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Mathematics LibreTexts

4.5E: Exercises

  • Page ID
    18279
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    In Exercises \((4.5E.1)\) to \((4.5E.12)\), find the general solution.

    Exercise \(\PageIndex{1}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ 3 & 4 \\ {-1} & 7 \end{array} \right] {\bf y}}\)

    Answer

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    Exercise \(\PageIndex{2}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ 0 & {-1} \\ 1 & {-2} \end{array} \right] {\bf y}}\)

    Answer

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    Exercise \(\PageIndex{3}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-7} & 4 \\ {-1} & {-11} \end{array} \right] {\bf y}}\)

    Answer

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    Exercise \(\PageIndex{4}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ 3 & 1 \\ {-1} & 1 \end{array} \right] {\bf y}}\)

    Answer

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    Exercise \(\PageIndex{5}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ 4 & {12} \\ {-3} & {-8} \end{array} \right] {\bf y}}\)

    Answer

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    Exercise \(\PageIndex{6}\)

    \(\displaystyle{{\bf y'} = \left[ \begin{array} \\ {-10} & 9 \\ {-4} & 2 \end{array} \right] {\bf y}}\)

    Answer

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    Exercise \(\PageIndex{7}\)

    \(\displaystyle{{\bf y'} = \left[ \begin{array} \\ {-13} & {16} \\ {-9} & {11} \end{array} \right] {\bf y}}\)

    Answer

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    Exercise \(\PageIndex{8}\)

    \(\displaystyle{{\bf y'} = \left[ \begin{array} \\ 0 & 2 & 1 \\ {-4} & 6 & 1 \\ 0 & 4 & 2 \end{array} \right] {\bf y}}\)

    Answer

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    Exercise \(\PageIndex{9}\)

    \(\displaystyle{{\bf y}' = {1\over3} \left[ \begin{array} \\ 1 & 1 & {-3} \\ {-4} & {-4} & 3 \\ {-2} & 1 & 0 \end{array} \right] {\bf y}}\)

    Answer

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    Exercise \(\PageIndex{10}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-1} & 1 & {-1} \\ {-2} & 0 & 2 \\ {-1} & 3 & {-1} \end{array} \right] {\bf y}}\)

    Answer

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    Exercise \(\PageIndex{11}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ 4 & {-2} & {-2} \\ {-2} & 3 & {-1} \\ 2 & {-1} & 3 \end{array} \right] {\bf y}}\)

    Answer

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    Exercise \(\PageIndex{12}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ 6 & {-5} & 3 \\ 2 & {-1} & 3 \\ 2 & 1 & 1 \end{array} \right] {\bf y}}\)

    Answer

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    In Exercises \((4.5E.13)\) to \((4.5E.23)\), solve the initial value problem.

    Exercise \(\PageIndex{13}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-11} & 8 \\ {-2} & {-3} \end{array} \right] {\bf y} , \quad {\bf y}(0) = \left[ \begin{array} \\ 6 \\ 2 \end{array} \right]}\)

    Answer

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    Exercise \(\PageIndex{14}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ {15} & {-9} \\ {16} & {-9} \end{array} \right] {\bf y} , \quad {\bf y}(0) = \left[ \begin{array} \\ 5 \\ 8 \end{array} \right] }\)

    Answer

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    Exercise \(\PageIndex{15}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-3} & {-4} \\ 1 & {-7} \end{array} \right] {\bf y} , \quad {\bf y}(0) = \left[ \begin{array} \\ 2 \\ 3 \end{array} \right]}\)

    Answer

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    Exercise \(\PageIndex{16}\)

    \( {\bf y}' = \left[ \begin{array} \\ -7 & 24 \\ -6 & 17 \end{array} \right] {\bf y} , \quad {\bf y}(0) = \left[ \begin{array} \\ 3 \\ 1 \end{array} \right] \)

    Answer

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    Exercise \(\PageIndex{17}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-7} & 3 \\ {-3} & {-1} \end{array} \right] {\bf y} , \quad {\bf y}(0) = \left[ \begin{array} \\ 0 \\ 2 \end{array} \right]}\)

    Answer

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    Exercise \(\PageIndex{18}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-1} & 1 & 0 \\ 1 & {-1} & {-2} \\ {-1} & {-1} & {-1} \end{array} \right] {\bf y} , \quad {\bf y}(0) = \left[ \begin{array} \\ 6 \\ 5 \\ {-7} \end{array} \right] }\)

    Answer

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    Exercise \(\PageIndex{19}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-2} & 2 & 1 \\ {-2} & 2 & 1 \\ {-3} & 3 & 2 \end{array} \right] {\bf y} , \quad {\bf y}(0) = \left[ \begin{array} \\ {-6} \\ {-2} \\ 0 \end{array} \right] }\)

    Answer

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    Exercise \(\PageIndex{20}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-7} & {-4} & 4 \\ {-1} & 0 & 1 \\ {-9} & {-5} & 6 \end{array} \right] {\bf y} , \quad \bf {\bf y}(0) = \left[ \begin{array} \\ {-6} \\ 9 \\ {-1} \end{array} \right]}\)

    Answer

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    Exercise \(\PageIndex{21}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-1} & {-4} & {-1} \\ 3 & 6 & 1 \\ {-3} & {-2} & 3 \end{array} \right] {\bf y} , \quad \bf y(0) = \left[ \begin{array} \\ {-2} \\ 1 \\ 3 \end{array} \right]}\)

    Answer

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    Exercise \(\PageIndex{22}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ 4 & {-8} & {-4} \\ {-3} & {-1} & {-3} \\ 1 & {-1} & 9 \end{array} \right] {\bf y} , \quad {\bf y}(0) = \left[ \begin{array} \\ {-4} \\ 1 \\ {-3} \end{array} \right]}\)

    Answer

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    Exercise \(\PageIndex{23}\)

    \(\displaystyle{{\bf y}'= \left[ \begin{array} \\ {-5} & {-1} & {11} \\ {-7} & 1 & {13} \\ {-4} & 0 & 8 \end{array} \right] {\bf y} , \quad {\bf y}(0) = \left[ \begin{array} \\ 0 \\ 2 \\ 2 \end{array} \right] }\)

    Answer

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    The coefficient matrices in Exercises \((4.5E.24)\) to \((4.5E.32)\) have eigenvalues of multiplicity \(3\). Find the general solution.

    Exercise \(\PageIndex{24}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ 5 & {-1} & 1 \\ {-1} & 9 & {-3} \\ {-2} & 2 & 4 \end{array} \right] {\bf y}}\)

    Answer

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    Exercise \(\PageIndex{25}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ 1 & {10} & {-12} \\ 2 & 2 & 3 \\ 2 & {-1} & 6 \end{array} \right] {\bf y}}\)

    Answer

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    Exercise \(\PageIndex{26}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-6} & {-4} & {-4} \\ 2 & {-1} & 1 \\ 2 & 3 & 1 \end{array} \right] {\bf y}}\)

    Answer

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    Exercise \(\PageIndex{27}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ 0 & 2 & {-2} \\ {-1} & 5 & {-3} \\ 1 & 1 & 1 \end{array} \right] {\bf y}}\)

    Answer

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    Exercise \(\PageIndex{28}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-2} & {-12} & {10} \\ 2 & {-24} & {11} \\ 2 & {-24} & 8 \end{array} \right] {\bf y}}\)

    Answer

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    Exercise \(\PageIndex{29}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-1} & {-12} & 8 \\ 1 & {-9} & 4 \\ 1 & {-6} & 1 \end{array} \right] {\bf y}}\)

    Answer

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    Exercise \(\PageIndex{30}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-4} & 0 & {-1} \\ {-1} & {-3} & {-1} \\ 1 & 0 & {-2} \end{array} \right] {\bf y}}\)

    Answer

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    Exercise \(\PageIndex{31}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-3} & {-3} & 4 \\ 4 & 5 & {-8} \\ 2 & 3 & {-5} \end{array} \right] \bf y}\)

    Answer

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    Exercise \(\PageIndex{32}\)

    \({\bf y}' = { \left[ \begin{array} \\ {-3} & {-1} & 0 \\ 1 & {-1} & 0 \\ {-1} & {-1} & {-2} \end{array} \right]}{\bf y}\)

    Answer

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    Exercise \(\PageIndex{33}\)

    Under the assumptions of Theorem \((4.5.1)\), suppose \({\bf u}\) and \(\hat{\bf u}\) are vectors such that

    \begin{eqnarray*}
    (A - \lambda_1I) {\bf u} = {\bf x} \quad \mbox{and} \quad (A - \lambda_1I) \hat {\bf u} = {\bf x},
    \end{eqnarray*}

    and let

    \begin{eqnarray*}
    {\bf y}_2 = {\bf u} e^{\lambda_1t} + {\bf x} t e^{\lambda_1t} \quad \mbox{and} \quad \hat {\bf y}_2 = \hat {\bf u} e^{\lambda_1t} + {\bf x} te^{\lambda_1t}.
    \end{eqnarray*}

    Show that \({\bf y}_2-\hat{\bf y}_2\) is a scalar multiple of \({\bf y}_1={\bf x}e^{\lambda_1t}\).

    Answer

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    Exercise \(\PageIndex{34}\)

    Under the assumptions of Theorem \((4.5.2)\), let

    \begin{eqnarray*}
    {\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{eqnarray*}

    Complete the proof of Theorem \((4.5.2)\) 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.

    Answer

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    Exercise \(\PageIndex{35}\)

    Suppose the matrix

    \begin{eqnarray*}
    A = \left[ \begin{array} \\ a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right]
    \end{eqnarray*}

    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 \({\bf x}\).

    Answer

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    In Exercises \((4.5E.36)\) to \((4.5E.45)\), plot trajectories of the given system.

    Exercise \(\PageIndex{36}\)

    \({\bf y}'=\displaystyle{ \left[ \begin{array} \\ {-3} & {-1} \\ 4 & 1 \end{array} \right] }{\bf y}\)

    Answer

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    Exercise \(\PageIndex{37}\)

    \({\bf y}' = \displaystyle{ \left[ \begin{array} \\ 2 & {-1} \\ 1 & 0 \end{array} \right]}{\bf y}\)

    Answer

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    Exercise \(\PageIndex{38}\)

    \({\bf y}' = \displaystyle{ \left[ \begin{array} \\ {-1} & {-3} \\ 3 & 5 \end{array} \right] }{\bf y}\)

    Answer

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    Exercise \(\PageIndex{39}\)

    \({\bf y}' = \displaystyle{ \left[ \begin{array} \\ {-5} & 3 \\ {-3} & 1 \end{array} \right] }{\bf y}\)

    Answer

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    Exercise \(\PageIndex{40}\)

    \({\bf y}' = \displaystyle{ \left[ \begin{array} \\ {-2} & {-3} \\ 3 & 4 \end{array} \right] }{\bf y}\)

    Answer

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    Exercise \(\PageIndex{41}\)

    \({\bf y}' = \displaystyle { \left[ \begin{array} \\ {-4} & {-3} \\ 3 & 2 \end{array} \right] }{\bf y}\)

    Answer

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    Exercise \(\PageIndex{42}\)

    \({\bf y}' = \displaystyle{ \left[ \begin{array} \\ 0 & {-1} \\ 1 & {-2} \end{array} \right] }{\bf y}\)

    Answer

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    Exercise \(\PageIndex{43}\)

    \({\bf y}' = \displaystyle{ \left[ \begin{array} \\ 0 & 1 \\ {-1} & 2 \end{array} \right] }{\bf y}\)

    Answer

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    Exercise \(\PageIndex{44}\)

    \({\bf y}' = \displaystyle{ \left[ \begin{array} \\ {-2} & 1 \\ {-1} & 0 \end{array} \right] }{\bf y}\)

    Answer

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    Exercise \(\PageIndex{45}\)

    \({\bf y}' = \displaystyle{ \left[ \begin{array} \\ 0 & {-4} \\ 1 & {-4} \end{array} \right] }{\bf y}\)

    Answer

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