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Mathematics LibreTexts

10.5E: Constant Coefficient Homogeneous Systems II (Exercises)

  • Page ID
    18301
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    In Exercises [exer:10.5.1}– [exer:10.5.12} find the general solution. [exer:10.5.1] \(\

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    [exer:10.5.31] \(\{{\bf y}' =\threebythree{-3}{-3}445{-8}23{-5}\bf y}\) [exer:10.5.32] \({\bf y}'={\threebythree{-3}{-1}01{-1}0{-1}{-1}{-2}}{\bf y}\) [exer:10.5.33] Under the assumptions of Theorem [thmtype:10.5.1}, suppose \({\bf u}\) and \(\hat{\bf u}\) are vectors such that

    \[(A-\lambda_1I){\bf u}={\bf x}\quad\mbox{and }\quad (A-\lambda_1I)\hat{\bf u}={\bf x},\]

    and let

    \[{\bf y}_2={\bf u}e^{\lambda_1t}+{\bf x}te^{\lambda_1t} \quad\mbox{and }\quad \hat{\bf y}_2=\hat{\bf u}e^{\lambda_1t}+{\bf x}te^{\lambda_1t}.\]

    Show that \({\bf y}_2-\hat{\bf y}_2\) is a scalar multiple of \({\bf y}_1={\bf x}e^{\lambda_1t}\). [exer:10.5.34] Under the assumptions of Theorem [thmtype:10.5.2}, let

    \[\begin{aligned} {\bf y}_1 &=&{\bf x} e^{\lambda_1t},\\ {\bf y}_2&=&{\bf u}e^{\lambda_1t}+{\bf x} te^{\lambda_1t},\mbox{ and }\\ {\bf y}_3&=&{\bf v}e^{\lambda_1t}+{\bf u}te^{\lambda_1t}+{\bf x} {t^2e^{\lambda_1t}\over2}.\end{aligned}\]

    Complete the proof of Theorem [thmtype:10.5.2} by showing that \({\bf y}_3\) is a solution of \({\bf y}'=A{\bf y}\) and that \(\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}\) is linearly independent. [exer:10.5.35] Suppose the matrix ?has a repeated eigenvalue \(\lambda_1\) and the associated eigenspace is one-dimensional. Let \({\bf x}\) be a \(\lambda_1\)-eigenvector of \(A\). Show that if \((A-\lambda_1I){\bf u}_1={\bf x}\) and \((A-\lambda_1I){\bf u}_2={\bf x}\), then \({\bf u}_2-{\bf u}_1\) is parallel to \({\bf x}\). Conclude from this that all vectors \({\bf u}\) such that \((A-\lambda_1I){\bf u}={\bf x}\) define the same positive and negative half-planes with respect to the line \(L\) through the origin parallel to \({\bf x}\). [exer:10.5.36] \({\bf y}'=\{\twobytwo{-3}{-1}41}{\bf y}\)& [exer:10.5.37]   \({\bf y}'=\{\twobytwo2{-1}10}{\bf y}\) [exer:10.5.38]   \({\bf y}'=\{\twobytwo{-1}{-3}35}{\bf y}\)& [exer:10.5.39]   \({\bf y}'=\{\twobytwo{-5}3{-3}1}{\bf y}\) [exer:10.5.40]   \({\bf y}'=\{\twobytwo{-2}{-3}34}{\bf y}\)& [exer:10.5.41]   \({\bf y}'=\{\twobytwo{-4}{-3}32}{\bf y}\) [exer:10.5.42]   \({\bf y}'=\{\twobytwo0{-1}1{-2}}{\bf y}\)&

    [exer:10.5.43] \({\bf y}'=\{\twobytwo01{-1}2}{\bf y}\)

    [exer:10.5.44]   \({\bf y}'=\{\twobytwo{-2}1{-1}0}{\bf y}\)&

    [exer:10.5.45]   \({\bf y}'=\{\twobytwo0{-4}1{-4}}{\bf y}\)