Problem 64 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

The solution is: \( y(t) = 1 - e^{-t} \) for \( 0 \leq t < 1 \), and \( y(t) = -1 - e^{-t} + 2e^{-(t-1)} \) for \( t \geq 1 \).

Step 1: Take the Laplace Transform

First, apply the Laplace transform to the differential equation. The Laplace of the derivative term is \( \mathcal{L}\{y'\} = sY(s) - y(0) \). The given initial value is \( y(0) = 0 \). Thus, the Laplace transform of the equation becomes: \( sY(s) + Y(s) = F(s) \), where \( F(s) \) is the Laplace transform of \( f(t) \).

Step 2: Determine the Laplace Transform of f(t)

Since \( f(t) \) is a piecewise function, write it using unit step functions: \( f(t) = 1 - 2u(t-1) \), where \( u(t-1) \) is the unit step function at \( t = 1 \). The Laplace transform is: \( F(s) = \frac{1}{s} - \frac{2e^{-s}}{s} \).

Step 3: Solve for Y(s) in the Transformed Equation

We now have \( sY(s) + Y(s) = \frac{1}{s} - \frac{2e^{-s}}{s} \). Factor out \( Y(s) \): \[ (s + 1)Y(s) = \frac{1}{s} - \frac{2e^{-s}}{s} \]. Solve for \( Y(s) \): \[ Y(s) = \frac{1}{s(s+1)} - \frac{2e^{-s}}{s(s+1)} \].

Step 4: Perform Partial Fraction Decomposition

For the first term, decompose \( \frac{1}{s(s+1)} \) into partial fractions: \( \frac{1}{s} - \frac{1}{s+1} \). The second term remains: \( \frac{2e^{-s}}{s(s+1)} \). So, \[ Y(s) = (\frac{1}{s} - \frac{1}{s+1}) - 2\frac{e^{-s}}{s(s+1)} \].

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

Apply the inverse Laplace transform: \( \mathcal{L}^{-1}\{\frac{1}{s}\} = 1 \), and \( \mathcal{L}^{-1}\{\frac{1}{s+1}\} = e^{-t} \). For \( \mathcal{L}^{-1}\{\frac{2e^{-s}}{s(s+1)}\} \), use the shifting theorem: \( 2u(t-1)(1 - e^{-(t-1)}) \). Thus, \[ y(t) = 1 - e^{-t} - 2u(t-1)(1 - e^{-(t-1)}) \].

Step 6: Simplify the Solution

Simplify the expression: For \( 0 \leq t < 1 \), \( y(t) = 1 - e^{-t} \); for \( t \geq 1 \), \( y(t) = 1 - e^{-t} - 2(1 - e^{-(t-1)}) \), which simplifies to \( y(t) = 1 - e^{-t} + 2e^{-(t-1)} - 2 \).

Key Concepts

Initial Value Problem

An Initial Value Problem (IVP) is a type of differential equation along with specific values at a starting point. In this case, we have the equation \( y^{\prime} + y = f(t) \) with the initial condition \( y(0) = 0 \). This means at time \( t = 0 \), the value of \( y \) is specified to be 0. IVPs are crucial in determining unique solutions to differential equations, as they establish a baseline condition for solving the system. Without an initial condition, the solution could drift to any number of possibilities. By knowing the initial value, we can find a unique trajectory of the solution over time.

Piecewise Function

A piecewise function is a function composed of multiple sub-functions, with each sub-function applying to a certain interval of the main function's domain. In this IVP, \( f(t) \) is defined piecewise as:

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

This means that \( f(t) \) behaves differently before and after \( t = 1 \). Such functions are helpful in modeling systems that experience a change at a specific point in time. Here, it's represented using unit step functions to convert it into a form suitable for the Laplace transform, allowing us to handle the different behaviors effectively.

Inverse Laplace Transform

The inverse Laplace transform is the process of converting a function in the frequency domain, \( Y(s) \), back to the time domain, \( y(t) \). Once we've transformed and manipulated our equations in the \( s \)-domain, we need to go back to the original time domain to find our solution, \( y(t) \). For example, using the known inverse Laplace transforms:

• \( \mathcal{L}^{-1}\{\frac{1}{s}\} = 1 \)
• \( \mathcal{L}^{-1}\{\frac{1}{s+1}\} = e^{-t} \)

In the problem, the shifting theorem is also employed to find the inverse Laplace transform of terms with exponentials, showing how different parts of \( Y(s) \) are mapped back to time functions. Only by performing this can one fully grasp the dynamics tackled by the original problem.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to express a complex fraction as a sum of simpler fractions, making it easier to perform operations like the inverse Laplace transform. In this exercise, the term \( \frac{1}{s(s+1)} \) is decomposed into \( \frac{1}{s} - \frac{1}{s+1} \). This simplification allows us to directly apply inverse Laplace transforms to each part separately. By doing so, we turn a difficult problem into manageable pieces, streamlining the process of converting \( Y(s) \) back into \( y(t) \). Partial fractions are essential when dealing with rational functions in Laplace transforms, especially when inverse transforming products of simpler transforms.