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. $$ y^{\prime \prime}+2 y^{\prime}+y=\left\\{\begin{array}{cl} e^{t}, & 0 \leq t<1, \\ e^{t}-1, & t \geq 1, \end{array} \quad y(0)=3, y^{\prime}(0)=-1\right. $$

Short answer

Question: Solve the initial value problem involving a second order linear differential equation using the Laplace transform, and provide the graph of the solution. Given differential equation: \(y'' + 2y' + y = f(t)\), where \(f(t) = e^t\) for \(0\leq t<1,\) and \(f(t) = e^t -1\) for \(t \geq 1.\) Initial conditions: \(y(0)=3\) and \(y'(0)=-1\).

Step-by-step solution

Apply Laplace transform to the given differential equation

To start, apply the Laplace transform to the given differential equation: $$ \mathcal{L}\{y''\}+\mathcal{L}\{2y'\}+\mathcal{L}\{y\}=\mathcal{L}\{f(t)\}, $$ where \(f(t) = e^t\) for \(0\leq t<1,\) and \(f(t) = e^t -1\) for \(t \geq 1.\)

Apply Laplace transform properties to simplify the equation

Next, apply the respective Laplace transform properties to simplify the equation: $$ s^2Y(s)-sy(0)-y'(0)+2(sY(s)-y(0)) + Y(s) = F(s), $$ where \(Y(s) = \mathcal{L}\{y(t)\}\) and \(F(s)=\mathcal{L}\{f(t)\}\) Now, plug in the initial conditions \(y(0)=3\) and \(y'(0)=-1\): $$ s^2Y(s)-3s+1+6-6s + Y(s) = F(s). $$

Solve for \(Y(s)\) and separate the equation into two cases

Solve for \(Y(s)\) and separate the equation into two cases for the two parts of \(f(t)\): $$ Y(s) = \frac{F(s) + 3s-1-6+6s}{s^2 + 2s + 1}, $$ with: - Case 1: \(f(t) = e^t\) for \(0\leq t<1\), - Case 2: \(f(t) = e^t-1\) for \(t \geq 1\).

Find the inverse Laplace transform for each case

Next, find the inverse Laplace transform for both cases: - Case 1: \(y_1(t) = \mathcal{L}^{-1}\{Y_1(s)\}\) - Case 2: \(y_2(t) = \mathcal{L}^{-1}\{Y_2(s)\}\)

Use the Heaviside function to combine the solutions

Since there's a piecewise function, use Heaviside's function to combine both solutions: $$ y(t) = y_1(t)u_1(t) + y_2(t)u_2(t). $$ where \(u_1(t)=1-t\) for \(0\leq t<1\) and \(u_2(t)=t\) for \(t \geq 1.\)

Graph the solution

Finally, plot the graph of the solution, taking into account the two different cases and their respective Laplace transforms: - For \(0\leq t<1\), graph \(y_1(t)\). - For \(t \geq 1\), graph \(y_2(t)\). Make sure to clearly mark the graphs with labels to differentiate between the two cases.

Key concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They are fundamental in describing the rates at which variables change and are widely used in fields such as physics, engineering, and economics.
When we talk about an initial value problem (IVP) in the context of differential equations, we are referring to a problem where we need to find a function that satisfies a differential equation and also meets specified initial conditions. These conditions are given values for the function and its derivatives at a certain point, which in our problem happens to be when the time, denoted by 't', is equal to zero.
Solving an IVP typically requires finding a solution that satisfies both the differential equation and the initial conditions exactly. This process often involves integrating the differential equation, which can be complex, especially when dealing with higher-order derivatives or piecewise functions, as seen in the given exercise.

Inverse Laplace Transform

The Laplace transform is a powerful tool for solving differential equations, especially when dealing with linear time-invariant systems. It converts a function of time, often a difficult differential equation, into a function of a complex variable, simplifying the equation to algebraic form.
Once we have the algebraic equation in the Laplace domain, the inverse Laplace transform is used to revert back to the time domain, retrieving the solution to the original differential equation. This step is crucial as it translates the problem back into a language that can be understood in terms of the original variables.

Applying the Inverse Laplace Transform

In the given exercise, after applying the Laplace transform to the initial value problem, we obtained an expression for Y(s), which we now need to convert back using the inverse Laplace transform to find the function y(t). The inversion process relies on known transforms of elementary functions or using partial fraction decomposition if needed.

Heaviside Function

The Heaviside function, also known as the unit step function, is crucial when working with piecewise functions in differential equations. It effectively allows us to 'turn on' or 'turn off' parts of a function at specific points, and it's especially useful when dealing with discontinuities or sudden changes in a system.
Given by u(t - a), where 'a' is the step location, the Heaviside function is 0 for t < a and 1 for t >= a.

Using Heaviside Function

In the context of our problem, we're dealing with a function that changes at t = 1. The Heaviside function enables us to describe the behavior of the differential equation's solutions before and after this point, combining them into a single expression that captures the whole scenario. This approach simplifies working with piecewise functions and is pivotal for obtaining the overall solution.

Piecewise Continuous Functions

Piecewise continuous functions can be defined by different expressions over different intervals. They are widespread in practical scenarios where a system's behavior changes at certain points or conditions.
Solving differential equations involving piecewise continuous functions can be more complex, as it typically requires addressing each piece of the function separately before combining them into a comprehensive solution.

Addressing Piecewise Functions

To manage such functions, we often employ the Heaviside function to facilitate the solution process, as seen in the exercise. By defining distinct functions for intervals of interest and utilizing the Heaviside function, we can construct a single function that we can work with more conveniently. This is precisely the strategy we used in our differential equation's piecewise function, which changes its form at t = 1.