*
\[ x'' + 9x = \cos(3t) \]
*Initial Conditions:*
\[ x(0) = 0, \quad x'(0) = 2 \]
**Task:**
*Solve for \( x(s) \) and STOP! Express \( x(s) \) as a single fraction.*
**Step-by-Step Solution:**
1. **Apply the Laplace Transform to both sides of the differential equation.**
- Recall that:
\[ \mathcal{L} \{ x''(t) \} = s^2 X(s) - sx(0) - x'(0) \]
\[ \mathcal{L} \{ x'(t) \} = s X(s) - x(0) \]
\[ \mathcal{L} \{ x(t) \} = X(s) \]
\[ \mathcal{L} \{ \cos(3t) \} = \frac{s}{s^2 + 9} \]
2. **Substitute the Laplace Transforms into the differential equation:**
Given:
\[ x'' + 9x = \cos(3t) \]
After applying Laplace Transforms, we get:
\[ s^2 X(s) - sx(0) - x'(0) + 9X(s) = \frac{s}{s^2 + 9} \]
3. **Substitute the initial conditions \( x(0) = 0 \) and \( x'(0) = 2 \):**
\[ s^2 X(s) - 0 - 2 + 9X(s) = \frac{s}{s^2 + 9} \]
Simplifies to:
\[ (s^2 + 9)X(s) - 2 = \frac{s}{s^2 + 9} \]
4. **Solve for \( X(s) \):**
Move \(-2\) to the other side:
\[ (s
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