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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![**Solving a Differential Equation Using Laplace Transform (Educational Example)**
### Problem Statement
Solve the differential equation using the Laplace transform:
\[ y'' - y' = e^t \cos t, \quad y(0) = 0, \quad y'(0) = 0 \]
### Explanation and Solution:
To solve the given differential equation using Laplace transform, we follow these steps:
1. **Take the Laplace Transform of both sides of the equation:**
\[ \mathcal{L}\{ y'' - y'\} = \mathcal{L}\{ e^t \cos t \} \]
2. **Use the properties of the Laplace transform on the left side:**
Recall the transforms:
\[
\mathcal{L}\{ y'' \} = s^2 Y(s) - sy(0) - y'(0)
\]
\[
\mathcal{L}\{ y' \} = s Y(s) - y(0)
\]
Since \( y(0) = 0 \) and \( y'(0) = 0 \), this simplifies to:
\[
\mathcal{L}\{ y'' \} = s^2 Y(s)
\]
\[
\mathcal{L}\{ y' \} = s Y(s)
\]
Therefore, the left side transforms to:
\[
s^2 Y(s) - s Y(s) = (s^2 - s) Y(s)
\]
3. **Laplace Transform of the right-hand side:**
Use the known Laplace Transform:
\[
\mathcal{L}\{ e^t \cos t \} = \frac{s - 1}{(s - 1)^2 + 1}
\]
4. **Equate the transformed equations:**
\[
(s^2 - s) Y(s) = \frac{s - 1}{(s - 1)^2 + 1}
\]
5. **Solve for \(Y(s)\):**
\[
Y(s) = \frac{\frac{s - 1}{(s - 1)^2 + 1}}{s^2 - s}
\]
6. **Simplify and solve the inverse](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fab6a44fb-472d-4115-ae99-90593ad92ba0%2F7b18f925-5909-406e-a086-3d69798e0374%2F0c5ryt23_processed.png&w=3840&q=75)
Transcribed Image Text:**Solving a Differential Equation Using Laplace Transform (Educational Example)**
### Problem Statement
Solve the differential equation using the Laplace transform:
\[ y'' - y' = e^t \cos t, \quad y(0) = 0, \quad y'(0) = 0 \]
### Explanation and Solution:
To solve the given differential equation using Laplace transform, we follow these steps:
1. **Take the Laplace Transform of both sides of the equation:**
\[ \mathcal{L}\{ y'' - y'\} = \mathcal{L}\{ e^t \cos t \} \]
2. **Use the properties of the Laplace transform on the left side:**
Recall the transforms:
\[
\mathcal{L}\{ y'' \} = s^2 Y(s) - sy(0) - y'(0)
\]
\[
\mathcal{L}\{ y' \} = s Y(s) - y(0)
\]
Since \( y(0) = 0 \) and \( y'(0) = 0 \), this simplifies to:
\[
\mathcal{L}\{ y'' \} = s^2 Y(s)
\]
\[
\mathcal{L}\{ y' \} = s Y(s)
\]
Therefore, the left side transforms to:
\[
s^2 Y(s) - s Y(s) = (s^2 - s) Y(s)
\]
3. **Laplace Transform of the right-hand side:**
Use the known Laplace Transform:
\[
\mathcal{L}\{ e^t \cos t \} = \frac{s - 1}{(s - 1)^2 + 1}
\]
4. **Equate the transformed equations:**
\[
(s^2 - s) Y(s) = \frac{s - 1}{(s - 1)^2 + 1}
\]
5. **Solve for \(Y(s)\):**
\[
Y(s) = \frac{\frac{s - 1}{(s - 1)^2 + 1}}{s^2 - s}
\]
6. **Simplify and solve the inverse
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