Solve using laplace transform (D^2 + 4D+5)y = 5

To solve the differential equation \((D^2 + 4D + 5)y = 5\) using the Laplace transform, we follow these steps:

1. **Take the Laplace transform of both sides of the equation.**
2. **Solve for \(Y(s)\).**
3. **Take the inverse Laplace transform to find \(y(t)\).**

### Step 1: Take the Laplace transform

Given the differential equation:

\[ (D^2 + 4D + 5)y = 5 \]

Using the Laplace transform properties:

\[ \mathcal{L}\{D^2y\} = s^2 Y(s) - s y(0) - y'(0) \]
\[ \mathcal{L}\{Dy\} = s Y(s) - y(0) \]

Let \( \mathcal{L}\{y(t)\} = Y(s) \). Applying the Laplace transform to both sides of the equation:

\[ \mathcal{L}\{D^2y + 4Dy + 5y\} = \mathcal{L}\{5\} \]

\[ \mathcal{L}\{D^2y\} + 4\mathcal{L}\{Dy\} + 5\mathcal{L}\{y\} = \frac{5}{s} \]

Substitute the Laplace transform terms:

\[ (s^2 Y(s) - s y(0) - y'(0)) + 4(s Y(s) - y(0)) + 5Y(s) = \frac{5}{s} \]

### Step 2: Solve for \(Y(s)\)

Combine the terms involving \(Y(s)\):

\[ s^2 Y(s) - s y(0) - y'(0) + 4s Y(s) - 4y(0) + 5Y(s) = \frac{5}{s} \]

\[ (s^2 + 4s + 5)Y(s) - s y(0) - y'(0) - 4y(0) = \frac{5}{s} \]

Assuming initial conditions \(y(0) = 0\) and \(y'(0) = 0\):

\[ (s^2 + 4s + 5)Y(s) = \frac{5}{s} \]

Solve for \(Y(s)\):

\[ Y(s) = \frac{5}{s(s^2 + 4s + 5)} \]

### Step 3: Take the inverse Laplace transform

First, decompose \(Y(s)\):

\[ Y(s) = \frac{5}{s(s^2 + 4s + 5)} \]

To perform partial fraction decomposition, we can rewrite the denominator:

\[ s(s^2 + 4s + 5) = s(s+2)^2 + 1 \]

The partial fraction decomposition becomes:

\[ \frac{5}{s(s^2 + 4s + 5)} = \frac{A}{s} + \frac{Bs + C}{s^2 + 4s + 5} \]

Multiply both sides by the denominator \(s(s^2 + 4s + 5)\):

\[ 5 = A(s^2 + 4s + 5) + s(Bs + C) \]

Solve for \(A\), \(B\), and \(C\):

1. Set \(s = 0\):
   \[ 5 = A(5) \Rightarrow A = 1 \]

2. Equate coefficients of \(s^2\) and \(s\):
   \[ 0 = A + B \Rightarrow B = -1 \]
   \[ 0 = 4A + C \Rightarrow C = -4 \]

Thus, the decomposition is:

\[ Y(s) = \frac{1}{s} + \frac{-s - 4}{s^2 + 4s + 5} \]

Now, take the inverse Laplace transform:

1. \(\mathcal{L}^{-1}\left\{\frac{1}{s}\right\} = 1\)

For the second term, rewrite it using completing the square on the quadratic term:

\[ s^2 + 4s + 5 = (s+2)^2 + 1 \]

So,

\[ \frac{-s - 4}{(s+2)^2 + 1} = \frac{-(s+2) - 2}{(s+2)^2 + 1} = \frac{-(s+2)}{(s+2)^2 + 1} - \frac{2}{(s+2)^2 + 1} \]

Then,

\[ \mathcal{L}^{-1}\left\{\frac{-(s+2)}{(s+2)^2 + 1}\right\} = -e^{-2t} \cos(t) \]
\[ \mathcal{L}^{-1}\left\{\frac{-2}{(s+2)^2 + 1}\right\} = -2e^{-2t} \sin(t) \]

So,

\[ \mathcal{L}^{-1}\left\{Y(s)\right\} = 1 - e^{-2t} \cos(t) - 2e^{-2t} \sin(t) \]

### Final Solution

\[ y(t) = 1 - e^{-2t} \cos(t) - 2e^{-2t} \sin(t) \]

Do you want details or have any questions?

Here are some relative questions to expand on this information:

1. What is the Laplace transform?
2. How is the Laplace transform applied to differential equations?
3. What are the properties of the Laplace transform?
4. How do you solve initial value problems using the Laplace transform?
5. What is partial fraction decomposition, and how is it used in Laplace transforms?
6. How do you complete the square for a quadratic expression?
7. What is the inverse Laplace transform?
8. How do you handle complex roots in the context of the Laplace transform?

**Tip:** The Laplace transform is a powerful tool for solving linear differential equations, particularly when dealing with initial conditions and systems of differential equations.

Learn how to solve the differential equation (D^2 + 4D + 5)y = 5 using Laplace transform. This detailed step-by-step solution covers Laplace transform properties, inverse Laplace transform, and solving differential equations. Suitable for college-level students and anyone studying differential equations.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation