Question

In Exercises \(1-20\) use the Laplace transform to solve the initial value problem. Where indicated by \(\mathrm{C} / \mathrm{G}\), graph the solution. $$ y^{\prime \prime}-5 y^{\prime}+4 y=\left\\{\begin{array}{rl} 1, & 0 \leq t<1 \\ -1, & 1 \leq t<2, \\ 0, & t \geq 2 \end{array} \quad y(0)=3, \quad y^{\prime}(0)=-5\right. $$

Short answer

Question: Solve the following initial value problem (IVP) involving a second-order linear differential equation with piecewise constant forcing function and provide the solution y(t): \(y''(t) - 5y'(t) + 4y(t) = \begin{cases}1, & 0 \le t < 1\\ -1, & 1 \le t < 2\\ 0, & t \ge 2 \end{cases}\), with \(y(0)=3\) and \(y'(0)=-5\). Solution: \(y(t) = -5e^{4t} + 8e^{t} + e^{-t}u(t-1) -e^{-2t}u(t-2)\), where u(t) is the unit step function.

Step-by-step solution

Apply the Laplace transform

First, we apply the Laplace transform to both sides of the given differential equation, recalling that the Laplace transform of a derivative is given by, \(\mathcal{L}(y') = sY(s)-y(0)\) and \(\mathcal{L}(y'') = s^2Y(s) - sy(0) - y'(0)\). The Laplace transform of the differential equation is then: $$\begin{aligned} s^2 Y(s) - sy(0) - y'(0) -5[sY(s)-y(0)]+4Y(s) = \mathcal{L}\\{1\\}(s) -\mathcal{L}\\{-1\\}(s)u_1(s)+\mathcal{L}\\{0\\}(s)u_2(s) \end{aligned}$$

Solve for Y(s)

Now, plug in the initial conditions \(y(0)=3\) and \(y'(0)=-5\), and use the known Laplace transforms for the constants. Rearrange the equation to solve for \(Y(s)\): $$\begin{aligned} Y(s) = \frac{1}{s^2-5s+4}\left(3s+8+\frac{e^{-s}}{s} - \frac{e^{-2s}}{s}\right) \end{aligned}$$

Find the inverse Laplace transform

Now find the inverse Laplace transform of \(Y(s)\) to obtain the solution y(t). First, perform partial fraction decomposition on the term \(\frac{3s+8}{s^2-5s+4}\): $$\frac{3s+8}{s^2-5s+4} = \frac{A}{s-4}+\frac{B}{s-1}$$ Finding A and B, we get: $$A = -5, \quad B = 8$$ The given inverse Laplace transforms can be computed using the theorems of inverse Laplace transforms for shifted exponential functions and shifted inverse Laplace transforms for constant functions: $$\begin{aligned} y(t) = -5e^{4t} + 8e^{t} + e^{-t}u(t-1) -e^{-2t}u(t-2) \end{aligned}$$

Plot the solution (if asked)

We were not asked to plot the solution in this problem, so we are done. If asked, we would sketch a graph of the expression obtained in Step 3, using the given conditions and the provided forcing function intervals.

Key concepts

Differential Equations

Differential equations are mathematical tools that describe relationships involving rates of change and initial values of physical or other quantities. They contain derivatives, which are measures of how a quantity changes as another one changes. For example, if we look at the exercise where the equation is given by y'' - 5y' + 4y = piecewise function, this describes how the function y(t), which could represent a physical process like temperature or position, evolves over time.

Solving the differential equation involves finding the function y(t) that satisfies this equation for all t. The solution typically requires initial conditions, which are the values of the function and its derivatives at the initial time. These conditions help determine the unique solution curve among the many that are theoretically possible. In our case, the initial conditions are y(0) = 3 and y'(0) = -5.

Inverse Laplace Transform

The inverse Laplace transform is a process applied to convert a given Laplace transform back into its original time-domain form. It's a crucial step for solving differential equations in the frequency or 's' domain. Essentially, the Laplace transform is used to simplify the differential equation by turning derivatives into polynomials, which are generally easier to handle. After solving these simpler equations for Y(s), we must revert this solution back into the time domain to interpret it in a physically meaningful way.

In our exercise, once Y(s) is found, we use the inverse Laplace transform to determine the function y(t). The computation often requires looking up Laplace pairs and potentially using complex mathematics such as partial fraction decomposition to break down fractions into simpler 'inverse-able' parts.

Initial Conditions

Initial conditions are specific values of a function and its derivatives at the start of the observation, which in many physics and engineering problems, is t = 0. These values are essential in defining a unique solution to a differential equation. Without initial conditions, you may end up with a family of solutions rather than a specific answer.

The term 'initial value problem' arises in the context of differential equations, where the problem is to find a solution that satisfies both the differential equation and the initial conditions. In the provided exercise, the initial conditions given are y(0) = 3, and y'(0) = -5. When using the Laplace transform to solve differential equations, these initial conditions help us create a solvable equation for Y(s) in the 's' domain before applying the inverse transform.

Partial Fraction Decomposition

Partial fraction decomposition is a technique to break down complex rational expressions into simpler, more manageable pieces. It's especially useful in the context of inverse Laplace transforms, as it allows us to transform complicated fractions into a sum of simpler terms each with its own inverse Laplace transform.

In the step-by-step solution, we decompose the expression (3s + 8) / (s^2 - 5s + 4) into a sum of fractions with numerators A and B to make it easier to apply the inverse Laplace transform. The values of A and B are determined by the method of equating coefficients or by plugging in convenient values of s to form a system of linear equations. Doing this allows us to express Y(s) in a form where the inverse Laplace can be directly applied to each term.