Problem 29 - Advanced Engineering Mathematics

Question

Use the Laplace transform to solve the given initial-value problem. $$ y^{\prime \prime}-y^{\prime}=e^{t} \cos t, \quad y(0)=0, \quad y^{\prime}(0)=0 $$

Step-by-Step Solution

Answer

Use Laplace Transform to find \(Y(s)\), perform partial fraction decomposition, and then find \(y(t)\) by inverse transformation.

Step 1: Take the Laplace Transform

Begin by taking the Laplace transform of both sides of the differential equation.\[ \mathcal{L}\{y'' - y'\} = \mathcal{L}\{e^t \cos t\}\]Using linearity and properties of the Laplace transform, we have:\[ s^2Y(s) - sy(0) - y'(0) - (sY(s) - y(0)) = \mathcal{L}\{e^t \cos t\}\]Since \(y(0)=0\) and \(y'(0)=0\), this simplifies to:\[ (s^2 - s)Y(s) = \mathcal{L}\{e^t \cos t\}\]

Step 2: Find Laplace Transform of the Right-Hand Side

We use the formula for the Laplace transform of \(e^{at}\cos(bt)\), which is:\[ \mathcal{L}\{e^{at} \cos(bt)\} = \frac{s-a}{(s-a)^2 + b^2}\]For \(e^t \cos t\), \(a = 1\) and \(b = 1\). Thus:\[ \mathcal{L}\{e^t \cos t\} = \frac{s-1}{(s-1)^2 + 1^2} = \frac{s-1}{s^2 - 2s + 2}\]

Step 3: Solve for Y(s)

Substitute the result from Step 2 into the equation from Step 1:\[ (s^2 - s)Y(s) = \frac{s-1}{s^2 - 2s + 2}\]Divide both sides by \((s^2 - s)\):\[ Y(s) = \frac{s-1}{(s^2 - 2s + 2)(s^2 - s)}\]

Step 4: Perform Partial Fraction Decomposition

To perform partial fraction decomposition, express \(Y(s)\) as a sum of simpler fractions:\[ \frac{s-1}{(s^2 - 2s + 2)(s^2 - s)} = \frac{A}{s} + \frac{B}{s^2} + \frac{Cs + D}{s^2 - 2s + 2}\]Multiply through by the denominator and solve for \(A, B, C,\) and \(D\) by equating coefficients or substituting convenient values of \(s\).

Step 5: Inverse Laplace Transform

Once you have the partial fraction decomposition, take the inverse Laplace transform of each term separately. Each term corresponds to a standard formula in the Laplace transform table.Sum the individual terms to obtain \(y(t)\), recognizing and applying initial conditions.

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down a complex rational expression into simpler, more manageable parts. This is particularly useful when dealing with difficult Laplace transforms. By expressing a complicated rational function as a sum of simpler fractions, we can more easily take the inverse Laplace transform.Here’s how partial fraction decomposition generally works:

• Identify the complex rational function. In our example, this is the expression for \( Y(s) \) derived from the equation \( \frac{s-1}{(s^2 - 2s + 2)(s^2 - s)} \).
• Break it down into a sum of simpler fractions. Each factor in the denominator gets its own fractional term. In our example, we use terms like \( \frac{A}{s}, \frac{B}{s^2}, \frac{Cs + D}{s^2 - 2s + 2} \).
• Multiply through by the common denominator to eliminate fractions, which allows us to find the coefficients \( A, B, C, \) and \( D \).
• Use strategic values for \(s \) or equate coefficients to solve for these unknowns. This often involves some algebraic manipulation and calculations.

Once decomposed, these simpler fractions can be individually transformed back from the Laplace domain, making the reconstruction of the solution much simpler.

Inverse Laplace Transform

The inverse Laplace transform is used to convert a function from the Laplace domain back into the time domain. It is an essential step in solving differential equations using the Laplace technique. After applying the Laplace transform to the original equation, we work within the transformed domain to simplify and solve.When ready to interpret the solution in the context of the original problem, we apply the inverse. Here’s how it works:

• Start by expressing the solution in a form suitable for inverse transformation, often achieved via partial fraction decomposition, which simplifies the terms.
• Use known inverse transforms from standard tables to convert each simple fraction back to a time domain function. Each fraction corresponds to a straightforward inverse transform.
• Add the transformed terms to build the solution in the time domain, ensuring all initial conditions are met. This reconstitution step provides the final solution \( y(t) \).

Having the ability to go back and forth between domains allows us to leverage mathematical tools only available in one domain. This parallel transition reveals the design of the solution, allowing us to solve complex differential equations.

Differential Equations

Differential equations are mathematical equations that form a crucial part of modeling real-world phenomena, involving derivatives that reflect the rate of change. They describe how one or more functions depend on their derivatives and are often used in fields such as physics, engineering, and economics.Here’s how differential equations relate to the problem:

• With initial conditions \( y(0) = 0 \) and \( y'(0) = 0 \), the differential equation \( y'' - y' = e^t \cos t \) dictates the behavior of the system over time.
• Solve these equations using tools like the Laplace transform, which simplifies handling differential components by converting them into algebraic expressions in the Laplace domain.
• Returning to the time domain after solving yields a function \( y(t) \) that satisfies both the differential equation and the initial conditions, thus providing the complete solution.

Understanding how to manipulate these equations enables us to predict and analyze dynamic behaviors, serving various applications such as control systems and signal processing. By mastering differential equations, one can unlock the ability to explore complex systems composed of changing elements.