Question

Consider the equation $$ a y^{\prime \prime}+b y^{\prime}+c y=0 $$ $$ \begin{array}{l}{\text { where } a, b, \text { and } c \text { are constants. In Chapter } 3 \text { it was shown that the general solution depended }} \\\ {\text { on the roots of the characteristic equation }}\end{array} $$ $$ a r^{2}+b r+c=0 $$ $$ \begin{array}{l}{\text { (a) Transform Eq. (i) into a system of first order equations by letting } x_{1}=y, x_{2}=y^{\prime} . \text { Find }} \\ {\text { the system of equations } x^{\prime}=A x \text { satisfied by } x=\left(\begin{array}{l}{x_{1}} \\ {x_{2}} \\ {x_{2}}\end{array}\right)} \\\ {\text { (b) Find the equation that determines the eigenvalues of the coefficient matrix } \mathbf{A} \text { in part (a). }} \\ {\text { Note that this equation is just the characteristic equation (ii) of Eq. (i). }}\end{array} $$

Short answer

Question: Determine the eigenvalues of the coefficient matrix A and show that the equation for finding these eigenvalues is the same as the characteristic equation of the original second-order differential equation given by \(a y'' + b y' + c y = 0\).

Step-by-step solution

Transforming the second-order equation into a system of first-order equations

Let \(x_{1}=y\) and \(x_{2}=y'\), then \(x_{1}' = y'=x_{2}\) and \(x_{2}' = y''\). Now, substitute \(x_1\) and \(x_2\) into the given second-order differential equation: $$ a y'' + b y' + c y = a x_2' + b x_1' + c x_1 = 0 $$ Now, we have two first-order equations: $$ x_1' = x_2 $$ $$ x_2' = -\frac{b}{a} x_1 - \frac{c}{a} x_2 $$

Finding the system of equations in the form \(x'=Ax\)

Write down the expression for \(x'\): $$ x' = \begin{pmatrix} x_1' \\ x_2' \end{pmatrix} = \begin{pmatrix} x_2 \\ -\frac{b}{a} x_1 - \frac{c}{a} x_2 \end{pmatrix} $$ The matrix A can be found by comparing the expressions for \(x_1'\) and \(x_2'\) with the matrix multiplication \(Ax\): $$ A = \begin{pmatrix} 0 & 1 \\ -\frac{c}{a} & -\frac{b}{a} \end{pmatrix} $$ Now, we have the system of equations \(x'=Ax\).

Finding the equation for the eigenvalues of matrix A

Find the characteristic equation for the matrix A, which is given by: $$ \text{Det}(A-\lambda I) = \begin{vmatrix} -\lambda & 1 \\ -\frac{c}{a} & -\frac{b}{a}-\lambda \end{vmatrix} = (-\lambda)(-\frac{b}{a}-\lambda) - (-\frac{c}{a}) = 0 $$ Multiplying through by 'a', we get: $$ a\lambda^2 + b\lambda + c = 0 $$ This is the same as the characteristic equation for the original second-order linear homogeneous differential equation. Thus, we have transformed the given equation into a system of first-order equations and found that the equation for the eigenvalues of the coefficient matrix A is the same as the original equation's characteristic equation.

Key concepts

characteristic equation

In the context of linear differential equations, the characteristic equation is a pivotal concept. It emerges when we handle differential equations of the form \(a y'' + b y' + c y = 0\). This equation cannot be straightaway solved using simple arithmetic or algebra but can be tackled by breaking it into smaller, manageable problems. The characteristic equation provides the roots that are essential for finding the general solution of the differential equation. In our example, we have the characteristic equation given by \(a r^2 + b r + c = 0\), where \(r\) represents the roots or solutions that solve this polynomial equation. The nature of these roots — whether they are real and distinct, real and repeated, or complex — significantly influences the form of the general solution of the differential equation.

first-order system

Transforming a second-order differential equation into a system of first-order equations is a common technique used to simplify complex problems. This involves introducing new variables, such as \(x_1 = y\) and \(x_2 = y'\) from the given equation \(a y'' + b y' + c y = 0\). Subsequently, the system can be represented as \(x_1' = x_2\) and \(x_2' = -\frac{b}{a} x_1 - \frac{c}{a} x_2\). This change of variables helps in reformulating the second-order equation into a paired system of first-order equations, which are generally easier to analyze and solve. Such systems are not only crucial in solving differential equations but also play significant roles in fields like control systems, physics, and engineering.

eigenvalues

Eigenvalues play a crucial role when dealing with systems of linear equations or differential equations expressed in matrix form. They are vital in determining the stability and behavior of solutions. For the matrix \(A\) from our transformed system of first-order equations \(x' = Ax\), finding the eigenvalues involves solving the characteristic equation derived from the determinant \(\text{Det}(A - \lambda I)\). Here, \(I\) is the identity matrix and \(\lambda\) represents the eigenvalues to be determined. In our example, this leads to the quadratic equation \(a\lambda^2 + b\lambda + c = 0\), identical to the characteristic equation for the original second-order problem. These eigenvalues contain information about the oscillatory features and exponential growth or decay represented by the solutions to our differential equation.

coefficient matrix

The coefficient matrix \(A\) is an important element in converting a second-order linear differential equation into a system of first-order equations. In our scenario, this matrix is \(\begin{pmatrix} 0 & 1 \ -\frac{c}{a} & -\frac{b}{a} \end{pmatrix}\). Each entry of the matrix aligns with the relationships between the derived variables \(x_1\) and \(x_2\). The matrix helps to describe how each function or variable in the system changes in relation to the others, essentially capturing the dynamics of the system. By examining the coefficient matrix, we can extract crucial information, such as eigenvalues, which provide insights into the behavior of solutions. It effectively distills the differential equation's core properties into a form that is simpler for analysis and computation.