Problem 69 - Advanced Engineering Mathematics

Question

In Problems, use the Laplace transform to solve the given initial-value problem. $$ \begin{gathered} y^{\prime \prime}+y=f(t), \quad y(0)=0, y^{\prime}(0)=1, \text { where } \\ f(t)=\left\\{\begin{array}{ll} 0, & 0 \leq t<\pi \\ 1, & \pi \leq t<2 \pi \\ 0, & t \geq 2 \pi \end{array}\right. \end{gathered} $$

Step-by-Step Solution

Answer

The solution is a piecewise function: \( y(t) = \sin t \) for \( 0 \leq t < \pi \), \( 0 \) for \( \pi \leq t < 2\pi \), and \( \sin(t-2\pi) \) for \( t \geq 2\pi \).

Step 1: Take the Laplace Transform

Take the Laplace transforms of both sides of the differential equation. The Laplace transform of the zero initial conditions are as follows: \[ \mathcal{L} \{ y'' \} = s^2 Y(s) - s y(0) - y'(0) = s^2 Y(s) - 1. \] Since the initial conditions are provided as \(y(0) = 0\) and \(y'(0) = 1\). Also, \[ \mathcal{L} \{ y \} = Y(s). \] The Laplace transform of the RHS is \(\mathcal{L} \{ f(t) \} \).

Step 2: Express Piecewise Function Using Heaviside Functions

Express the function \(f(t)\) in terms of Heaviside step functions. \[ f(t) = u_\pi(t) - u_{2\pi}(t), \] where \(u_a(t)\) is the Heaviside step function that shifts at \(t=a\).

Step 3: Laplace Transform of Heaviside Functions

Find the Laplace transform of \(f(t)\). Using properties of the Laplace transform, we know:\[ \mathcal{L}\{u_\pi(t) \} = \frac{e^{-\pi s}}{s}, \] \[ \mathcal{L}\{u_{2\pi}(t) \} = \frac{e^{-2\pi s}}{s}. \] Thus, \[ \mathcal{L}\{f(t)\} = \frac{e^{-\pi s}}{s} - \frac{e^{-2\pi s}}{s}. \]

Step 4: Solve for Y(s) using Algebraic Manipulation

Set up the equation from the transforms:\[ s^2 Y(s) + Y(s) = \mathcal{L} \{ f(t) \} + 1, \]\[ (s^2 + 1) Y(s) = \frac{e^{-\pi s}}{s} - \frac{e^{-2\pi s}}{s} + 1. \]Solve for \( Y(s) \):\[ Y(s) = \frac{\frac{e^{-\pi s}}{s} - \frac{e^{-2\pi s}}{s} + 1}{s^2 + 1}. \]

Step 5: Inverse Laplace Transform to Find y(t)

Perform the inverse Laplace transform to find \(y(t)\).1. Calculate the inverse transform: \[ \mathcal{L}^{-1}\left( \frac{1}{s^2 + 1} \right) = \sin t. \] 2. Use the linearity of the Laplace transform to break the expression into simpler parts and use the shifting theorem.3. Transform results to obtain \(y(t)=\left(\sin(t-\pi) \cdot u_{\pi}(t) - \sin(t-2\pi) \cdot u_{2\pi}(t) \right) + \sin t.\) Overall, after combining terms:\[ y(t) = \begin{cases}\sin t, & 0 \leq t < \pi,\0, & \pi \leq t < 2\pi,\\sin(t-2\pi), & t \geq 2\pi.\end{cases} \]

Key Concepts

Heaviside step functions

Heaviside step functions are invaluable in solving differential equations, particularly when dealing with functions that switch values at certain points in time. These functions are essentially a way to represent a piecewise function in a mathematical form that is more convenient for analysis, especially in the context of Laplace transforms. The Heaviside step function, denoted as \(u_a(t)\), changes value from zero to one at the point \(t = a\).

In our exercise, we express the given function \(f(t)\) as a combination of Heaviside step functions:

• \(u_\pi(t)\) accounts for the function starting at \(t = \pi\).
• \(u_{2\pi}(t)\) accounts for the function returning to zero at \(t = 2\pi\).

Using Heaviside step functions simplifies the use of Laplace transforms, as it allows us to handle complex piecewise functions using simple algebraic expressions.

Inverse Laplace transform

The inverse Laplace transform is a critical step in solving differential equations using Laplace transforms. This process converts a function in the Laplace domain back to the time domain, where it can be more easily interpreted and utilized. In the given exercise, after solving for \( Y(s) \) in the Laplace domain, the next step is to apply the inverse Laplace transform to find the solution \( y(t) \).

The inverse transform involves using known transforms:

• For example, \( \mathcal{L}^{-1}\left( \frac{1}{s^2 + 1} \right) = \sin t \).

Additionally, properties such as the shifting theorem are used to handle exponential terms that arise from translating steps in the original function. By separating \( Y(s) \) into manageable parts and applying inverse transformations individually, we convert \( Y(s) \) into \( y(t) \). As seen in the example, this results in a time-domain function expressed in terms quickly understood by translating shifts through sine functions.

Initial-value problems

Initial-value problems are a class of problems involving differential equations where the solution must satisfy specific initial conditions. In the context of Laplace transforms, these conditions are crucial. They allow us to transform differential equations into algebraic equations that are easier to solve. The given exercise provides two initial conditions:

• \( y(0) = 0 \)
• \( y'(0) = 1 \)

These are used to simplify the Laplace transform of the differential equation by eliminating terms associated with initial values from the transform of \( y'' \).

Once the Laplace-transformed equation is solved for \( Y(s) \), the inverse Laplace transform is used to revert this solution back to the time domain, giving us the particular solution \( y(t) \) that satisfies the given initial conditions. Initial-value problems thus guide the solution's behavior at the outset and ensure that the mathematical model reflected by the solution accurately aligns with the real-world scenario being described.