Problem 19 - Differential Equations and Linear Algebra

Question

Use the Laplace transform to solve the given initial-value problem. \(y^{\prime \prime}-9 y=13 \sin 2 t, \quad y(0)=3, \quad y^{\prime}(0)=1\).

Step-by-Step Solution

Answer

The solution to the given initial-value problem using the Laplace transform is: \(y(t) = \frac{13}{9}\sinh(3t) - \frac{13}{9}\sin(2t) - 3\cosh(3t)\).

Step 1: Calculate the Laplace transform of the given equation

Given the equation: \(y^{\prime\prime} - 9y = 13\sin(2t)\), We know that the Laplace transforms of y', y'', and sin(2t) are sY(s) - y(0), s^2Y(s) - sy(0) - y'(0), and \(\frac{13}{s^2 + 4}\), respectively. Taking the Laplace transform of both sides, we get: \[(s^2Y(s) - sy(0) - y'(0)) - 9Y(s) = \frac{13}{s^2 + 4}\]

Step 2: Substitute the initial values and solve for Y(s)

Now, substitute the initial values y(0) = 3 and y'(0) = 1 into the Laplace transformed equation: \[(s^2Y(s) - s(3) - 1) - 9Y(s) = \frac{13}{s^2 + 4}\] Now, we'll try to isolate Y(s): \[(s^2 - 9)Y(s) = -3s - 1 + \frac{13}{s^2 + 4}\] Finally, we will write the equation in terms of Y(s): \[Y(s) = \frac{-3s - 1}{s^2 - 9} + \frac{13}{(s^2 - 9)(s^2 + 4)}\]

Step 3: Find the inverse Laplace transform to solve for y(t)

Now, we find the inverse Laplace transform of Y(s) to determine y(t). The inverse Laplace transform can be computed using partial fraction decomposition: 1. As the first term is already decomposed, we only need to work on the second term: \[Y(s) = \frac{-3s - 1}{s^2 - 9} + \frac{13}{(s^2 - 9)(s^2 + 4)}\] We can rewrite this term like this: \[\frac{13}{(s^2-9)(s^2+4)}=\frac{As+B}{s^2-9}+\frac{Cs+D}{s^2+4}\] 2. Solve for A, B, C, and D using partial fraction decomposition. You should find that: \(A = 0, B = \frac{13}{9}, C = 0, \) and \(D = \frac{-13}{9}\) 3. Now plug them back into the Y(s) equation: \(Y(s) = \frac{13}{9}\frac{1}{s^2 - 9} - \frac{13}{9}\frac{1}{s^2 + 4} - \frac{3s}{s^2 - 9}\) 4. Finally, find the inverse Laplace transform for each term: \(y(t) = L^{-1}\{Y(s)\} = \frac{13}{9}\sinh(3t) - \frac{13}{9}\sin(2t) - 3\cosh(3t)\) So, the solution to the given initial-value problem is: \(y(t) = \frac{13}{9}\sinh(3t) - \frac{13}{9}\sin(2t) - 3\cosh(3t)\)

Key Concepts

Initial-Value Problem

An initial-value problem in mathematics involves solving differential equations with specific initial conditions. This means you not only solve the equation but also find a specific solution that satisfies given values at the start of the problem. The differential equation here is a second order linear equation:

• The equation: \(y'' - 9y = 13\sin 2t\).
• Initial conditions are: \(y(0) = 3\) and \(y'(0) = 1\).

The goal is to find a function \(y(t)\) that satisfies both the equation and the initial conditions. When solving such problems using Laplace transforms, initial conditions become initial terms that help simplify the Laplace-transformed equation. This practical approach allows us to handle differentials and find solutions that naturally fit into applied situations. Breaking the problem down into smaller parts makes it easier to solve and understand the overall behavior of the equation.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to simplify complex rational expressions into simpler fractions that are easier to work with. When a rational expression, like \(\frac{13}{(s^2 - 9)(s^2 + 4)}\), is decomposed, it can be broken down into a sum of fractions. This is helpful for finding inverse Laplace transforms because the inverse is more straightforward for simpler fractions.The decomposition process involves:

• Assuming a form: \(\frac{As + B}{s^2 - 9} + \frac{Cs + D}{s^2 + 4}\).
• Solving for coefficients \(A, B, C,\) and \(D\). For the given expression:
• Coefficients found were \(A = 0\), \(B = \frac{13}{9}\), \(C = 0\), and \(D = \frac{-13}{9}\).

By decomposing the fraction, each part can be individually transformed back into the time domain, facilitating the solution of differential equations. It reduces the complexity of the problem, making the calculation of inverse transforms doable thorough algebraic manipulation.

Inverse Laplace Transform

The inverse Laplace transform is a process that converts functions in the Laplace domain back into the time domain. When tackling differential equations, transforming equations using the Laplace method often results in a transformed function \(Y(s)\). Reverting \(Y(s)\) back to \(y(t)\) is necessary to find the time-dependent solution.For example, in this exercise, we have the Laplace-transformed function:

• \(Y(s) = \frac{13}{9}\frac{1}{s^2 - 9} - \frac{13}{9}\frac{1}{s^2 + 4} - \frac{3s}{s^2 - 9}\).

The inverse of each component term is computed separately by matching them with known Laplace pairs:

• \(\frac{1}{s^2 - 9}\) corresponds to \(\sinh(3t)\),
• \(\frac{1}{s^2 + 4}\) corresponds to \(\sin(2t)\),
• \(\frac{s}{s^2 - 9}\) corresponds to \(\cosh(3t)\).

Combining these inverses gives the final solution, \(y(t) = \frac{13}{9}\sinh(3t) - \frac{13}{9}\sin(2t) - 3\cosh(3t)\). This solution satisfies both the differential equation and the initial conditions, revealing the inherent behavior of the system over time.