Problem 35 - Advanced Engineering Mathematics

Question

Use the Laplace transform to solve the given equation. $$ y^{\prime \prime}+6 y^{\prime}+5 y=t-t 9(t-2), \quad y(0)=1, y^{\prime}(0)=0 $$

Step-by-Step Solution

Answer

Solve the equation by taking the Laplace transform, using partial fractions, and applying the inverse transform.

Step 1: Identify the Laplace Transform Components

Firstly, note the equation given is a second-order linear differential equation with constant coefficients. The non-homogeneous part is \(t - t \cdot 9(t-2)\). The initial conditions are \(y(0) = 1\) and \(y'(0) = 0\).

Step 2: Take the Laplace Transform of the Entire Equation

Apply the Laplace Transform to each term in the equation. For the derivatives:- \(\mathcal{L}\{y''\} = s^2Y(s) - sy(0) - y'(0)\).- \(\mathcal{L}\{y'\} = sY(s) - y(0)\).Where \(Y(s)\) is the Laplace Transform of \(y(t)\). The Laplace Transform of the right-hand side is: \(\mathcal{L}\{t\} = \frac{1}{s^2}\) and \(9 \cdot \mathcal{L}\{t \cdot 9(t-2)\} = \frac{9e^{-2s}}{s^2}\).

Step 3: Substitute Transforms into the Equation

Insert the Laplace Transforms into the equation:\[s^2Y(s) - sy(0) - y'(0) + 6(sY(s) - y(0)) + 5Y(s) = \frac{1}{s^2} - \frac{9e^{-2s}}{s^2}\]Substitute the initial conditions \(y(0) = 1\) and \(y'(0) = 0\) to get:\[s^2Y(s) - s + 6sY(s) - 6 + 5Y(s) = \frac{1}{s^2} - \frac{9e^{-2s}}{s^2}\]

Step 4: Simplify and Solve for \(Y(s)\)

Combine and group terms:\[(s^2 + 6s + 5)Y(s) - (s + 6) = \frac{1}{s^2} - \frac{9e^{-2s}}{s^2} \]Solve for \(Y(s)\):\[Y(s) = \frac{1}{s^2(s^2 + 6s + 5)} + \frac{9e^{-2s}}{s^2(s^2 + 6s + 5)} + \frac{s + 6}{s^2 + 6s + 5}\]

Step 5: Perform Partial Fraction Decomposition

Decompose the first fraction into simpler partial fractions. It is crucial for the inverse Laplace transformation:\[\frac{1}{s^2(s + 5)(s + 1)} = \frac{A}{s^2} + \frac{B}{s} + \frac{C}{s+1} + \frac{D}{s+5}\]Solve for \(A, B, C,\) and \(D\).

Step 6: Apply Inverse Laplace Transform

Calculate the inverse Laplace Transform to determine \(y(t)\). Utilize known transforms and properties to convert each term from the \(Y(s)\) function back into the time domain. You'll face distinct cases for each part of \(Y(s)\) derived in Step 5. The presence of \(e^{-2s}\) also calls for shifting properties in the inverse transformation.

Step 7: Final Solution for the Differential Equation

After applying the inverse Laplace Transform, combine the terms to form the solution, \(y(t)\), expressed entirely in the time domain, taking into account the initial conditions and applicability of heaviest terms.

Key Concepts

Second-Order Differential Equations

In many fields of science and engineering, differential equations are used to describe how quantities change. A second-order differential equation involves the second derivative of a function, which often represents acceleration in physical systems.
For a function \( y(t) \), a second-order differential equation typically has the form:

• \( y'' + a_1 y' + a_0 y = g(t) \)

In our original exercise, the equation \( y'' + 6y' + 5y = t - t\cdot9(t-2) \) is a second-order differential equation with constant coefficients. Here, \( a_1 = 6 \), \( a_0 = 5 \), and \( g(t) = t - t\cdot9(t-2) \) is a non-homogeneous term that influences the behavior of the system differently than a homogeneous equation would.
Understanding second-order differential equations involves:

• Recognizing initial conditions, which provide necessary information to determine a unique solution.
• Handling non-homogeneous terms, which can represent external forces in physical scenarios.

In practice, the Laplace transform is often used to simplify solving these equations by transforming them into algebraic equations, which are easier to handle. This makes solving more complex differential equations manageable.

Initial Value Problems

To ensure a differential equation has a unique solution, initial value problems (IVPs) are employed. When solving a differential equation, giving initial conditions such as \( y(0) = 1 \) and \( y'(0) = 0 \) helps determine a specific solution tailored to meet these initial states.
The initial conditions reflect the state of a system at \( t=0 \), providing essential parameters for the problem. These values help in solving the constant terms that arise when integrating, ensuring the solution fits the physical or theoretical scenario represented.
When solving second-order differential equations as initial value problems:

• Initial conditions substitute directly into the transformed equation, allowing specific constants in the solution to be calculated.
• They provide an exact starting point for the trajectory of a system's response over time.
• The Laplace transform technique is particularly effective for initial value problems because it easily incorporates these initial conditions during the transformation process.

For example, in our equation, after applying the Laplace transform and substituting initial conditions, it becomes more straightforward to manage and solve for functions in the complex \( s \)-domain.

Partial Fraction Decomposition

In the realm of solving differential equations using Laplace transforms, partial fraction decomposition becomes crucial in handling more complex algebraic expressions. Fundamentally, partial fraction decomposition involves breaking down complex rational expressions into simpler, easy-to-manage fractions.
For instance, solving:

• \( \frac{1}{s^2(s + 5)(s + 1)} \)

requires expressing it as a sum of simpler fractions like:

• \( \frac{A}{s^2} + \frac{B}{s} + \frac{C}{s+1} + \frac{D}{s+5} \)

By determining the constants \( A, B, C, \) and \( D \), we make it feasible to perform inverse Laplace transforms on each part. This process facilitates converting each fraction from the \( s \)-domain back to the time domain.
The steps typically involve:

• Setting up the equation by matching coefficients or solving systems of equations to find these constants.
• Utilizing known inverse Laplace transforms to handle each decomposed term quite easily.

This decomposition process is essential for simplifying and solving differential equations, allowing us to interpret and convert initial equations involving the Laplace transform back into practical time-domain solutions, making the approach even more powerful.