Problem 66 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

The solution is \(y(t) = \frac{1}{2} - \frac{1}{2}u(t-1) - \sin(2t)\).

Step 1: Convert the problem using Laplace Transforms

First, we'll take the Laplace transform of the entire differential equation. The Laplace transform of \(y''+4y=f(t)\) is \(Y(s)(s^2+4) - sy(0) - y'(0) = F(s)\). Given the initial conditions \(y(0)=0\) and \(y'(0)=-1\), we substitute these into the equation: \(Y(s)(s^2+4) + 1 = F(s).\)

Step 2: Find the Laplace Transform of the function

The function \(f(t)\) is a piecewise function. Its Laplace transform is \(F(s) = \int_0^1 e^{-st} \, dt\), because \(f(t)=1\) for \(0 \leq t < 1\) and 0 otherwise. Evaluating this integral gives \(F(s) = \frac{1}{s} (1 - e^{-s})\).

Step 3: Solve for Y(s)

Substitute the Laplace transform of the piecewise function \(F(s) = \frac{1}{s}(1-e^{-s})\) into the transformed differential equation: \[Y(s)(s^2+4) + 1 = \frac{1}{s}(1-e^{-s})\]. Solve for \(Y(s)\) to get \[ Y(s) = \frac{1}{s(s^2+4)}(1-e^{-s}) - \frac{1}{s^2+4} \].

Step 4: Simplify and take Inverse Laplace Transform

Rewrite \(Y(s)\) and use partial fraction decomposition if necessary: \[ Y(s) = \frac{1}{s(s^2+4)} - \frac{e^{-s}}{s(s^2+4)} - \frac{1}{s^2+4} \].Take the inverse Laplace transform of each term separately:- \(\mathcal{L}^{-1}\left\{\frac{1}{s(s^2+4)}\right\}\) is a combination of basic transforms, yielding the time function involving step functions.- \(\mathcal{L}^{-1}\left\{\frac{e^{-s}}{s(s^2+4)}\right\}\) gives a delayed response function.- \(\mathcal{L}^{-1}\left\{\frac{1}{s^2+4}\right\}\) is a sine function. Combine these results to find \(y(t)\).

Step 5: Combine and construct y(t)

Considering the results of each inverse transform, we reconstruct \(y(t)\) by combining any Heaviside functions and specific solutions. For the function given by \(f(t)\), the final solution sums up to: \[ y(t) = \frac{1}{2} - \frac{1}{2}u(t-1) - \sin(2t) \]where \(u(t-1)\) is the Heaviside step function that turns on at \(t=1\).

Key Concepts

Initial-Value Problem

An initial-value problem in mathematics is a type of differential equation with given conditions. The conditions provide specific information about the function and its derivatives at a certain point. For the problem here, you're given two conditions at \( t = 0 \):

• \(y(0) = 0\)
• \(y'(0) = -1\)

These values establish what the function and its rate of change are expected to be when time \( t = 0 \). Initial-value problems are crucial because they turn a general differential equation into a specific one that can be solved uniquely. It's like having a map and a starting point, which together define the exact path you need to follow to reach your destination. By incorporating these initial conditions, we can use methods like the Laplace transform to find a unique solution for \(y(t)\).

Piecewise Function

A piecewise function is a function composed of multiple sub-functions, each of which applies to a certain interval of the main function's domain. The function \( f(t) \) in this problem is an example:

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

This means that the function behaves one way for part of its domain and another way for the rest.
Piecewise functions are often used to represent real-world scenarios where conditions change at specific intervals. For example, a business might have different shipping charges based on the weight of an item. In the Laplace transform process, knowing how to deal with piecewise functions lets us correctly apply mathematical operations over specific intervals, allowing us to solve the differential equation and ultimately reconstruct the solution over time.

Inverse Laplace Transform

The inverse Laplace transform is essentially the process of transforming a function from the Laplace domain back to the time domain. It's like translating a language back to its original form. After converting a differential equation into the Laplace domain to simplify solving, you use the inverse Laplace transform to find the actual function \( y(t) \).
In this exercise, we decompose \( Y(s) \) into simpler parts to compute their inverse Laplace individually. This involves partial fraction decomposition when necessary, which simplifies complex functions into a sum of simpler ones. By utilizing properties of basic Laplace transforms, we derive the time-based solution. This process is crucial for solving initial-value problems with piecewise functions as it allows easy manipulation of functions defined by multiple conditions. When completed, you get the specific behavior of \( y(t) \) that satisfies the original differential equation and initial conditions.

Heaviside Step Function

The Heaviside step function, often symbolized as \( u(t-a) \), is a special function used to "turn on" or "activate" a certain behavior in a function at a specific point \( a \). In this problem, it is used to account for changes in the piecewise function \( f(t) \).
The Heaviside function is defined as:

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

This allows it to switch function behaviors. For example, in the solution \( y(t) = \frac{1}{2} - \frac{1}{2} u(t-1) - \sin(2t) \), the \( -\frac{1}{2} u(t-1) \) part indicates a subtractive effect that kicks in at \( t = 1 \). This aligns with the piecewise definition of \( f(t) \), which changes from 1 to 0 at \( t = 1 \). Using the Heaviside function makes handling these "switching" behaviors mathematically concise and prevents mistakes in defining intervals incorrectly.