Problem 66 - Advanced Engineering Mathematics

Question

In Problems, use the Laplace transform to solve the given initial-value problem. $$ \begin{gathered} y^{\prime \prime}+4 y=f(t), \quad y(0)=0, y^{\prime}(0)=-1, \text { where } \\\ f(t)=\left\\{\begin{array}{lr} 1, & 0 \leq t<1 \\ 0, & t \geq 1 \end{array}\right. \end{gathered} $$

Step-by-Step Solution

Answer

Apply Laplace transforms, solve for \(Y(s)\), and use inverse Laplace to find \(y(t)\).

Step 1: Apply the Laplace Transform

Begin by applying the Laplace transform to both sides of the differential equation \(y'' + 4y = f(t)\). The Laplace transform of \(y''\) is \(s^2Y(s) - sy(0) - y'(0)\), and \(f(t)\) as a piecewise function can be written in Laplace terms.

Step 2: Substitute Initial Conditions

Substitute the initial conditions \(y(0) = 0\) and \(y'(0) = -1\) into the Laplace transformed equation. This leads to \((s^2Y(s) - 0 + 1) + 4Y(s) = \mathcal{L}\{f(t)\}\). Simplify this to \(s^2Y(s) + 4Y(s) = \mathcal{L}\{f(t)\} - 1\).

Step 3: Solve for Y(s)

Factor out \(Y(s)\) from the left-hand side to get \(Y(s)(s^2 + 4) = \mathcal{L}\{f(t)\} - 1\). Solve for \(Y(s)\) by dividing both sides by \(s^2 + 4\), resulting in \(Y(s) = \frac{\mathcal{L}\{f(t)\} - 1}{s^2 + 4}\).

Step 4: Determine \(\mathcal{L}\{f(t)\}\)

Determine the Laplace transform of the piecewise function \(f(t)\). Recognize that \(f(t)\) can be expressed using Heaviside step functions: \(f(t) = 1 - u_1(t)\) where \(u_1(t)\) is the step function that turns on at \(t = 1\). \(\mathcal{L}\{f(t)\} = \frac{1}{s} - \frac{e^{-s}}{s}\).

Step 5: Substitute \(\mathcal{L}\{f(t)\}\) into Y(s)

Substitute the expression for \(\mathcal{L}\{f(t)\}\) into the expression for \(Y(s)\), yielding \(Y(s) = \frac{(\frac{1}{s} - \frac{e^{-s}}{s}) - 1}{s^2 + 4}\). Simplify the numerator to get \(\frac{1 - e^{-s} - s}{s}\).

Step 6: Simplify Y(s)

Simplify the expression for \(Y(s)\), resulting in \(Y(s) = \frac{1}{s(s^2 + 4)} - \frac{e^{-s}}{s(s^2 + 4)} - \frac{1}{s^2 + 4}\).

Step 7: Apply the Inverse Laplace Transform

Take the inverse Laplace transform of each term in \(Y(s)\). The inverse Laplace of \(\frac{1}{s(s^2 + 4)}\) is a convolution, the inverse of \(\frac{1}{s^2 + 4}\) yields \(\sin(2t)\), and the inverse for the term involving \(e^{-s}\) is shifted. Combine all this information to find \(y(t)\).

Step 8: Write the Final Solution for y(t)

Combine all components to express \(y(t)\) in its piecewise form with respect to \(f(t)\). This will involve using properties of convolutions and Heaviside step functions to appropriately express \(y(t)\) in terms of \(t\).

Key Concepts

Initial-Value Problem

An initial-value problem in mathematics is a type of differential equation along with a specified condition, known as an initial condition. This defines the value of the solution at a certain point. In our exercise, we are given:

• The differential equation: \( y'' + 4y = f(t) \)
• Initial conditions: \( y(0) = 0 \) and \( y'(0) = -1 \)

This means that at time \( t = 0 \), the function \( y(t) \) starts at 0 and its initial velocity (or first derivative) is -1.
Solving an initial-value problem involves finding the function \( y(t) \) that satisfies both the differential equation and the initial conditions at the starting point, enabling us to describe the system’s behavior from initial state conditions through time.

Piecewise Function

A piecewise function is a type of function that has different expressions or rules for different intervals of its domain. In this particular problem, our piecewise function is defined as:

• \( f(t) = 1 \) for \( 0 \leq t < 1 \)
• \( f(t) = 0 \) for \( t \geq 1 \)

Here, the function \( f(t) \) changes behavior at \( t=1 \). This type of function is crucial in many real-world scenarios where a system can have different behaviors over different time periods or conditions.
In mathematical computations, piecewise functions require special consideration, particularly when finding a standard form expression for transformations, such as Laplace transforms.

Heaviside Step Function

The Heaviside step function, often denoted as \( u(t-a) \), is a mathematical function that represents an instantaneous jump or "step" starting at a specified time \( a \). It is defined as:

• \( u(t-a) = 0 \) for \( t < a \)
• \( u(t-a) = 1 \) for \( t \ge a \)

In our exercise, the transformation of a piecewise function into a Laplace-friendly form often involves the usage of Heaviside functions. The function \( f(t) = 1 - u_1(t) \) indicates that the function is initially 1 and then switches to 0 after \( t = 1 \). This is extremely useful for dealing with problems in physics and engineering, where systems may turn on and off abruptly.

Inverse Laplace Transform

The inverse Laplace transform is a mathematical process for finding the original function from its Laplace transform. It transforms functions back from the frequency domain to the time domain, which is critical in solving differential equations that are presented in Laplace form.
In this exercise, once we find \( Y(s) \), we need to convert it back to \( y(t) \). The inverse Laplace transform of terms is handled as follows:

• The inverse of \( \frac{1}{s(s^2 + 4)} \) involves convolution operations.
• \( \frac{1}{s^2 + 4} \) directly transforms to \( \sin(2t) \).
• Terms like \( \frac{e^{-s}}{s(s^2 + 4)} \) will require application of shifting theorems.

Utilizing these techniques, we effectively transition back to the time domain and construct the full solution for \( y(t) \). Understanding inverse Laplace transforms is essential for interpreting and solving time-dependent behavior in various applications.