Solve the following initial value problem using Laplace transform method: y' +6y=e4t; y (0) = 2.

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ISBN:9780470458365
Author:Erwin Kreyszig
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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
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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