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 \(6.3 .\) $$ y^{\prime \prime}+y=\left\\{\begin{array}{ll}{t,} & {0 \leq t<1,} \\ {0,} & {1 \leq t<\infty ; \quad y(0)=0, \quad y^{\prime}(0)=0}\end{array}\right. $$

Short answer

In summary, we solved the given initial value problem in the Laplace transform domain and found that: $$ Y(s) = \frac{1}{(s^2 + 1)s^2}(1-e^{-s}) = \frac{1}{(s^2 + 1)s^2} -\frac{e^{-s}}{(s^2 + 1)s^2}. $$ To find the solution of the initial value problem, we can apply the inverse Laplace transform to $Y(s)$ using the method developed in Section 6.3.

Step-by-step solution

Write out the given IVP

First, let's write out the given initial value problem: $$ y^{\prime \prime} + y = f(t), \quad y(0) = 0, \quad y^{\prime}(0) = 0, $$ where $$ f(t) =\left\{\begin{array}{ll}{t,} & {0 \leq t<1,} \\\ {0,} & {1 \leq t<\infty .}\end{array}\right. $$

Find the Laplace transform of the given IVP

Now we find the Laplace transform of the given IVP. Recall that the Laplace transform is a linear operator and can be distributed over the sum and derivative operations in the IVP. Moreover, note that \(\mathcal{L}[y^{\prime \prime}] = s^2Y(s) - sy(0) - y^{\prime}(0)\), \(\mathcal{L}[y] = Y(s)\) and \(\mathcal{L}[t] = \frac{1}{s^2}\). Thus, we have $$ s^2Y(s) - 0 - 0 + Y(s) = \frac{1}{s^2} - \frac{1}{s^2}e^{-s}. $$ This simplifies to $$ \left(s^2 + 1\right)Y(s) = \frac{1}{s^2}(1-e^{-s}). $$

Solve the equation for Y(s)

Now, let's solve the equation for \(Y(s)\), we can do this by dividing both sides by \((s^2 + 1)\): $$ Y(s) = \frac{1}{(s^2 + 1)s^2}(1-e^{-s}). $$

Split the term with e^{-s} into separate fractions

To help with finding the inverse Laplace transform, rewrite the expression for \(Y(s)\) as: $$ Y(s) =\frac{1}{(s^2 + 1)s^2} -\frac{e^{-s}}{(s^2 + 1)s^2}. $$

Mention the method to find the inverse Laplace transform

Now that we have found \(Y(s)\), we can use the method developed in Section 6.3 to find the inverse Laplace transform and obtain the solution of the initial value problem in the form \(y(t) = \mathcal{L}^{-1}[Y(s)]\).

Key concepts

Inverse Laplace Transform

The inverse Laplace transform is a powerful tool in solving problems involving differential equations. It's the process of transforming a function in the Laplace domain back to the time domain. Consider it as the reverse operation of the Laplace transform.
In the context of the given exercise, we first applied the Laplace transform to convert the differential equation into an algebraic equation in terms of a function, here denoted as \(Y(s)\). Having found \(Y(s)\), obtaining the solution in the time domain involves finding the inverse Laplace transform \(y(t) = \mathcal{L}^{-1}[Y(s)]\).
To find the inverse, the method mentioned in Section 6.3 can be used. This usually involves using predefined tables of Laplace transforms or mathematical operations such as partial fraction decomposition, which convert complex fractions into simpler terms for which the inverse transforms are known. By applying these techniques, one effectively reverses the process, translating back the components of \(Y(s)\) to their original time-domain solutions.

Initial Value Problems

Initial value problems (IVPs) are a common type of problem in differential equations where the solution is determined by initial conditions. In these problems, you know the value of the function and its derivatives at a specific point in time, typically \(t = 0\).
In our exercise, the initial conditions provided are \(y(0) = 0\) and \(y'(0) = 0\). These conditions tell us the starting point for our differential equation's solution. They are crucial because they help in uniquely determining the solution for the differential equation.
When applying the Laplace transform to such problems, these initial conditions are directly applied. For instance, in the Laplace domain, the conditions appear as constants in our transformed algebraic equation, simplifying the process of finding \(Y(s)\). Thus, these initial values guide us through the correct pathway to find the intended solution of the differential equation.

Differential Equations

Differential equations are equations that involve an unknown function and its derivatives. They are used to describe various physical phenomena such as motion, heat, and electricity.
In this exercise, we are dealing with a second-order linear differential equation: \(y'' + y = f(t)\). Here, \(y(t)\) is the unknown function, and \(f(t)\) represents an external force or input. For specific segments of the timeline, \(f(t)\) is a piecewise function, starting with \(t\) for \(0 \leq t < 1\) and zero for \(t \geq 1\).
Solving such differential equations involves understanding both the given conditions, like initial values, and the dynamic defined by \(f(t)\). The Laplace transform simplifies the process by converting the problem into the algebraic realm, where calculus becomes algebra. This transition makes finding solutions more straightforward, especially when combined with the inverse Laplace transform to return to the time domain.