Question

In each of the following exercises, use the Laplace transform to find the solution of the given linear system that satisfies the given initial conditions. \(x^{\prime \prime}-3 x^{\prime}+y^{\prime}+2 x-y=0\) \(x^{\prime}+y^{\prime}-2 x+y=0\) \(x(0)=0, y(0)=-1, x^{\prime}(0)=0\)

Short answer

The solution to the given system of linear differential equations using Laplace transform is \( x(t) = -te^t \) and \( y(t) = (1 - t)e^t - 1 \).

Step-by-step solution

Laplace transform the equations

First, let's take the Laplace transform of each equation: \( \mathcal{L} \{ x^{\prime\prime}(t) - 3x^{\prime}(t) + y^{\prime}(t) + 2x(t) - y(t) = 0 \} \) \( \mathcal{L} \{ x^{\prime}(t) + y^{\prime}(t) - 2x(t) + y(t) = 0 \} \)

Apply Laplace transform properties and initial conditions

Applying the properties of Laplace transform and the given initial conditions, we have: \( s^2X(s) - sx(0) - x'(0) - 3[sX(s) - x(0)] + sY(s) - y(0) + 2X(s) - Y(s) = 0 \) \( sX(s) - x(0) + sY(s) - y(0) - 2X(s) + Y(s) = 0 \) Plugging in the initial conditions \( x(0) = 0 \), \( y(0) = -1 \), and \( x'(0) = 0 \), we get: \( s^2X(s) - 3sX(s) + sY(s) + Y(s) + 2X(s) = 0 \) \( sX(s) + sY(s) + Y(s) - 2X(s) = -1 \)

Solve for X(s) and Y(s)

Combine terms and rearrange to get the transformed system of equations: \( (s^2 - 3s + 2)X(s) + (s + 1)Y(s) = 0 \) \( (-2 + s)X(s) + (s + 1)Y(s) = -1 \) We can use the method of substitution, elimination, or matrices to solve this system of equations. Let's solve it using the substitution method. From the first equation, we can get the expression for Y(s): \( Y(s) = -\frac{(s^2 - 3s + 2)X(s)}{s + 1} \) Substitute this into the second equation: \( (-2 + s)X(s) + (s + 1)\left(-\frac{(s^2 - 3s + 2)X(s)}{s + 1}\right) = -1 \) Solve for X(s): \( X(s) = \frac{-1}{s^2 - 2s + 1} \) Now substitute the expression for X(s) back into the equation for Y(s): \( Y(s) = -\frac{(s^2 - 3s + 2)\left(\frac{-1}{s^2 - 2s + 1}\right)}{s + 1} \) \( Y(s) = \frac{s^2 - 3s + 2}{s^2 - 2s + 1} \)

Inverse Laplace transform of X(s) and Y(s)

Now, we just need to find the inverse Laplace transforms of X(s) and Y(s) to get the functions x(t) and y(t). \( x(t) = \mathcal{L}^{-1} \left\{ \frac{-1}{(s-1)^2} \right\} \) \( y(t) = \mathcal{L}^{-1} \left\{ \frac{s^2 - 3s + 2}{(s-1)^2} \right\} \) Using the inverse Laplace transform properties and tables, we get: \( x(t)= -te^{t} \) \( y(t) = (1-t)e^{t} - 1 \) So, the solution to the given system of linear differential equations is: \( x(t) = -te^t \) and \( y(t) = (1 - t)e^t - 1 \)

Key concepts

Linear Differential Equations

Linear differential equations are fundamental in describing systems that depend on an independent variable, typically time, and their derivatives. They can model a variety of phenomena in engineering, physics, and other sciences. These equations make use of functions like \( x(t) \) and \( y(t) \), and their derivatives to describe system behavior over time.

Linear differential equations come in various forms, but the ones involved in our exercise are linear ordinary differential equations (ODEs). In linear ODEs:

• Each term is either a constant or a product of a constant and the first power of the variable.
• They can be first order, involving the first derivative, or higher order, involving second derivatives and beyond.

Solving these equations often requires leveraging methodologies like the Laplace transform, which simplifies the process by transforming these differential equations into algebraic ones in the complex frequency domain.

Initial Value Problems

An initial value problem (IVP) specifies the value of the function and possibly its derivatives at the start of the problem domain. In differential equations, it means providing initial conditions to find a unique solution for the variables involved.

In the context of the provided exercise:

• We have two functions, \( x(t) \) and \( y(t) \), with initial conditions given as \( x(0) = 0 \), \( y(0) = -1 \), and \( x'(0) = 0 \).
• These conditions help in determining a unique solution and directly influence the Laplace transform approach by providing specific values that will be applied when using properties like shifting theorem or partial fraction expansions.

Initial conditions act as anchors that guide the solution of a differential equation system to a specific trajectory, eliminating arbitrary constants and ambiguities sometime typical in solving differential equations without such conditions.

Inverse Laplace Transform

The inverse Laplace transform is instrumental in converting functions from the Laplace domain back to the time domain. This is a core step in solving differential equations using the Laplace transform technique.

Once we perform a Laplace transform on a differential equation and manipulate it into a simpler algebraic form, finding the inverse transform allows us to retrieve the solution in its original time-based form. In most applications:

• We use known inverse Laplace transforms found in tables for common functions.
• Inverse transforms often use partial fraction decomposition or the convolution theorem to simplify more complex expressions back into time functions.

For example, in our exercise, the transformed functions \( X(s) \) and \( Y(s) \) were converted back to \( x(t) \) and \( y(t) \) as functions of time, resulting in solutions like \( x(t) = -te^t \) and \( y(t) = (1-t)e^t - 1 \). This step bridges the gap between frequency domain techniques and practical, time-domain interpretations of differential problems.

Substitution Method

The substitution method is an algebraic technique used to solve systems of equations, either linear or non-linear. It involves solving one of the equations for one variable and substituting it into another.

In the context of Laplace transformed equations, this method simplifies the process of finding specific solutions for transformed variables, such as \( X(s) \) and \( Y(s) \). Here's how it typically works:

• Choose one equation and express one variable in terms of the others using this equation.
• Substitute that expression into another equation to reduce the number of variables.
• Continue the process until you isolate one variable and solve for it directly.

Applying the substitution method in our original system allowed for the isolation of \( Y(s) \) in terms of \( X(s) \), which was then substituted back to obtain an expression solely for \( X(s) \). This strategic substitution simplifies the steps needed to ultimately find the inverse Laplace transform of these expressions.