Use the Laplace transform to solve the initial-value problem y' + 3y = 13 sin2t, y(0) = 6
Use the Laplace transform to solve the initial-value problem y' + 3y = 13 sin2t, 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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![**Using the Laplace Transform to Solve the Initial-Value Problem**
*Problem Statement:*
Given the differential equation
\[ y' + 3y = 13 \sin(2t), \]
with the initial condition
\[ y(0) = 6, \]
solve for \( y(t) \) using the Laplace transform.
### Steps to Solve:
1. **Take the Laplace Transform of Both Sides:**
The Laplace transform of \( y' \) is \( sY(s) - y(0) \), where \( Y(s) \) is the Laplace transform of \( y(t) \) and \( y(0) = 6 \).
Applying the Laplace transform to the given equation:
\[
\mathcal{L}\{y'\} + 3\mathcal{L}\{y\} = \mathcal{L}\{13\sin(2t)\}
\]
This results in:
\[
sY(s) - 6 + 3Y(s) = 13 \cdot \frac{2}{s^2 + 4}
\]
2. **Simplify the Equation:**
Combine the terms involving \( Y(s) \):
\[
(s + 3)Y(s) - 6 = \frac{26}{s^2 + 4}
\]
Then isolate \( Y(s) \):
\[
(s + 3)Y(s) = \frac{26}{s^2 + 4} + 6
\]
3. **Solve for \( Y(s) \):**
Rewrite to clearly solve for \( Y(s) \):
\[
Y(s) = \frac{26}{(s^2 + 4)(s + 3)} + \frac{6}{s + 3}
\]
4. **Inverse Laplace Transform:**
To find \( y(t) \), we need the inverse Laplace transform of \( Y(s) \). We can split the fractions and apply partial fraction decomposition if necessary:
\[
Y(s) = \frac{M_1}{s + 3} + \frac{Ns + P}{s^2 + 4}
\]
Identify constants \( M_1, N, \) and \(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd1d700aa-072a-4ff7-be1d-cdca0f10aa9d%2Fddeacc33-8abe-417d-a57b-a1dd02173589%2F8qosup_processed.png&w=3840&q=75)
Transcribed Image Text:**Using the Laplace Transform to Solve the Initial-Value Problem**
*Problem Statement:*
Given the differential equation
\[ y' + 3y = 13 \sin(2t), \]
with the initial condition
\[ y(0) = 6, \]
solve for \( y(t) \) using the Laplace transform.
### Steps to Solve:
1. **Take the Laplace Transform of Both Sides:**
The Laplace transform of \( y' \) is \( sY(s) - y(0) \), where \( Y(s) \) is the Laplace transform of \( y(t) \) and \( y(0) = 6 \).
Applying the Laplace transform to the given equation:
\[
\mathcal{L}\{y'\} + 3\mathcal{L}\{y\} = \mathcal{L}\{13\sin(2t)\}
\]
This results in:
\[
sY(s) - 6 + 3Y(s) = 13 \cdot \frac{2}{s^2 + 4}
\]
2. **Simplify the Equation:**
Combine the terms involving \( Y(s) \):
\[
(s + 3)Y(s) - 6 = \frac{26}{s^2 + 4}
\]
Then isolate \( Y(s) \):
\[
(s + 3)Y(s) = \frac{26}{s^2 + 4} + 6
\]
3. **Solve for \( Y(s) \):**
Rewrite to clearly solve for \( Y(s) \):
\[
Y(s) = \frac{26}{(s^2 + 4)(s + 3)} + \frac{6}{s + 3}
\]
4. **Inverse Laplace Transform:**
To find \( y(t) \), we need the inverse Laplace transform of \( Y(s) \). We can split the fractions and apply partial fraction decomposition if necessary:
\[
Y(s) = \frac{M_1}{s + 3} + \frac{Ns + P}{s^2 + 4}
\]
Identify constants \( M_1, N, \) and \(
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