solve the given IVP using Laplace transforms and the following substitutions. Given: L{x"} = s² - x(s) − x(0) - s - x'(0) L{x} = $x(s) — x(0) L{x} = x(s) 3. Find the correct form of x(s) and stop. State x(s) as a single fraction. x" + 7x' + 6x = 2t x(0) = − 1, x'(0) = 2
solve the given IVP using Laplace transforms and the following substitutions. Given: L{x"} = s² - x(s) − x(0) - s - x'(0) L{x} = $x(s) — x(0) L{x} = x(s) 3. Find the correct form of x(s) and stop. State x(s) as a single fraction. x" + 7x' + 6x = 2t x(0) = − 1, x'(0) = 2
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 the Given Initial Value Problem (IVP) using Laplace Transforms
#### Substitutions:
Laplace transforms will be used with the following substitutions:
\[ \mathcal{L}\{x''\} = s^2 \cdot x(s) - x(0) \cdot s - x'(0) \]
\[ \mathcal{L}\{x'\} = s \cdot x(s) - x(0) \]
\[ \mathcal{L}\{x\} = x(s) \]
#### Problem Statement:
1. **Given Equation:**
\[ x'' + 7x' + 6x = 2t \]
2. **Initial Conditions:**
\[ x(0) = -1, \quad x'(0) = 2 \]
3. **Task:**
Find the correct form of \( x(s) \) and stop. State \( x(s) \) as a single fraction.
---
### Solution:
To solve this problem, follow these steps:
1. Apply the Laplace transform to both sides of the given differential equation.
2. Substitute the initial conditions into the resulting equation.
3. Solve for \( x(s) \).
---
**Applying Laplace Transform:**
\[ \mathcal{L}\{x'' + 7x' + 6x\} = \mathcal{L}\{2t\} \]
Using the substitutions provided:
\[ \mathcal{L}\{x''\} + 7\mathcal{L}\{x'\} + 6\mathcal{L}\{x\} = \mathcal{L}\{2t\} \]
\[ (s^2 x(s) - s x(0) - x'(0)) + 7(s x(s) - x(0)) + 6x(s) = \frac{2}{s^2} \]
Substituting the initial conditions \( x(0) = -1 \) and \( x'(0) = 2 \):
\[ (s^2 x(s) - s(-1) - 2) + 7(s x(s) - (-1)) + 6x(s) = \frac{2}{s^2} \]
Simplify:
\[ s^2 x(s) + s - 2 + 7s x(s) + 7](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5dfcde0a-cc8e-4f7e-b5e2-9ec6ad36a3c9%2Fde544165-f356-46dc-9e72-9b07cea133c6%2Fl1x12c_processed.png&w=3840&q=75)
Transcribed Image Text:### Solving the Given Initial Value Problem (IVP) using Laplace Transforms
#### Substitutions:
Laplace transforms will be used with the following substitutions:
\[ \mathcal{L}\{x''\} = s^2 \cdot x(s) - x(0) \cdot s - x'(0) \]
\[ \mathcal{L}\{x'\} = s \cdot x(s) - x(0) \]
\[ \mathcal{L}\{x\} = x(s) \]
#### Problem Statement:
1. **Given Equation:**
\[ x'' + 7x' + 6x = 2t \]
2. **Initial Conditions:**
\[ x(0) = -1, \quad x'(0) = 2 \]
3. **Task:**
Find the correct form of \( x(s) \) and stop. State \( x(s) \) as a single fraction.
---
### Solution:
To solve this problem, follow these steps:
1. Apply the Laplace transform to both sides of the given differential equation.
2. Substitute the initial conditions into the resulting equation.
3. Solve for \( x(s) \).
---
**Applying Laplace Transform:**
\[ \mathcal{L}\{x'' + 7x' + 6x\} = \mathcal{L}\{2t\} \]
Using the substitutions provided:
\[ \mathcal{L}\{x''\} + 7\mathcal{L}\{x'\} + 6\mathcal{L}\{x\} = \mathcal{L}\{2t\} \]
\[ (s^2 x(s) - s x(0) - x'(0)) + 7(s x(s) - x(0)) + 6x(s) = \frac{2}{s^2} \]
Substituting the initial conditions \( x(0) = -1 \) and \( x'(0) = 2 \):
\[ (s^2 x(s) - s(-1) - 2) + 7(s x(s) - (-1)) + 6x(s) = \frac{2}{s^2} \]
Simplify:
\[ s^2 x(s) + s - 2 + 7s x(s) + 7
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