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.