Solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem below. y" +9y = 5t³, y(0) = 0, y'(0) = 0
Solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem below. y" +9y = 5t³, y(0) = 0, y'(0) = 0
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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Help for Laplace transform
![### Solving for Y(s), the Laplace Transform of y(t)
To find Y(s), the Laplace transform of the solution \( y(t) \) to the given initial value problem, follow the steps below.
#### Initial Value Problem
The differential equation and initial conditions provided are:
\[ y'' + 9y = 5t^3 \]
\[ y(0) = 0 \]
\[ y'(0) = 0 \]
#### Steps to Solve:
1. **Take the Laplace Transform of Both Sides.**
The Laplace transform of the left-hand side is:
\[
\mathcal{L}\{y''(t)\} + 9\mathcal{L}\{y(t)\}
\]
Applying the Laplace transform to each term:
\[
\mathcal{L}\{y''(t)\} = s^2Y(s) - sy(0) - y'(0)
\]
\[
\mathcal{L}\{y(t)\} = Y(s)
\]
Given the initial conditions \( y(0) = 0 \) and \( y'(0) = 0 \), the equation simplifies to:
\[
s^2Y(s) + 9Y(s)
\]
2. **Taking the Laplace Transform of the Right-Hand Side:**
\[
\mathcal{L}\{5t^3\}
\]
The Laplace transform of \( t^n \) is \( \frac{n!}{s^{n+1}} \). For \( t^3 \), it is:
\[
\mathcal{L}\{t^3\} = \frac{3!}{s^4} = \frac{6}{s^4}
\]
Therefore:
\[
\mathcal{L}\{5t^3\} = 5 \cdot \frac{6}{s^4} = \frac{30}{s^4}
\]
3. **Combine and Solve for \( Y(s) \):**
\[
s^2Y(s) + 9Y(s) = \frac{30}{s^4}
\]
Factor out \( Y(s) \):
\[
Y(s)(s](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbbba0909-7b4f-4e80-8e24-06dfda2d061e%2F0ce65d51-0afb-42bf-a666-8327360cc2ea%2Ftn4kvwl8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Solving for Y(s), the Laplace Transform of y(t)
To find Y(s), the Laplace transform of the solution \( y(t) \) to the given initial value problem, follow the steps below.
#### Initial Value Problem
The differential equation and initial conditions provided are:
\[ y'' + 9y = 5t^3 \]
\[ y(0) = 0 \]
\[ y'(0) = 0 \]
#### Steps to Solve:
1. **Take the Laplace Transform of Both Sides.**
The Laplace transform of the left-hand side is:
\[
\mathcal{L}\{y''(t)\} + 9\mathcal{L}\{y(t)\}
\]
Applying the Laplace transform to each term:
\[
\mathcal{L}\{y''(t)\} = s^2Y(s) - sy(0) - y'(0)
\]
\[
\mathcal{L}\{y(t)\} = Y(s)
\]
Given the initial conditions \( y(0) = 0 \) and \( y'(0) = 0 \), the equation simplifies to:
\[
s^2Y(s) + 9Y(s)
\]
2. **Taking the Laplace Transform of the Right-Hand Side:**
\[
\mathcal{L}\{5t^3\}
\]
The Laplace transform of \( t^n \) is \( \frac{n!}{s^{n+1}} \). For \( t^3 \), it is:
\[
\mathcal{L}\{t^3\} = \frac{3!}{s^4} = \frac{6}{s^4}
\]
Therefore:
\[
\mathcal{L}\{5t^3\} = 5 \cdot \frac{6}{s^4} = \frac{30}{s^4}
\]
3. **Combine and Solve for \( Y(s) \):**
\[
s^2Y(s) + 9Y(s) = \frac{30}{s^4}
\]
Factor out \( Y(s) \):
\[
Y(s)(s
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