Solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem below. y" +9y = 5t²-6, y(0) = 0, y'(0) = - 8 Click here to view the table of Laplace transforms. Click here to view the table of properties of Laplace transforms. *** Y(s) =
Solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem below. y" +9y = 5t²-6, y(0) = 0, y'(0) = - 8 Click here to view the table of Laplace transforms. Click here to view the table of properties of Laplace transforms. *** Y(s) =
Advanced Engineering Mathematics
10th Edition
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Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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6 help Laplace problem
![### Solving Initial Value Problems Using the Laplace Transform
#### Problem Statement:
Solve for \( Y(s) \), the Laplace transform of the solution \( y(t) \) to the initial value problem below.
\[ y'' + 9y = 5t^2 - 6, \quad y(0) = 0, \quad y'(0) = -8 \]
#### Links for Reference:
1. [Click here to view the table of Laplace transforms.](#)
2. [Click here to view the table of properties of Laplace transforms.](#)
#### Approach:
1. **Transform the Differential Equation:**
Use the Laplace transform on both sides of the differential equation \( y'' + 9y = 5t^2 - 6 \) to convert it from the time domain into the s-domain.
2. **Initial Conditions:**
Apply the given initial conditions \( y(0) = 0 \) and \( y'(0) = -8 \).
3. **Solve Algebraically:**
Solve the algebraic equation obtained after applying the Laplace transform.
#### Solution Outline:
Given \( y'' + 9y = 5t^2 - 6 \):
1. Taking the Laplace transform of both sides:
\[ \mathcal{L}\{y''(t)\} + 9 \mathcal{L}\{y(t)\} = \mathcal{L}\{5t^2 - 6\} \]
2. Using the properties of the Laplace transform:
\[ s^2 Y(s) - s y(0) - y'(0) + 9 Y(s) = \frac{5}{s^3} - \frac{6}{s} \]
3. Substitute the initial conditions \( y(0) = 0 \) and \( y'(0) = -8 \):
\[ s^2 Y(s) + 8 + 9 Y(s) = \frac{5}{s^3} - \frac{6}{s} \]
4. Combine like terms:
\[ (s^2 + 9) Y(s) + 8 = \frac{5}{s^3} - \frac{6}{s} \]
5. Solve for \( Y(s) \):
\[ Y(s) (s^2 + 9) = \frac{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbbba0909-7b4f-4e80-8e24-06dfda2d061e%2Ff7cda98c-9182-489e-9647-71a36318719a%2Fec7wrq8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Solving Initial Value Problems Using the Laplace Transform
#### Problem Statement:
Solve for \( Y(s) \), the Laplace transform of the solution \( y(t) \) to the initial value problem below.
\[ y'' + 9y = 5t^2 - 6, \quad y(0) = 0, \quad y'(0) = -8 \]
#### Links for Reference:
1. [Click here to view the table of Laplace transforms.](#)
2. [Click here to view the table of properties of Laplace transforms.](#)
#### Approach:
1. **Transform the Differential Equation:**
Use the Laplace transform on both sides of the differential equation \( y'' + 9y = 5t^2 - 6 \) to convert it from the time domain into the s-domain.
2. **Initial Conditions:**
Apply the given initial conditions \( y(0) = 0 \) and \( y'(0) = -8 \).
3. **Solve Algebraically:**
Solve the algebraic equation obtained after applying the Laplace transform.
#### Solution Outline:
Given \( y'' + 9y = 5t^2 - 6 \):
1. Taking the Laplace transform of both sides:
\[ \mathcal{L}\{y''(t)\} + 9 \mathcal{L}\{y(t)\} = \mathcal{L}\{5t^2 - 6\} \]
2. Using the properties of the Laplace transform:
\[ s^2 Y(s) - s y(0) - y'(0) + 9 Y(s) = \frac{5}{s^3} - \frac{6}{s} \]
3. Substitute the initial conditions \( y(0) = 0 \) and \( y'(0) = -8 \):
\[ s^2 Y(s) + 8 + 9 Y(s) = \frac{5}{s^3} - \frac{6}{s} \]
4. Combine like terms:
\[ (s^2 + 9) Y(s) + 8 = \frac{5}{s^3} - \frac{6}{s} \]
5. Solve for \( Y(s) \):
\[ Y(s) (s^2 + 9) = \frac{
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