Question

Find the Laplace transform \(Y(s)=\mathcal{L}[y]\) of the solution of the given initial value problem. A method of determining the inverse transform is developed in Section \(6.3 .\) $$ y^{\prime \prime}+4 y=\left\\{\begin{array}{ll}{1,} & {0 \leq t<\pi,} \\ {0,} & {\pi \leq t<\infty}\end{array} \quad y(0)=1, \quad y^{\prime}(0)=0\right. $$

Short answer

In this exercise, we found the Laplace transform of the solution to the given initial value problem. We first found the Laplace transform of the equation with the given forcing function and incorporated the initial conditions. Next, we solved for the Laplace transform Y(s) in both conditions (for 0 ≤ t < π and for π ≤ t < ∞). Our final result was a piecewise function: $$ Y(s)=\left\{\begin{array}{ll}{\frac{1+s}{s(s^2+4)},} & {0 \leq t<\pi,} \\\ {0,} & {\pi \leq t<\infty}\end{array}\right. $$ This represents the Laplace transform of the solution to the initial value problem. The inverse Laplace transform to find the function y(t) will be covered in a later section of the textbook.

Step-by-step solution

Find the Laplace transform of the equation

First, we'll find the Laplace transform of the entire equation. Recall that the Laplace transform of a derivative is given by: $$ \mathcal{L}[y'](s)=sY(s) - y(0) $$ And for the second derivative, the Laplace transform is: $$ \mathcal{L}[y''(s)]=s^2 Y(s) - sy(0) - y'(0) $$ Now apply the Laplace transform to the given equation: $$ s^2 Y(s) - sy(0) - y'(0) + 4Y(s) = \mathcal{L}[1] \text{ for } 0\leq t < \pi \\ s^2 Y(s) - sy(0) - y'(0) + 4Y(s) = \mathcal{L}[0] \text{ for } \pi\leq t < \infty $$

Apply the initial conditions and simplify

Now let's plug in the initial conditions: \(y(0)=1\) and \(y'(0)=0\). Thus, we get: $$ s^2 Y(s) - s + 4Y(s) = \frac{1}{s} \text{ for } 0\leq t < \pi \\ s^2 Y(s) + 4Y(s) = 0 \text{ for } \pi\leq t <\infty $$

Solve both equations for \(Y(s)\)

Now we can solve both equations for \(Y(s)\): $$ Y(s)=\frac{1+s}{s(s^2+4)} \text{ for }0\leq t < \pi\\ Y(s)=\frac{0}{(s^2+4)}=0 \text{ for } \pi\leq t <\infty $$ As we have both conditions for \(Y(s)\), we can combine them into a single piecewise function: $$ Y(s)=\left\\{\begin{array}{ll}{\frac{1+s}{s(s^2+4)},} & {0 \leq t<\pi,} \\\ {0,} & {\pi \leq t<\infty}\end{array}\right. $$ Now we have found the Laplace transform \(Y(s)\) of the solution for the given initial value problem. Note that the method for determining the inverse Laplace transform, which would give the function \(y(t)\), is not shown in this explanation since it will be covered in Section 6.3 of the textbook as mentioned in the exercise.

Key concepts

Differential Equations

Differential equations are mathematical equations that involve an unknown function and its derivatives. They are used to model relationships and dynamics in various fields, like physics, engineering, and economics. A differential equation can be classified by its order, which is determined by the highest derivative present in the equation. In our given exercise, we are dealing with a second-order differential equation because the highest derivative is a second derivative, denoted as \(y''\).
The purpose of solving a differential equation is to find the unknown function, usually denoted as \(y(t)\), that satisfies the equation. In the context of this problem, the equation \(y'' + 4y\) involves finding a function \(y(t)\) that describes how the system behaves over time. Solving this differential equation provides insight into the system's dynamics, which can then be analyzed to understand or predict behavior in real-world scenarios, such as oscillating springs or circuits.

Initial Value Problem

An initial value problem (IVP) is a type of differential equation accompanied by additional information, which specifies the values of the unknown function and its derivatives at a given point, usually at \(t = 0\). This makes it possible to find a unique solution to the differential equation.
In our exercise, the initial value problem is defined by the conditions \(y(0) = 1\) and \(y'(0) = 0\). These conditions inform us about the initial state of our system at \(t = 0\).

• The condition \(y(0) = 1\) means that at time zero, the value of our function is 1.
• The condition \(y'(0) = 0\) indicates that initially, the rate of change of \(y(t)\) is zero, suggesting no initial movement or change.

Using these conditions, we can determine the constants that arise when solving the differential equation, leading us to the specific solution that fits the system's original state.

Piecewise Functions

Piecewise functions are functions that have different definitions or expressions for different intervals in their domain. They are particularly useful for modeling situations where a system behaves differently in different scenarios or time periods.
In the original problem, the function defined as part of the differential equation is piecewise:

• For \(0 \leq t < \pi\), the function is constant and equal to 1.
• For \(\pi \leq t < \infty\), the function becomes zero.

This type of function effectively captures changes over time, similar to switching signals or forces acting on a system. In the Laplace transform domain, we express the transformed function as a piecewise function for \(Y(s)\) to reflect the distinct behavior at different time intervals.

Inverse Laplace Transform

The inverse Laplace transform is the process of converting a function from the Laplace domain back into the time domain. In other words, it transforms \(Y(s)\), the Laplace transform of \(y(t)\), back into \(y(t)\) itself. This is crucial because it allows us to find the original function described by the differential equation.
The inverse Laplace transform involves applying complex mathematical techniques, often using tables of known transforms, partial fraction decomposition, and sometimes residue calculus for complex functions. However, as the exercise mentions, developing the method for determining the inverse is covered in Section 6.3, but the key idea is to transition back from \(Y(s)\) to \(y(t)\) in order to find the time-dependent behavior of the system. It is a vital step in completely solving an initial value problem where time-domain solutions are required.