Solve the following initial value problem using Laplace transform method: y' +6y=e4t; y (0) = 2.
Solve the following initial value problem using Laplace transform method: y' +6y=e4t; y (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 an Initial Value Problem using the Laplace Transform
In this section, we will solve an initial value problem using the Laplace transform method. Consider the following differential equation and initial condition:
\[ y' + 6y = e^{4t}, \]
\[ y(0) = 2. \]
Let's solve this step-by-step using the Laplace transform method.
#### Step 1: Apply the Laplace Transform
First, take the Laplace transform of both sides of the differential equation. Recall that the Laplace transform of a derivative \( y'(t) \) with initial condition \( y(0) \) is given by:
\[ \mathcal{L}\{y'(t)\} = sY(s) - y(0), \]
where \( Y(s) \) is the Laplace transform of \( y(t) \).
Applying the Laplace transform to the given differential equation:
\[ \mathcal{L}\{y'(t) + 6y(t)\} = \mathcal{L}\{e^{4t}\}. \]
#### Step 2: Use Laplace Transform Properties
Now, use the linearity of the Laplace transform and the initial condition \( y(0) = 2 \):
\[ sY(s) - y(0) + 6Y(s) = \frac{1}{s - 4}. \]
Substitute \( y(0) = 2 \):
\[ sY(s) - 2 + 6Y(s) = \frac{1}{s - 4}. \]
Combine like terms:
\[ (s + 6)Y(s) - 2 = \frac{1}{s - 4}. \]
#### Step 3: Solve for \( Y(s) \)
Isolate \( Y(s) \):
\[ (s + 6)Y(s) = \frac{1}{s - 4} + 2, \]
\[ Y(s) = \frac{1}{(s + 6)(s - 4)} + \frac{2}{s + 6}. \]
#### Step 4: Partial Fraction Decomposition
Perform partial fraction decomposition on \( Y(s) \):
\[ \frac{1}{(s + 6)(s - 4)} = \frac{A}{s + 6} + \frac{B}{s - 4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F20f91b20-384f-406f-a619-71f1b1f8c76e%2F97cbc8b7-c6af-4c05-b6a2-0a15c36e2768%2F7vcty5_processed.png&w=3840&q=75)
Transcribed Image Text:### Solving an Initial Value Problem using the Laplace Transform
In this section, we will solve an initial value problem using the Laplace transform method. Consider the following differential equation and initial condition:
\[ y' + 6y = e^{4t}, \]
\[ y(0) = 2. \]
Let's solve this step-by-step using the Laplace transform method.
#### Step 1: Apply the Laplace Transform
First, take the Laplace transform of both sides of the differential equation. Recall that the Laplace transform of a derivative \( y'(t) \) with initial condition \( y(0) \) is given by:
\[ \mathcal{L}\{y'(t)\} = sY(s) - y(0), \]
where \( Y(s) \) is the Laplace transform of \( y(t) \).
Applying the Laplace transform to the given differential equation:
\[ \mathcal{L}\{y'(t) + 6y(t)\} = \mathcal{L}\{e^{4t}\}. \]
#### Step 2: Use Laplace Transform Properties
Now, use the linearity of the Laplace transform and the initial condition \( y(0) = 2 \):
\[ sY(s) - y(0) + 6Y(s) = \frac{1}{s - 4}. \]
Substitute \( y(0) = 2 \):
\[ sY(s) - 2 + 6Y(s) = \frac{1}{s - 4}. \]
Combine like terms:
\[ (s + 6)Y(s) - 2 = \frac{1}{s - 4}. \]
#### Step 3: Solve for \( Y(s) \)
Isolate \( Y(s) \):
\[ (s + 6)Y(s) = \frac{1}{s - 4} + 2, \]
\[ Y(s) = \frac{1}{(s + 6)(s - 4)} + \frac{2}{s + 6}. \]
#### Step 4: Partial Fraction Decomposition
Perform partial fraction decomposition on \( Y(s) \):
\[ \frac{1}{(s + 6)(s - 4)} = \frac{A}{s + 6} + \frac{B}{s - 4
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