4.6E: Exercises

In Exercises \((4.6E.1)\) to \((4.6E.16)\), find the general solution.

Exercise \(\PageIndex{1}\)

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

Answer

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Exercise \(\PageIndex{2}\)

\(\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-11} & 4 \\ {-26} & 9 \end{array} \right] {\bf y} }\)

Answer

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Exercise \(\PageIndex{3}\)

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

Answer

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Exercise \(\PageIndex{4}\)

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

Answer

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Exercise \(\PageIndex{5}\)

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

Answer

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Exercise \(\PageIndex{6}\)

\(\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-3} & 3 & 1 \\ 1 & {-5} & {-3} \\ {-3} & 7 & 3 \end{array} \right] {\bf y}}\)

Answer

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Exercise \(\PageIndex{7}\)

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

Answer

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Exercise \(\PageIndex{8}\)

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

Answer

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Exercise \(\PageIndex{9}\)

\(\displaystyle{{\bf y}' = \left[ \begin{array} \\ 5 & {-4} \\ {10} & 1 \end{array} \right] {\bf y}}\)

Answer

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Exercise \(\PageIndex{10}\)

\(\displaystyle{{\bf y}' = {1\over3} \left[ \begin{array} \\ 7 & {-5} \\ 2 & 5 \end{array} \right] {\bf y}}\)

Answer

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Exercise \(\PageIndex{11}\)

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

Answer

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Exercise \(\PageIndex{12}\)

\(\displaystyle{{\bf y}' = \left[ \begin{array} \\ {34} & {52} \\ {-20} & {-30} \end{array} \right] {\bf y}}\)

Answer

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Exercise \(\PageIndex{13}\)

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

Answer

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Exercise \(\PageIndex{14}\)

\(\displaystyle{{\bf y}' = \left[ \begin{array} \\ 3 & {-4} & {-2} \\ {-5} & 7 & {-8} \\ {-10} & {13} & {-8} \end{array} \right] {\bf y}}\)

Answer

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Exercise \(\PageIndex{15}\)

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

Answer

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Exercise \(\PageIndex{16}\)

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

Answer

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In Exercises \((4.6E.17)\) to \((4.6E.24)\), solve the initial value problem.

Exercise \(\PageIndex{17}\)

\(\displaystyle{{\bf y}' = \left[ \begin{array} \\ 4 & {-6} \\ 3 & {-2} \end{array} \right] {\bf y}, \quad {\bf y}(0) = \left[ \begin{array} \\ 5 \\ 2 \end{array} \right] } \)

Answer

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Exercise \(\PageIndex{18}\)

\(\displaystyle{{\bf y}' = \left[ \begin{array} \\ 7 & {15} \\ {-3} & 1 \end{array} \right] {\bf y}, \quad {\bf y}(0) = \left[ \begin{array} \\ 5 \\ 1 \end{array} \right] } \)

Answer

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Exercise \(\PageIndex{19}\)

\(\displaystyle{{\bf y}' = \left[ \begin{array} \\ 7 & {-15} \\ 3 & {-5} \end{array} \right] {\bf y}, \quad {\bf y}(0) = \left[ \begin{array} \\ {17} \\ 7 \end{array} \right]} \)

Answer

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Exercise \(\PageIndex{20}\)

\(\displaystyle{{\bf y}' = {1\over 6} \left[ \begin{array} \\ 4 & {-2} \\ 5 & 2 \end{array} \right] {\bf y}, \quad {\bf y}(0) = \left[ \begin{array} \\ 1 \\ {-1} \end{array} \right] }\)

Answer

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Exercise \(\PageIndex{21}\)

\(\displaystyle{{\bf y}' = \left[ \begin{array} \\ 5 & 2 & {-1} \\ {-3} & 2 & 2 \\ 1 & 3 & 2 \end{array} \right] {\bf y}, \quad {\bf y}(0) = \left[ \begin{array} \\ 4 \\ 0 \\ 6 \end{array} \right]}\)

Answer

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Exercise \(\PageIndex{22}\)

\(\displaystyle{{\bf y}' = \left[ \begin{array} \\ 4 & 4 & 0 \\ 8 & {10} & {-20} \\ 2 & 3 & {-2} \end{array} \right] {\bf y}, \quad {\bf y}(0) = \left[ \begin{array} \\ 8 \\ 6 \\ 5 \end{array} \right] }\)

Answer

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Exercise \(\PageIndex{23}\)

\(\displaystyle{{\bf y}' = \left[ \begin{array} \\ 1 & {15} & {-15} \\ {-6} & {18} & {-22} \\ {-3} & {11} & {-15} \end{array} \right] {\bf y}, \quad {\bf y}(0) = \left[ \begin{array} \\ {15} \\ {17} \\ {10} \end{array} \right]}\)

Answer

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Exercise \(\PageIndex{24}\)

\(\displaystyle{{\bf y}' = \left[ \begin{array} \\ 4 & {-4} & 4 \\ {-10} & 3 & {15} \\ 2 & {-3} & 1 \end{array} \right] {\bf y}, \quad {\bf y}(0) = \left[ \begin{array} \\ {16} \\ {14} \\ 6 \end{array} \right] }\)

Answer

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Exercise \(\PageIndex{25}\)

Suppose an \(n\times n\) matrix \(A\) with real entries has a complex eigenvalue \(\lambda=\alpha+i\beta\) (\(\beta\ne0\)) with associated eigenvector \({\bf x}={\bf u}+i{\bf v}\), where \({\bf u}\) and \({\bf v}\) have real components. Show that \({\bf u}\) and \({\bf v}\) are both nonzero.

Answer

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Exercise \(\PageIndex{26}\)

Verify that

\begin{eqnarray*}
{\bf y}_1 = e^{\alpha t} ({\bf u} \cos \beta t - {\bf v} \sin \beta t) \quad \mbox{and} \quad {\bf y}_2 = e^{\alpha t} ({\bf u} \sin \beta t + {\bf v} \cos \beta t),
\end{eqnarray*}

are the real and imaginary parts of

\begin{eqnarray*}
e^{\alpha t} (\cos \beta t + i \sin \beta t) ({\bf u} + i {\bf v}).
\end{eqnarray*}

Answer

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Exercise \(\PageIndex{27}\)

Show that if the vectors \({\bf u}\) and \({\bf v}\) are not both \({\bf 0}\) and \(\beta\ne0\) then the vector functions

\begin{eqnarray*}
{\bf y}_1 = e^{\alpha t}({\bf u} \cos \beta t - {\bf v} \sin \beta t) \quad \mbox{and} \quad {\bf y}_2 = e^{\alpha t} ({\bf u} \sin \beta t + {\bf v} \cos \beta t)
\end{eqnarray*}

are linearly independent on every interval.

Hint: There are two cases to consider: \(\{{\bf u},{\bf v}\}\) linearly independent, and \(\{{\bf u},{\bf v}\}\) linearly dependent. In either case, exploit the the linear independence of \(\{\cos\beta t,\sin\beta t\}\) on every interval.

Answer

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Exercise \(\PageIndex{28}\)

Suppose \({\bf u}=\displaystyle{ \left[ \begin{array} \\ {u_1} \\ {u_2} \end{array} \right] }\) and \({\bf v}=\displaystyle{ \left[ \begin{array} \\ {v_1} \\ {v_2} \end{array} \right]}\) are not orthogonal; that is, \(({\bf u},{\bf v})\ne0\).

(a) Show that the quadratic equation

\begin{eqnarray*}
({\bf u}, {\bf v}) k^2 + (\| {\bf v} \|^2 - \| {\bf u} \|^2) k - ({\bf u}, {\bf v}) = 0
\end{eqnarray*}

has a positive root \(k_1\) and a negative root \(k_2=-1/k_1\).

(b) Let \({\bf u}_1^{(1)}={\bf u}-k_1{\bf v}\), \({\bf v}_1^{(1)}={\bf v}+k_1{\bf u}\), \({\bf u}_1^{(2)}={\bf u}-k_2{\bf v}\), and \({\bf v}_1^{(2)}={\bf v}+k_2{\bf u}\), so that \(({\bf u}_1^{(1)},{\bf v}_1^{(1)}) =({\bf u}_1^{(2)},{\bf v}_1^{(2)})=0\), from the discussion given above. Show that

\begin{eqnarray*}
{\bf u}_1^{(2)} = {{\bf v}_1^{(1)} \over k_1} \quad \mbox{and} \quad {\bf v}_1^{(2)} = -{{\bf u}_1^{(1)} \over k_1}.
\end{eqnarray*}

(c) Let \({\bf U}_1\), \({\bf V}_1\), \({\bf U}_2\), and \({\bf V}_2\) be unit vectors in the directions of \({\bf u}_1^{(1)}\), \({\bf v}_1^{(1)}\), \({\bf u}_1^{(2)}\), and \({\bf v}_1^{(2)}\), respectively. Conclude from part (a) that \({\bf U}_2={\bf V}_1\) and \({\bf V}_2=-{\bf U}_1\), and that therefore the counterclockwise angles from \({\bf U}_1\) to \({\bf V}_1\) and from \({\bf U}_2\) to \({\bf V}_2\) are both \(\pi/2\) or both \(-\pi/2\).

Answer

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In Exercises \((4.6E.29)\) to \((4.6E.32)\), find vectors \({\bf U}\) and \({\bf V}\) parallel to the axes of symmetry of the trajectories, and plot some typical trajectories.

Exercise \(\PageIndex{29}\)

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

Answer

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Exercise \(\PageIndex{30}\)

\(\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-15} & {10} \\ {-25} & {15} \end{array} \right] {\bf y}}\)

Answer

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Exercise \(\PageIndex{31}\)

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

Answer

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Exercise \(\PageIndex{32}\)

\(\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-3} & {-15} \\ 3 & 3 \end{array} \right] {\bf y}}\)

Answer

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In Exercises \((4.6E.33)\) to \((4.6E.40)\), find vectors \({\bf U}\) and \({\bf V}\) parallel to the axes of symmetry of the shadow trajectories, and plot a typical trajectory.

Exercise \(\PageIndex{33}\)

\(\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-5} & 6 \\ {-12} & 7 \end{array} \right] {\bf y}}\)

Answer

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Exercise \(\PageIndex{34}\)

\(\displaystyle{{\bf y}' = \left[ \begin{array} \\ 5 & {-12} \\ 6 & {-7} \end{array} \right] {\bf y}}\)

Answer

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Exercise \(\PageIndex{35}\)

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

Answer

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Exercise \(\PageIndex{36}\)

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

Answer

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Exercise \(\PageIndex{37}\)

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

Answer

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Exercise \(\PageIndex{38}\)

\(\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-1} & {-5} \\ {20} & {-1} \end{array} \right] {\bf y}}\)

Answer

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Exercise \(\PageIndex{39}\)

\(\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-7} & {10} \\ {-10} & 9 \end{array} \right] {\bf y}}\)

Answer

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Exercise \(\PageIndex{40}\)

\(\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-7} & 6 \\ {-12} & 5 \end{array} \right] {\bf y}}\)

Answer

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