Problem 18 - Differential Equations and Linear Algebra

Question

Convert the given linear differential equations to a first-order linear system. $$y^{\prime \prime}+a y^{\prime}+b y=F(t), \quad a, b \text { constants. }$$

Step-by-Step Solution

Answer

Introduce a new variable \(z(t)\) such that \(z(t) = y'(t)\), which changes the equation to \(z'(t) + a z(t) + by(t) = F(t)\). Now, we can write the first-order linear system as: \[\frac{d}{dt} \begin{bmatrix} y(t) \\ z(t) \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -b & -a \end{bmatrix} \begin{bmatrix} y(t) \\ z(t) \end{bmatrix} + \begin{bmatrix} 0 \\ F(t) \end{bmatrix}\]

Step 1: Introduce new variables for first order derivatives

Let us introduce a new variable \(z(t)\) such that: \[z(t) = y'(t)\] Now, our equation becomes: \[z'(t) + a z(t) + by(t) = F(t)\]

Step 2: Solve for the additional variable

We have the following system of equations: \[y'(t) = z(t)\] \[z'(t) = F(t) - a z(t) - b y(t)\] Now we can rewrite this system in matrix form as follows: \[\begin{bmatrix} y'(t) \\ z'(t) \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -b & -a \end{bmatrix} \begin{bmatrix} y(t) \\ z(t) \end{bmatrix} + \begin{bmatrix} 0 \\ F(t) \end{bmatrix}\]

Step 3: Write the first-order linear system

Now, we have successfully converted the second-order linear differential equation into a first-order linear system: \[\frac{d}{dt} \begin{bmatrix} y(t) \\ z(t) \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -b & -a \end{bmatrix} \begin{bmatrix} y(t) \\ z(t) \end{bmatrix} + \begin{bmatrix} 0 \\ F(t) \end{bmatrix}\]

Key Concepts

Linear Differential Equations

Understanding linear differential equations is crucial for many fields of study, including mathematics, engineering, and physics. These equations are a form of a differential equation which are characterized by the property that the function and its derivatives appear linearly in the equation. This means that each term of the equation is either a constant or a product of a constant with the function or its derivatives.

For example, the equation \(y'' + ay' + by = F(t)\), given in the exercise, is a second-order linear differential equation. Here, \(y''\) represents the second derivative of \(y\) with respect to \(t\), \(ay'\) is the first derivative multiplied by a constant \(a\), \(by\) is the function multiplied by constant \(b\), and \(F(t)\) is a function of \(t\) known as the non-homogeneous term, which can vary over time.
The solution to a linear differential equation represents the behavior of the system described by the equation. To facilitate finding solutions, techniques such as introducing new variables and transforming the equation into a system of first-order equations are employed.

Matrix Representation of Differential Equations

The matrix representation of differential equations offers a structured approach to deal with systems of equations. It is most commonly used when dealing with linear systems of equations, as it allows for the organization of multiple equations into a compact, manageable form.

In the context of the original exercise, the second-order linear differential equation is transformed into a system of first-order equations by introducing a new variable for the derivative. This process, depicted in the step-by-step solution, effectively reduces the complexity of the original equation.
For instance, we use \(z(t)\) as \(y'(t)\) and rewrite the equation system as:

\[y'(t) = z(t)\]\[z'(t) = F(t) - a z(t) - b y(t)\]

This system can then be represented in matrix form, which allows us to use matrix operations to simplify and solve the problem. The resulting matrix equation makes it easier to visualize and manipulate the system of equations, particularly when applying methods such as eigenvalue analysis or diagonalization to solve the system or analyze its stability.

System of Differential Equations

A system of differential equations consists of several differential equations involving the same set of variables. These systems arise naturally when modeling real-world phenomena that involve multiple interrelated processes.

When converting a higher-order differential equation into a first-order system, as seen in the exercise, each equation in the system is coupled with the others through shared variables. In our case, \(y'(t)\) and \(z'(t)\) depend on both \(y(t)\) and \(z(t)\). This interdependence reflects the underlying relationships between the physical quantities in the modeled phenomenon.
By examining the final form of the system in matrix notation:

\[\frac{d}{dt}\begin{bmatrix} y(t) \ z(t)\end{bmatrix}=\begin{bmatrix} 0 & 1 \ -b & -a \end{bmatrix}\begin{bmatrix} y(t)  \ z(t) \end{bmatrix}+\begin{bmatrix} 0  \ F(t) \end{bmatrix}\]

We observe that the first-order linear system is apt for methods such as the eigenvalue analysis or for obtaining a numerical solution using computational techniques. Furthermore, understanding the structure and behavior of such systems provides insight into the dynamics of the modeled system and is a significant step towards deriving a general solution.