Chapter 9: Problem 2
\(\mathbf{A}=\left[ \begin{array}{rr}{-2} & {-5} \\ {1} & {2}\end{array}\right]\)
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Chapter 9: Problem 2
\(\mathbf{A}=\left[ \begin{array}{rr}{-2} & {-5} \\ {1} & {2}\end{array}\right]\)
These are the key concepts you need to understand to accurately answer the question.
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Use the substitution \(x_{1}=y, x_{2}=y^{\prime}\) to convert the linear equation \(a y^{\prime \prime}+b y^{\prime}+c y=0,\) where \(a, b,\) and \(c\) are constants, into a normal system. Show that the characteristic equation for this system is the same as the auxiliary equation for the original equation.
To complete the proof of Theorem \(6,\) page \(527,\) assume the induction hypothesis that \(\mathbf{u}_{1}, \ldots, \mathbf{u}_{k}, 2 \leq k,\) are linearly independent. (a) Show that if \(c_{1} \mathbf{u}_{1}+\cdots+c_{k} \mathbf{u}_{k}+c_{k+1} \mathbf{u}_{k+1}=\mathbf{0}\) then \(c_{1}\left(r_{1}-r_{k+1}\right) \mathbf{u}_{1}+\cdots+c_{k}\left(n_{k}-r_{k+1}\right) \mathbf{u}_{k}=\mathbf{0}\) (b) Use the result of part (a) and the induction hypothesis to conclude that \(\mathbf{u}_{1}, \ldots, \mathbf{u}_{k+1}\) are linearly independent. The theorem follows by induction.
\(t \mathbf{x}^{\prime}(t)=\left[ \begin{array}{ll}{4} & {-3} \\ {8} & {-6}\end{array}\right] \mathbf{x}(t)+\left[ \begin{array}{c}{t} \\ {2 t}\end{array}\right]\)
(a) Show that the Cauchy-Euler equation \(a t^{2} y^{\prime \prime}+b t y^{\prime}+c y=0 \quad\) can \(\quad\) be written as a Cauchy-Euler system $$ t \mathbf{x}^{\prime}=\mathbf{A x} $$ with a constant coefficient matrix \(\mathbf{A},\) by setting \(x_{1}=y / t\) and \(x_{2}=y^{\prime}\) (b) Show that for \(t>0\) any system of the form \((25)\) with A an \(n \times n\) constant matrix has nontrivial solutions of the form \(\mathbf{x}(t)=t^{r} \mathbf{u}\) if and only if \(r\) is an eigenvalue of \(\mathbf{A}\) and \(\mathbf{u}\) is a corresponding eigenvector.
\(\mathbf{A}=\left[ \begin{array}{ll}{8} & {-4} \\ {4} & {-2}\end{array}\right], \quad \mathbf{f}(t)=\left[ \begin{array}{l}{t^{-2} / 2} \\ {t^{-2}}\end{array}\right]\)
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