**Problem Statement: Solving an Initial Value Problem with Laplace Transform**We are tasked with solving the following initial value problem using the Laplace transform method:\[ y'' + 4y = 4t \]Given the initial conditions:- $ y(0) = 1 $- $ y'(0) = 5 $### ExplanationThis problem involves a second-order linear differential equation with constant coefficients. To solve it using the Laplace transform, we will:1. Take the Laplace transform of both sides of the equation.2. Apply the initial conditions.3. Solve for the Laplace transform of the solution.4. Transform back to the time domain to obtain the solution for $ y(t) $.This approach leverages the power of the Laplace transform to handle initial conditions easily and convert the differential equation into an algebraic equation.

Answered: Solution the following initial Value…

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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**Problem Statement: Solving an Initial Value Problem with Laplace Transform**

We are tasked with solving the following initial value problem using the Laplace transform method:

\[ y'' + 4y = 4t \]

Given the initial conditions:
- $ y(0) = 1 $
- $ y'(0) = 5 $

### Explanation

This problem involves a second-order linear differential equation with constant coefficients. To solve it using the Laplace transform, we will:

1. Take the Laplace transform of both sides of the equation.
2. Apply the initial conditions.
3. Solve for the Laplace transform of the solution.
4. Transform back to the time domain to obtain the solution for $ y(t) $.

This approach leverages the power of the Laplace transform to handle initial conditions easily and convert the differential equation into an algebraic equation.

Expert Solution

Step 1

Given the initial value problem

y''+4y = 4t, y(0)=1, y'(0)=5

Solve it using Laplace transform.

Step 2

Transform both sides of the differential equation.

$L \{y'' + 4 y\} = 4 L \{t\}$

Since the operator $L$ is linear,

$L \{y''\} + 4 L \{y\} = 4 L \{t\} (s^{2} Y (s) - s y (0) - y' (0)) + 4 Y (s) = \frac{4}{s^{2}}$

Step 3

Plug the initial conditions y(0)=1 and y'(0)=5.

$(s^{2} Y (s) - s - 5) + 4 Y (s) = \frac{4}{s^{2}} \Rightarrow (s^{2} + 4) Y (s) = \frac{4}{s^{2}} + s + 5 \Rightarrow Y (s) = \frac{4}{s^{2} (s^{2} + 4)} + \frac{s + 5}{s^{2} + 4} = \frac{4 + s^{2} (s + 5)}{s^{2} (s^{2} + 4)} = \frac{s^{3} + 5 s^{2} + 4}{s^{2} (s^{2} + 4)}$

Step 4

By using partial fraction,

$\frac{s^{3} + 5 s^{2} + 4}{s^{2} (s^{2} + 4)} = \frac{A}{s} + \frac{B}{s^{2}} + \frac{C s + D}{s^{2} + 4} = \frac{A s (s^{2} + 4) + B (s^{2} + 4) + (C s + D) s^{2}}{s^{2} (s^{2} + 4)}$

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