Solving Y'' + 16y = 16t^2 - 32t + 66 Initial Value Problem Using Laplace Transforms

Introduction to Laplace Transforms

In the realm of differential equations, particularly when dealing with initial value problems, the Laplace transform emerges as a powerful tool. It offers a systematic approach to converting differential equations into algebraic equations, which are often easier to solve. This method is especially effective for linear differential equations with constant coefficients, making it a staple in various engineering disciplines. The Laplace transform technique is a cornerstone in electrical engineering, mechanical engineering, and control systems, where it simplifies the analysis of circuits, mechanical systems, and feedback control loops. For an initial value problem such as y″ + 16y = 16t² - 32t + 66, y(0) = 0, y′(0) = 18, the Laplace transform provides a structured way to find the solution by sidestepping the complexities of traditional methods. By transforming the differential equation into an algebraic one, we can apply algebraic manipulations to find the Laplace transform of the solution, and then use the inverse Laplace transform to obtain the solution in the original time domain. This approach not only simplifies the solving process but also provides insights into the system's behavior and stability.

Problem Statement

Our main goal is to solve the initial value problem given by the second-order linear differential equation: y″ + 16y = 16t² - 32t + 66. This equation is accompanied by two initial conditions: y(0) = 0 and y′(0) = 18. These conditions are crucial because they specify the state of the system at the initial time (t = 0) and are essential for determining a unique solution to the differential equation. The differential equation itself describes the relationship between the function y(t) and its derivatives, representing a dynamic system where the current state depends on both the present and past conditions. The forcing function, 16t² - 32t + 66, represents an external influence on the system. Solving this initial value problem involves finding a function y(t) that satisfies both the differential equation and the given initial conditions. This type of problem is common in physics and engineering, where it might represent the motion of a mass-spring system, the current in an electrical circuit, or the behavior of a control system. The Laplace transform method provides a systematic way to tackle such problems, offering an alternative to classical methods like undetermined coefficients or variation of parameters.

Applying the Laplace Transform

To begin, we apply the Laplace transform to each term of the differential equation y″ + 16y = 16t² - 32t + 66. Recall that the Laplace transform converts a function of time, y(t), into a function of a complex variable, s, denoted as Y(s). The key to this process lies in the properties of the Laplace transform, particularly how it handles derivatives. Specifically, the Laplace transform of y″(t) is given by s²Y(s) - sy(0) - y′(0), where y(0) and y′(0) are the initial conditions. Similarly, the Laplace transform of y(t) is simply Y(s). For the right-hand side of the equation, we need to transform the polynomial terms. The Laplace transform of t² is 2/s³, the Laplace transform of t is 1/s², and the Laplace transform of a constant c is c/s. Applying these rules to our equation, we get:

s²Y(s) - sy(0) - y′(0) + 16Y(s) = 16(2/s³) - 32(1/s²) + 66/s.

Next, we substitute the initial conditions y(0) = 0 and y′(0) = 18 into the transformed equation, which simplifies the equation significantly. This substitution is a critical step as it incorporates the initial state of the system into the algebraic equation, allowing us to solve for Y(s). The resulting equation is an algebraic equation in Y(s), which can be solved using standard algebraic techniques.

Solving for Y(s)

After substituting the initial conditions (y(0) = 0 and y′(0) = 18) into the transformed equation, we have: s²Y(s) - 18 + 16Y(s) = 32/s³ - 32/s² + 66/s. The next step involves rearranging the equation to isolate Y(s). We combine the terms involving Y(s) on one side and move all other terms to the other side. This gives us: (s² + 16)Y(s) = 32/s³ - 32/s² + 66/s + 18. Now, we divide both sides by (s² + 16) to solve for Y(s), resulting in: Y(s) = (32/s³ - 32/s² + 66/s + 18) / (s² + 16). This expression for Y(s) is a rational function in s, which represents the Laplace transform of the solution y(t). To find y(t), we need to compute the inverse Laplace transform of Y(s). However, before we can do that, it's often helpful to simplify Y(s) by combining the fractions and performing any possible cancellations. This simplification makes the subsequent inverse Laplace transform process easier and less prone to errors. The form of Y(s) now sets the stage for using partial fraction decomposition and inverse transform tables to find the solution in the time domain.

Partial Fraction Decomposition

To find the inverse Laplace transform of Y(s) = (32/s³ - 32/s² + 66/s + 18) / (s² + 16), we first simplify the expression and then apply partial fraction decomposition. Combining the terms in the numerator, we get:

Y(s) = (18s³ + 66s² - 32s + 32) / (s³(s² + 16)).

This rational function is now in a form suitable for partial fraction decomposition. The denominator has factors of s³ and (s² + 16), which means we need to express Y(s) as a sum of simpler fractions. The decomposition will take the form:

Y(s) = A/s + B/s² + C/s³ + (Ds + E) / (s² + 16),

where A, B, C, D, and E are constants that we need to determine. To find these constants, we multiply both sides of the equation by the denominator s³(s² + 16), which clears the fractions. This results in an equation where we can equate the coefficients of like powers of s. By solving the resulting system of equations, we find the values of A, B, C, D, and E. This process breaks down the complex fraction into simpler fractions, each of which has a known inverse Laplace transform. Partial fraction decomposition is a critical technique in solving differential equations using Laplace transforms, as it allows us to handle complex rational functions by expressing them in terms of simpler, more manageable components.

Finding the Constants

After clearing the fractions in the partial fraction decomposition, we obtain the equation: 18s³ + 66s² - 32s + 32 = As²(s² + 16) + Bs(s² + 16) + C(s² + 16) + (Ds + E)s³**. This equation allows us to solve for the constants A, B, C, D, and E by equating coefficients of like powers of s. Expanding the right side, we get: 18s³ + 66s² - 32s + 32 = As⁴ + 16As² + Bs³ + 16Bs + Cs² + 16C + Ds⁴ + Es³. Now, we group the terms by powers of s: 18s³ + 66s² - 32s + 32 = (A + D)s⁴ + (B + E)s³ + (16A + C)s² + 16Bs + 16C. By equating the coefficients of s⁴, s³, s², s, and the constant term, we obtain the following system of equations:

• A + D = 0
• B + E = 18
• 16A + C = 66
• 16B = -32
• 16C = 32

Solving this system, we find:

• A = 4
• B = -2
• C = 2
• D = -4
• E = 20

These constants allow us to rewrite Y(s) in its decomposed form, which is crucial for the next step: finding the inverse Laplace transform. This systematic approach of equating coefficients provides a reliable method for determining the constants in partial fraction decomposition, ensuring an accurate representation of the original function.

Inverse Laplace Transform

Now that we have the constants, we can rewrite Y(s) using the partial fraction decomposition as: Y(s) = 4/s - 2/s² + 2/s³ + (-4s + 20) / (s² + 16). The final step is to find the inverse Laplace transform of Y(s) to obtain the solution y(t). We apply the inverse Laplace transform to each term separately. Recall the following standard Laplace transforms:

• L⁻¹{1/s} = 1
• L⁻¹{1/s²} = t
• L⁻¹{2/s³} = t²
• L⁻¹{s / (s² + a²)} = cos(at)
• L⁻¹{a / (s² + a²)} = sin(at)

Using these, we can find the inverse Laplace transform of each term in Y(s). For the term (−4s + 20) / (s² + 16), we can rewrite it as -4[s / (s² + 16)] + 5[4 / (s² + 16)]. Applying the inverse Laplace transform to each term, we get:

• L⁻¹{4/s} = 4
• L⁻¹{-2/s²} = -2t
• L⁻¹{2/s³} = t²
• L⁻¹{-4s / (s² + 16)} = -4 cos(4t)
• L⁻¹{5 * 4 / (s² + 16)} = 5 sin(4t)

Combining these results, we obtain the solution y(t).

Solution

Combining the results from the inverse Laplace transform of each term in Y(s), we arrive at the solution to the initial value problem: y(t) = 4 - 2t + t² - 4 cos(4t) + 5 sin(4t). This function y(t) satisfies both the original differential equation, y″ + 16y = 16t² - 32t + 66, and the initial conditions, y(0) = 0 and y′(0) = 18. The solution consists of polynomial terms (4 - 2t + t²) and sinusoidal terms (-4 cos(4t) + 5 sin(4t)), which indicate that the system's response includes both a steady-state component (the polynomial terms) and an oscillatory component (the sinusoidal terms). This type of solution is common in second-order linear differential equations with constant coefficients, especially when there is a forcing function that includes polynomial and constant terms. The cosine and sine terms arise from the homogeneous part of the equation and reflect the natural oscillatory behavior of the system, while the polynomial terms are a response to the forcing function. The Laplace transform method has allowed us to systematically find this solution by converting the differential equation into an algebraic equation, solving for Y(s), and then using the inverse Laplace transform to return to the time domain. This solution provides a complete description of the system's behavior over time, given the initial conditions and the external forcing function.