Solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem below. y" -8y' +12y = 4te 2¹, y(0) = 6, y'(0) = -6
Solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem below. y" -8y' +12y = 4te 2¹, y(0) = 6, y'(0) = -6
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Solving for the Laplace Transform of an Initial Value Problem
**Problem:**
Solve for \( Y(s) \), the Laplace transform of the solution \( y(t) \) to the initial value problem below.
\[ y'' - 8y' + 12y = 4t e^{2t}, \quad y(0) = 6, \quad y'(0) = -6 \]
**Solution Steps:**
To solve this differential equation using the Laplace transform, follow these steps:
1. **Take the Laplace transform of both sides of the differential equation.**
2. **Apply the initial conditions \( y(0) \) and \( y'(0) \).**
3. **Solve for \( Y(s) \), the Laplace transform of \( y(t) \).**
The provided links will be useful:
- [Table of Laplace transforms](#): This table will provide the Laplace transforms of commonly encountered functions.
- [Table of properties of Laplace transforms](#): This table will provide the properties and rules for the Laplace transform, such as linearity, differentiation, and convolution.
**Formulating the Laplace Transform Equation:**
1. **Differentiation Property of Laplace Transform:**
- \( \mathcal{L}\{y'(t)\} = sY(s) - y(0) \)
- \( \mathcal{L}\{y''(t)\} = s^2Y(s) - sy(0) - y'(0) \)
2. **Apply these properties to the equation \( y'' - 8y' + 12y = 4t e^{2t} \):**
- \( \mathcal{L}\{y''\} - 8 \mathcal{L}\{y'\} + 12 \mathcal{L}\{y\} = \mathcal{L}\{4t e^{2t}\} \)
3. **Insert the initial conditions:**
- \( \mathcal{L}\{y''\} = s^2 Y(s) - sy(0) - y'(0) = s^2 Y(s) - 6s + 6 \)
- \( \mathcal{L}\{y'\} = s Y(s) - y(0) = s Y(s) - 6](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbbba0909-7b4f-4e80-8e24-06dfda2d061e%2Fe9a83468-08a7-4f48-8cf1-e6853c938706%2Fpwqtnx_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Solving for the Laplace Transform of an Initial Value Problem
**Problem:**
Solve for \( Y(s) \), the Laplace transform of the solution \( y(t) \) to the initial value problem below.
\[ y'' - 8y' + 12y = 4t e^{2t}, \quad y(0) = 6, \quad y'(0) = -6 \]
**Solution Steps:**
To solve this differential equation using the Laplace transform, follow these steps:
1. **Take the Laplace transform of both sides of the differential equation.**
2. **Apply the initial conditions \( y(0) \) and \( y'(0) \).**
3. **Solve for \( Y(s) \), the Laplace transform of \( y(t) \).**
The provided links will be useful:
- [Table of Laplace transforms](#): This table will provide the Laplace transforms of commonly encountered functions.
- [Table of properties of Laplace transforms](#): This table will provide the properties and rules for the Laplace transform, such as linearity, differentiation, and convolution.
**Formulating the Laplace Transform Equation:**
1. **Differentiation Property of Laplace Transform:**
- \( \mathcal{L}\{y'(t)\} = sY(s) - y(0) \)
- \( \mathcal{L}\{y''(t)\} = s^2Y(s) - sy(0) - y'(0) \)
2. **Apply these properties to the equation \( y'' - 8y' + 12y = 4t e^{2t} \):**
- \( \mathcal{L}\{y''\} - 8 \mathcal{L}\{y'\} + 12 \mathcal{L}\{y\} = \mathcal{L}\{4t e^{2t}\} \)
3. **Insert the initial conditions:**
- \( \mathcal{L}\{y''\} = s^2 Y(s) - sy(0) - y'(0) = s^2 Y(s) - 6s + 6 \)
- \( \mathcal{L}\{y'\} = s Y(s) - y(0) = s Y(s) - 6
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