4.5E: Exercises
- Page ID
- 18279
This page is a draft and is under active development.
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In Exercises \((4.5E.1)\) to \((4.5E.12)\), find the general solution.
Exercise \(\PageIndex{1}\)
\(\displaystyle{{\bf y}' = \left[ \begin{array}{c}3 & 4 \\ {-1} & 7 \end{array} \right] {\bf y}}\)
Exercise \(\PageIndex{2}\)
\(\displaystyle{{\bf y}' = \left[ \begin{array}{c}0 & {-1} \\ 1 & {-2} \end{array} \right] {\bf y}}\)
- Answer
-
\( \left|\begin{array}{cc} 0 - \lambda & -1 \\ 1 & -2 - \lambda \end{array} \right| = (\lambda + 1)^2 .\) Hence, \( \lambda_1 = -1 \). The eigenvectors satisfy \( \left[ \begin{array}{cc} 1 & -1 \\ 1 & -1 \end{array} \right] \left[ \begin{array}{c} x_1 \\ x_2 \end{array} \right] = \left[ \begin{array}{c} 0 \\ 0 \end{array} \right] \) so \( x_1 = x_2 \). Taking \( x_2 = 1 \) yields \( {\bf y}_1 = \left[ \begin{array}{c} 1 \\ 1 \end{array} \right] e^{-t}. \) For a second solution, we need a vector \( {\bf u} \) such that \( \left[ \begin{array}{cc} 1 & -1 \\ 1 & -1 \end{array} \right] \left[ \begin{array}{c} u_1 \\ u_2 \end{array} \right] = \left[ \begin{array}{c} 1 \\ 1 \end{array} \right]. \) Let \( u_1 = 1 \) and \( u_2 = 0 \). Then, \( {\bf y}_2 = \left[ \begin{array}{c} 1 \\ 0 \end{array} \right] e^{-t} + \left[ \begin{array}{c} 1 \\ 1 \end{array} \right] t e^{-t}. \) The general solution is \( {\bf y} = c_1 \left[ \begin{array}{c} 1 \\ 1 \end{array} \right] e^{-t} + c_2 \left( \left[ \begin{array}{c} 1 \\ 0 \end{array} \right] e^{-t} + \left[ \begin{array}{c} 1 \\ 1 \end{array} \right] t e^{-t} \right). \)
Exercise \(\PageIndex{3}\)
\(\displaystyle{{\bf y}' = \left[ \begin{array}{c}{-7} & 4 \\ {-1} & {-11} \end{array} \right] {\bf y}}\)
Exercise \(\PageIndex{4}\)
\(\displaystyle{{\bf y}' = \left[ \begin{array}{c}3 & 1 \\ {-1} & 1 \end{array} \right] {\bf y}}\)
- Answer
-
\(
\left|\begin{array}{cc}3- \lambda & -1 \\ 1 & 1- \lambda \end{array} \right| = (\lambda - 2)^2
\)
Hence, \( \lambda_1 = 2 \).The eigenvectors satisfy
\(
\begin{pmatrix} 1 & -1 \\ -1 & -1 \end{pmatrix}
\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} =
\begin{pmatrix} 0 \\ 0 \end{pmatrix},
\)
so \( x_1 = -x_2 \). Taking \( x_2 = 1 \) yields
\(
{\bf y}_1 = \begin{pmatrix} -1 \\ 1 \end{pmatrix} e^{2t}.
\)For a second solution, we need a vector \( {\bf u} \) such that
\(
\begin{pmatrix} 1 & -1 \\ -1 & -1 \end{pmatrix}
\begin{pmatrix} u_1 \\ u_2 \end{pmatrix} =
\begin{pmatrix} -1 \\ 1 \end{pmatrix}.
\)
Let \( u_1 = -1 \) and \( u_2 = 0 \). Then,
\(
{\bf y}_2 = \begin{pmatrix} -1 \\ 0 \end{pmatrix} e^{2t} +
\begin{pmatrix} -1 \\ 1 \end{pmatrix} t e^{2t}.
\)The general solution is
\(
{\bf y} = c_1 \begin{pmatrix} -1 \\ 1 \end{pmatrix} e^{2t} +
c_2 \left( \begin{pmatrix} -1 \\ 0 \end{pmatrix} e^{2t} +
\begin{pmatrix} -1 \\ 1 \end{pmatrix} t e^{2t} \right).
\)
Exercise \(\PageIndex{5}\)
\(\displaystyle{{\bf y}' = \left[ \begin{array}{c}4 & {12} \\ {-3} & {-8} \end{array} \right] {\bf y}}\)
Exercise \(\PageIndex{6}\)
\(\displaystyle{{\bf y'} = \left[ \begin{array}{c}{-10} & 9 \\ {-4} & 2 \end{array} \right] {\bf y}}\)
- Answer
-
\(\left|\begin{array}{cc}{-10- \lambda} &9\\{-4} & 2- \lambda \end{array} \right| =(\lambda+4)^2\).
Hence \( \lambda_1=-4\).
Eigenvectors satisfy
\( \displaystyle{\left[\begin{array}{cc}{-6}&9\\{-4}&6 \end{array} \right] \left[ \begin{array}{c}{x_1}\\{x_2}\end{array} \right] = \left[ \begin{array}{c}0\\0\end{array} \right]} \),
so \( x_1={3\over2}x_2\). Taking \( x_2=2\) yields
\( {\bf y}_1=\displaystyle{ \left[ \begin{array}{c}3\\2\end{array} \right] }e^{-4t}\).
For a second solution we need a vector \( {\bf u}\)
such that \( \displaystyle{\left[\begin{array}{cc}{-6}&9\\{-4}&6 \end{array} \right] \left[ \begin{array}{c} {u_1}\\{u_2} \end{array} \right]
= \left[ \begin{array}{c} 3\\2\end{array} \right]} \). Let \( u_1=-{1\over2}\) and \( u_2=0\). Then
\( {\bf y}_2=\displaystyle{ \left[ \begin{array}{c}{-1}\\0\end{array} \right] }{e^{-4t}\over2}+\displaystyle{ \left[ \begin{array}{c}3\\2\end{array} \right] }te^{-4t}\).
The general solution is
\( {\bf y}=\displaystyle{c_1 \left[ \begin{array}{c}3\\2\end{array} \right] e^{-4t}+
c_2\left( \left[ \begin{array}{c}{-1}\\0\end{array} \right] {e^{-4t}\over2}+ \left[ \begin{array}{c}3\\2\end{array} \right] te^{-4t}\right)}\).
Exercise \(\PageIndex{7}\)
\(\displaystyle{{\bf y'} = \left[ \begin{array}{c}{-13} & {16} \\ {-9} & {11} \end{array} \right] {\bf y}}\)
Exercise \(\PageIndex{8}\)
\(\displaystyle{{\bf y'} = \left[ \begin{array}{c}0 & 2 & 1 \\ {-4} & 6 & 1 \\ 0 & 4 & 2 \end{array} \right] {\bf y}}\)
- Answer
-
\(\left|\begin{array}{ccc}0-\lambda&2&1\\{-4}&6-\lambda&1\\0&4-\lambda&2 \end{array} \right|
=-\lambda(\lambda-4)^2\).
Hence \( \lambda_1=0\) and \( \lambda_2=\lambda_3=4\).
The eigenvectors associated
with \( \lambda_1=0\) satisfy the system with augmented matrix
\( \displaystyle{\left[\begin{array}{rrrcr}0&2&1&\vdots&0\\{-4}&6&1&
\vdots&0\\0&4&2&\vdots&0\end{array}\right]}\),
which is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&0&{1\over2}&\vdots&0\\0&1&{1\over2}&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \( x_1=-\displaystyle{1\over2}x_3\) and \( x_2=-\displaystyle{1\over2}x_3\). Taking \( x_3=2\)
yields
\( {\bf y}_1=\displaystyle{\left[\begin{array}{c}{-1}\\{-1}\\2 \end{array}\right] }\).
The eigenvectors associated
with \( \lambda_2=4\) satisfy the system with augmented matrix
\( \displaystyle{\left[\begin{array}{rrrcr}-4&2&1&\vdots&0\\-4&2&1&
\vdots&0\\0&4&-2&\vdots&0\end{array}\right]}\),
which is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&0&-{1\over2}&\vdots&0\\0&1&-{1\over2}&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \( x_1=\displaystyle{1\over2}x_3\) and \( x_2=\displaystyle{1\over2}x_3\). Taking
\( x_3=2\) yields
\( {\bf y}_2=\displaystyle{\left[\begin{array}{c}1\\1\\2\end{array}\right]}e^{4t}\).
For a third solution we need a vector \( {\bf u}\) such that
\( \displaystyle{\left[\begin{array}{ccc}{-4}&2&1\\{-4}&2&1\\0&4&{-2}\end{array}\right]\left[\begin{array}{c}{u_1}\\{u_2}\\{u_3}\end{array}\right]}=\displaystyle{\left[\begin{array}{c}1\\1\\2\end{array}\right]}\).
The augmented matrix of this system is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&0&-{1\over2}&\vdots&0\\0&1&-{1\over2}&
\vdots&{1\over2}\\0&0&0&\vdots&0\end{array}\right]}\).
Let \( u_3=0\), \( u_1=0\), and \( u_2=\displaystyle{1\over2}\). Then
\( {\bf
y}_3=\displaystyle{\left[\begin{array}{c}0\\1\\0\end{array}\right]}\displaystyle{e^{4t}\over2}+\displaystyle{\left[\begin{array}{c}1\\1\\2\end{array}\right]}te^{4t}\).
The general solution is
$$
{\bf y}=\displaystyle{
c_1\left[\begin{array}{c}{-1}\\{-1}\\2\end{array}\right]+c_2\left[\begin{array}{c}1\\1\\2\end{array}\right]e^{4t}+
c_3\left(\left[\begin{array}{c}0\\1\\0\end{array}\right]{e^{4t}\over2}+\left[\begin{array}{c}1\\1\\2\end{array}\right]te^{4t}\right)}.
$$
Exercise \(\PageIndex{9}\)
\(\displaystyle{{\bf y}' = {1\over3} \left[ \begin{array}{c}1 & 1 & {-3} \\ {-4} & {-4} & 3 \\ {-2} & 1 & 0 \end{array} \right] {\bf y}}\)
Exercise \(\PageIndex{10}\)
\(\displaystyle{{\bf y}' = \left[ \begin{array}{c}{-1} & 1 & {-1} \\ {-2} & 0 & 2 \\ {-1} & 3 & {-1} \end{array} \right] {\bf y}}\)
- Answer
-
\(\left|\begin{array}{ccc}{-1-\lambda}&1&{-1}\\{-2}&{-\lambda}&2\\{-1}&3&{-1-\lambda}\end{array} \right|
=-(\lambda-2)(\lambda+2)^2\).
Hence \( \lambda_1=2\) and \( \lambda_2=\lambda_3=-2\).
The eigenvectors associated
with \( \lambda_1=2\) satisfy the system with augmented matrix
\( \displaystyle{\left[\begin{array}{rrrcr}-3&1&-1&\vdots&0\\-2&-2&2&
\vdots&0\\-1&3&-3&\vdots&0\end{array}\right]}\),
which is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&0&0&\vdots&0\\0&1&-1&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \( x_1=0\) and \( x_2=x_3\). Taking \( x_3=1\)
yields
\( {\bf y}_1=\displaystyle{\left[\begin{array}{c}1\\1 \end{array}\right] }e^{2t}\).
The eigenvectors associated
with \( \lambda_2=-2\) satisfy the system with augmented matrix
\( \displaystyle{\left[\begin{array}{rrrcr}1&1&-1&\vdots&0\\-2&2&2&
\vdots&0\\-1&3&1&\vdots&0\end{array}\right]}\),
which is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&0&-1&\vdots&0\\0&1&0&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \( x_1=x_3\) and \( x_2=0\). Taking \( x_3=1\)
yields
\( {\bf y}_2=\displaystyle{\left[\begin{array}{c}1\\0\\1 \end{array}\right] }e^{-2t}\).
For a third solution we need a vector \( {\bf u}\) such that
\( \displaystyle{\left[\begin{array}{ccc}1&1&{-1}\\{-2}&2&2\\{-1}&3&1 \end{array}\right] \left[\begin{array}{c}{u_1}\\{u_2}\\{u_3} \end{array}\right] }=\displaystyle{\left[\begin{array}{c}1\\0\\1 \end{array}\right] }\).
The augmented matrix of this system is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&0&-1&\vdots&{1\over2}\\0&1&0&
\vdots&{1\over2}\\0&0&0&\vdots&0\end{array}\right]}\).
Let \( u_3=0\), \( u_1=\displaystyle{1\over2}\), and \( u_2=\displaystyle{1\over2}\). Then \( {\bf
y}_3=\displaystyle{\left[\begin{array}{c}1\\1\\0 \end{array}\right] }\displaystyle{e^{-2t}\over2}+\displaystyle{\left[\begin{array}{c}1\\0\\1 \end{array}\right] }te^{-2t}\).
The general solution is
$$
{\bf y}=\displaystyle{
c_1\left[\begin{array}{c}0\\1\\1 \end{array}\right] e^{2t}+c_2\left[\begin{array}{c}1\\0\\1 \end{array}\right] e^{-2t}+c_3\left
(\left[\begin{array}{c}1\\1\\0 \end{array}\right] {e^{-2t}\over2}+\left[\begin{array}{c}1\\0\\1 \end{array}\right] te^{-2t}\right)}.
$$
Exercise \(\PageIndex{11}\)
\(\displaystyle{{\bf y}' = \left[ \begin{array}{c}4 & {-2} & {-2} \\ {-2} & 3 & {-1} \\ 2 & {-1} & 3 \end{array} \right] {\bf y}}\)
Exercise \(\PageIndex{12}\)
\(\displaystyle{{\bf y}' = \left[ \begin{array}{c}6 & {-5} & 3 \\ 2 & {-1} & 3 \\ 2 & 1 & 1 \end{array} \right] {\bf y}}\)
- Answer
-
\(\left|\begin{array}{ccc}6-\lambda&{-5}&3\\2&{-1-\lambda}&3\\2&1&1-\lambda\end{array} \right|
=-(\lambda+2)(\lambda-4)^2\).
Hence \( \lambda_1=-2\) and \( \lambda_2=\lambda_3=4\).
The eigenvectors associated
with \( \lambda_1=-2\) satisfy the system with augmented matrix
\( \displaystyle{\left[\begin{array}{rrrcr}8&-5&3&\vdots&0\\2&1&3&
\vdots&0\\2&1&3&\vdots&0\end{array}\right]}\),
which is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&0&1&\vdots&0\\0&1&1&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \( x_1=-x_3\) and \( x_2=-x_3\). Taking \( x_3=1\)
yields
\( {\bf y}_1=\displaystyle{\left[\begin{array}{c}{-1}\\{-1}\\1 \end{array} \right]}e^{-2t}\).
The eigenvectors associated
with \( \lambda_2=4\) satisfy the system with augmented matrix
\( \displaystyle{\left[\begin{array}{rrrcr}2&-5&3&\vdots&0\\2&-5&3&
\vdots&0\\2&1&-3&\vdots&0\end{array}\right]}\),
which is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&0&-1&\vdots&0\\0&1&-1&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \( x_1=x_3\) and \( x_2=x_3\). Taking \( x_3=1\)
yields
\( {\bf y}_2=\displaystyle{\left[\begin{array}{c}1\\1\\1\end{array}\right]}e^{4t}\).
For a third solution we need a vector \( {\bf u}\) such that
\( \displaystyle{\left[\begin{array}{ccc}2&{-5}&3\\2&{-5}&3\\2&1&{-3}\end{array}\right]\left[\begin{array}{c}{u_1}\\{u_2}\\{u_3}\end{array}\right]}=
\displaystyle{\left[\begin{array}{c}1\\1\\1\end{array}\right]}\).
The augmented matrix of this system is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&0&-1&\vdots&{1\over2}\\0&1&-1&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Let \( u_3=0\), \( u_1=\displaystyle{1\over2}\), and \( u_2=0\). Then \( {\bf
y}_3=\displaystyle{\left[\begin{array}{c}1\\0\\0\end{array}\right]}{e^{4t}\over2}+\displaystyle{\left[\begin{array}{c}1\\1\\1\end{array}\right]}te^{4t}\). The
general solution is
\( {\bf y}=\displaystyle{
c_1\left[\begin{array}{c}{-1}\\{-1}\\1\end{array}\right]e^{-2t}+c_2\left[\begin{array}{c}1\\1\\1\end{array}\right]e^{4t}+c_3\left
(\left[\begin{array}{c}1\\0\\0\end{array}\right]{e^{4t}\over2}+\left[\begin{array}{c}1\\1\\1\end{array}\right]te^{4t}\right)}\).
In Exercises \((4.5E.13)\) to \((4.5E.23)\), solve the initial value problem.
Exercise \(\PageIndex{13}\)
\(\displaystyle{{\bf y}' = \left[ \begin{array}{c}{-11} & 8 \\ {-2} & {-3} \end{array} \right] {\bf y} , \quad {\bf y}(0) = \left[ \begin{array}{c}6 \\ 2 \end{array} \right]}\)
Exercise \(\PageIndex{14}\)
\(\displaystyle{{\bf y}' = \left[ \begin{array}{c}{15} & {-9} \\ {16} & {-9} \end{array} \right] {\bf y} , \quad {\bf y}(0) = \left[ \begin{array}{c}5 \\ 8 \end{array} \right] }\)
- Answer
-
\(\left|\begin{array}{cc}{15-\lambda}&{-9}\\{16}&{-9-\lambda}\end{array} \right| =(\lambda-3)^2\).
Hence \( \lambda_1=3\).
Eigenvectors satisfy
\( \displaystyle{\left[\begin{array}{cc}{12}&{-9}\\{16}&{-12} \end{array} \right] \left[ \begin{array}{c}{x_1}\\{x_2} \end{array} \right]= \left[ \begin{array}{c}0\\0 \end{array} \right]}\),
so \( x_1=\displaystyle{3\over4}x_2\). Taking \( x_2=4\) yields
\( {\bf y}_1=\displaystyle{ \left[ \begin{array}{c}3\\4 \end{array} \right]}e^{3t}\).
For a second solution we need a vector \( {\bf u}\)
such that \( \displaystyle{\left[\begin{array}{cc}{12}&{-9}\\{16}&{-12} \end{array} \right] \left[ \begin{array}{c}{u_1}\\{u_2} \end{array} \right]
= \left[ \begin{array}{c}3\\4 \end{array} \right]}\). Let \( u_1=\displaystyle{1\over4}\) and \( u_2=0\). Then
\( {\bf y}_2=\displaystyle{ \left[ \begin{array}{c}1\\0 \end{array} \right]}{e^{3t}\over4}+\displaystyle{ \left[ \begin{array}{c}3\\4 \end{array} \right]}te^{3t}\).
The general solution is
\( {\bf y}=c_1\displaystyle{ \left[ \begin{array}{c}3\\4 \end{array} \right]}e^{3t}+
c_2\left(\displaystyle{ \left[ \begin{array}{c}1\\0 \end{array} \right]}{e^{3t}\over4}+\displaystyle{ \left[ \begin{array}{c}3\\4 \end{array} \right]}te^{3t}\right)\).
Now \( {\bf y}(0)=\displaystyle{ \left[ \begin{array}{c}5\\8 \end{array} \right]}\Rightarrow
c_1\displaystyle{ \left[ \begin{array}{c}3\\4 \end{array} \right]}+
c_2\displaystyle{ \left[ \begin{array}{c}{1\over4}\\0 \end{array} \right]}=\displaystyle{ \left[ \begin{array}{c}5\\8 \end{array} \right]}\), so \(c_1=2\)
and \(c_2=-4\). Therefore,
\( {\bf y}=\displaystyle{ \left[ \begin{array}{c}5\\8 \end{array} \right]e^{3t}- \left[ \begin{array}{c}{12}\\{16} \end{array} \right]te^{3t}}\).
Exercise \(\PageIndex{15}\)
\(\displaystyle{{\bf y}' = \left[ \begin{array}{c}{-3} & {-4} \\ 1 & {-7} \end{array} \right] {\bf y} , \quad {\bf y}(0) = \left[ \begin{array}{c}2 \\ 3 \end{array} \right]}\)
Exercise \(\PageIndex{16}\)
\( {\bf y}' = \left[ \begin{array}{c}-7 & 24 \\ -6 & 17 \end{array} \right] {\bf y} , \quad {\bf y}(0) = \left[ \begin{array}{c}3 \\ 1 \end{array} \right] \)
- Answer
-
\(\left|\begin{array}{cc}{-7}{24}{-6}{17}\end{array} \right| =(\lambda-5)^2\).
Hence \( \lambda_1=5\).
Eigenvectors satisfy
\( \displaystyle{\left[\begin{array}{cc}{-12}&{24}\\{-6}&{12} \end{array} \right]\left[ \begin{array}{c} {x_1}\\{x_2}\end{array} \right]= \left[ \begin{array}{c} 0\\0\end{array} \right]}\),
so \( x_1=2x_2\). Taking \( x_2=1\) yields
\( {\bf y}_1=\displaystyle{ \left[ \begin{array}{c}2\\1\end{array} \right]}e^{5t}\).
For a second solution we need a vector \( {\bf u}\)
such that \( \displaystyle{\left[\begin{array}{cc}{-12}&{24}\\{-6}&{12} \end{array} \right]\left[ \begin{array}{c}{u_1}\\{u_2}\end{array} \right]}
=\displaystyle{ \left[ \begin{array}{c}2\\1\end{array} \right]}\). Let \( u_1=\displaystyle{1\over6}\) and \( u_2=0\). Then
\( {\bf y}_2=\displaystyle{ \left[ \begin{array}{c}1\\0\end{array} \right]}{e^{5t}\over6}+\displaystyle{ \left[ \begin{array}{c}2\\1\end{array} \right]}te^{5t}\).
The general solution is
\( {\bf y}=c_1\displaystyle{ \left[ \begin{array}{c}2\\1\end{array} \right]}e^{5t}
+c_2\left(\displaystyle{ \left[ \begin{array}{c}1\\0\end{array} \right]}{e^{5t}\over6}+\displaystyle{ \left[ \begin{array}{c}2\\1\end{array} \right]}te^{5t}\right)\).
Now
\( {\bf y}(0)=\displaystyle{ \left[ \begin{array}{c}3\\1\end{array} \right]}\Rightarrow c_1\displaystyle{ \left[ \begin{array}{c}2\\1\end{array} \right]}
+c_2{ \left[ \begin{array}{c}{1\over6}\\0\end{array} \right]}=\displaystyle{ \left[ \begin{array}{c}3\\1\end{array} \right]}\), so \(c_1=1\)
and \(c_2=6\). Therefore,
\( {\bf y}=\displaystyle{ \left[ \begin{array}{c}3\\1\end{array} \right]e^{5t}- \left[ \begin{array}{c}{12}\\6\end{array} \right]te^{5t}}\).
Exercise \(\PageIndex{17}\)
\(\displaystyle{{\bf y}' = \left[ \begin{array}{c}{-7} & 3 \\ {-3} & {-1} \end{array} \right] {\bf y} , \quad {\bf y}(0) = \left[ \begin{array}{c}0 \\ 2 \end{array} \right]}\)
Exercise \(\PageIndex{18}\)
\(\displaystyle{{\bf y}' = \left[ \begin{array}{c}{-1} & 1 & 0 \\ 1 & {-1} & {-2} \\ {-1} & {-1} & {-1} \end{array} \right] {\bf y} , \quad {\bf y}(0) = \left[ \begin{array}{c}6 \\ 5 \\ {-7} \end{array} \right] }\)
- Answer
-
\(\left|\begin{array}{ccc}{-1}101{-1}{-2}{-1}{-1}{-1}\end{array} \right|
=-(\lambda-1)(\lambda+2)^2\).
Hence \( \lambda_1=1\) and \( \lambda_2=\lambda_3=-2\).
The eigenvectors associated
with \( \lambda_1=1\) satisfy the system with augmented matrix
\( \displaystyle{\left[\begin{array}{rrrcr}{-2}&1&0&\vdots&0\\1&-2&-2&
\vdots&0\\-1&-1&-2&\vdots&0\end{array}\right]}\),
which is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&0&{2\over3}&\vdots&0\\0&1&{4\over3}&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \( x_1=-\displaystyle{2\over3}x_3\) and \( x_2=-\displaystyle{4\over3}x_3\). Taking
\( x_3=3\) yields
The eigenvectors associated
with \( \lambda_2=-2\) satisfy the system with augmented matrix
\( \displaystyle{\left[\begin{array}{rrrcr}1&1&0&\vdots&0\\1&1&-2&
\vdots&0\\-1&-1&1&\vdots&0\end{array}\right]}\),
which is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&1&0&\vdots&0\\0&0&1&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \( x_1=-x_2\) and \( x_3=0\). Taking \( x_2=1\)
yields
\( {\bf y}_2=\displaystyle{\left[\begin{array}{c}{-1}\\10 \end{array} \right] }e^{-2t}\).
For a third solution we need a vector \( {\bf u}\) such that
\( \displaystyle{\left[\begin{array}{ccc}1&1&0\\1&1&{-2}\\{-1}&{-1}&1 \end{array} \right] \left[\begin{array}{c}{u_1}\\{u_2}\\{u_3} \end{array} \right] }=
\displaystyle{\left[\begin{array}{c}{-1}\\1\\0 \end{array} \right] }\).
The augmented matrix of this system is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&1&0&\vdots&-1\\0&0&1&
\vdots&-1\\0&0&0&\vdots&0\end{array}\right]}\).
Let \( u_2=0\), \( u_1=-1\), and \( u_3=-1\). Then \( {\bf
y}_3=\displaystyle{\left[\begin{array}{c}{-1}\\0\\{-1} \end{array} \right] }e^{-2t}+\displaystyle{\left[\begin{array}{c}{-1}\\1\\0 \end{array} \right] }te^{-2t}\).
The general solution is
\( {\bf y}=c_1\displaystyle{\left[\begin{array}{c}{-2}\\{-4}\\3 \end{array} \right] }e^t+
c_2\displaystyle{\left[\begin{array}{c}{-1}\\1\\0 \end{array} \right] }e^{-2t}
+c_3\left(\displaystyle{\left[\begin{array}{c}{-1}\\0\\{-1} \end{array} \right] }e^{-2t}+
\displaystyle{\left[\begin{array}{c}{-1}\\1\\0 \end{array} \right] }te^{-2t}\right)\).
Now \( {\bf y}(0)=\displaystyle{\left[\begin{array}{c}6\\5\\{-7} \end{array} \right] }\Rightarrow
c_1\displaystyle{\left[\begin{array}{c}{-2}\\{-4}\\3 \end{array} \right] }+ c_2\displaystyle{\left[\begin{array}{c}{-1}\\1\\0 \end{array} \right] }
+c_3\displaystyle{\left[\begin{array}{c}{-1}\\0\\{-1} \end{array} \right] }=\displaystyle{\left[\begin{array}{c}6\\5\\{-7} \end{array} \right] }\),
so \(c_1=-2\), \(c_2=-3\), and \(c_3=1\). Therefore,
\( {\bf y}=\displaystyle{\left[\begin{array}{c}4\\8\\{-6} \end{array} \right] e^t+\left[\begin{array}{c}2\\{-3}\\{-1} \end{array} \right] e^{-2t}+
\left[\begin{array}{c}{-1}\\1\\0 \end{array} \right] te^{-2t}}\).
Exercise \(\PageIndex{19}\)
\(\displaystyle{{\bf y}' = \left[ \begin{array}{c}{-2} & 2 & 1 \\ {-2} & 2 & 1 \\ {-3} & 3 & 2 \end{array} \right] {\bf y} , \quad {\bf y}(0) = \left[ \begin{array}{c}{-6} \\ {-2} \\ 0 \end{array} \right] }\)
Exercise \(\PageIndex{20}\)
\(\displaystyle{{\bf y}' = \left[ \begin{array}{c}{-7} & {-4} & 4 \\ {-1} & 0 & 1 \\ {-9} & {-5} & 6 \end{array} \right] {\bf y} , \quad \bf {\bf y}(0) = \left[ \begin{array}{c}{-6} \\ 9 \\ {-1} \end{array} \right]}\)
- Answer
-
\(\left|\begin{array}{ccc}{-7}{-4}4{-1}01{-9}{-5}6\end{array} \right|
=-(\lambda+3)(\lambda-1)^2\).
Hence \( \lambda_1=-3\) and \( \lambda_2=\lambda_3=1\).
The eigenvectors associated
with \( \lambda_1=-3\) satisfy the system with augmented matrix
\( \displaystyle{\left[\begin{array}{rrrcr}-4&-4&4&\vdots&0\\-1&3&1&
\vdots&0\\-9&-5&9&\vdots&0\end{array}\right]}\),
which is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&0&-1&\vdots&0\\0&1&0&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \( x_1=x_3\) and \( x_2=0\). Taking \( x_3=1\)
yields
\( {\bf y}_1=\displaystyle{\left[\begin{array}{c}1\\0\\1 \end{array} \right] }e^{-3t}\).
The eigenvectors associated
with \( \lambda_2=1\) satisfy the system with augmented matrix
\( \displaystyle{\left[\begin{array}{rrrcr}-8&-4&4&\vdots&0\\-1&-1&1&
\vdots&0\\-9&-5&-5&\vdots&0\end{array}\right]}\),
which is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&0&0&\vdots&0\\0&1&-1&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \( x_1=0\) and \( x_2=x_3\). Taking \( x_3=1\)
yields
\( {\bf y}_2=\displaystyle{\left[\begin{array}{c}0\\1\\1 \end{array} \right] }e^t\).
For a third solution we need a vector \( {\bf u}\) such that
\( \displaystyle{\left[\begin{array}{ccc}{-8}&{-4}&4\\{-1}&{-1}&1\\{-9}&{-5}&5 \end{array} \right] \left[\begin{array}{c}{u_1}\\{u_2}\\{u_3} \end{array} \right] }
=\displaystyle{\left[\begin{array}{c}0\\1\\1 \end{array} \right] }\).
The augmented matrix of this system is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&0&0&\vdots&1\\0&1&-1&
\vdots&-2\\0&0&0&\vdots&0\end{array}\right]}\).
Let \( u_3=0\), \( u_1=1\), and \( u_2=-2\). Then \( {\bf
y}_3=\displaystyle{\left[\begin{array}{c}1\\{-2}\\0 \end{array} \right] }e^t+\displaystyle{\left[\begin{array}{c}0\\1\\1 \end{array} \right] }te^{t}\). The general
solution is
\( {\bf y}=c_1\displaystyle{\left[\begin{array}{c}1\\0\\1 \end{array} \right] }e^{-3t}+ c_2\displaystyle{\left[\begin{array}{c}0\\1\\1 \end{array} \right] }e^t+
c_3\left(\displaystyle{\left[\begin{array}{c}1\\{-2}\\0 \end{array} \right] }e^t+\displaystyle{\left[\begin{array}{c}0\\1\\1 \end{array} \right] }te^{t}\right)\).
Now \( {\bf y}(0)=\displaystyle{\left[\begin{array}{c}{-6}\\9\\{-1} \end{array} \right] }\Rightarrow
c_1\displaystyle{\left[\begin{array}{c}1\\0\\1 \end{array} \right] }+ c_2\displaystyle{\left[\begin{array}{c}0\\1\\1 \end{array} \right] }+
c_3\displaystyle{\left[\begin{array}{c}1\\{-2}\\0 \end{array} \right] }=\displaystyle{\left[\begin{array}{c}{-6}\\9\\{-1} \end{array} \right] }\),
so \(c_1=-2\), \(c_2=1\), and \(c_3=-4\). Therefore,
\( {\bf y}=\displaystyle{-\left[\begin{array}{c}2\\0\\2 \end{array} \right] e^{-3t}+\left[\begin{array}{c}{-4}\\9\\1 \end{array} \right] e^t-
\left[\begin{array}{c}0\\4\\4 \end{array} \right] te^t}\).
Exercise \(\PageIndex{21}\)
\(\displaystyle{{\bf y}' = \left[ \begin{array}{c}{-1} & {-4} & {-1} \\ 3 & 6 & 1 \\ {-3} & {-2} & 3 \end{array} \right] {\bf y} , \quad \bf y(0) = \left[ \begin{array}{c}{-2} \\ 1 \\ 3 \end{array} \right]}\)
Exercise \(\PageIndex{22}\)
\(\displaystyle{{\bf y}' = \left[ \begin{array}{c}4 & {-8} & {-4} \\ {-3} & {-1} & {-3} \\ 1 & {-1} & 9 \end{array} \right] {\bf y} , \quad {\bf y}(0) = \left[ \begin{array}{c}{-4} \\ 1 \\ {-3} \end{array} \right]}\)
- Answer
-
\(\left|\begin{array}{ccc}4{-8}{-4}{-3}{-1}{-3}1{-1}9\end{array} \right|
=-(\lambda+4)(\lambda-8)^2\).
Hence \( \lambda_1=-4\) and \( \lambda_2=\lambda_3=8\).
The eigenvectors associated
with \( \lambda_1=-4\) satisfy the system with augmented matrix
\( \displaystyle{\left[\begin{array}{rrrcr}8&-8&-4&\vdots&0\\-3&3&-3&
\vdots&0\\1&-1&13&\vdots&0\end{array}\right]}\),
which is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&-1&0&\vdots&0\\0&0&1&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \( x_1=x_2\) and \( x_3=0\). Taking \( x_2=1\)
yields
\( {\bf y}_1=\displaystyle{\left[\begin{array}{c}1\\1\\0 \end{array} \right] }e^{t}\).
The eigenvectors associated
with \( \lambda_2=8\) satisfy the system with augmented matrix
\( \displaystyle{\left[\begin{array}{rrrcr}-4&-8&-4&\vdots&0\\-3&-9&-3&
\vdots&0\\1&-1&1&\vdots&0\end{array}\right]}\),
which is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&0&1&\vdots&0\\0&1&0&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \( x_1=-x_3\) and \( x_2=0\). Taking \( x_3=1\)
yields
\( {\bf y}_2=\displaystyle{\left[\begin{array}{c}{-1}\\0\\1 \end{array} \right] }e^{8t}\).
For a third solution we need a vector \( {\bf u}\) such that
\( \displaystyle{\left[\begin{array}{ccc}{-4}&{-8}&{-4}\\{-3}&{-9}&{-3}\\1&{-1}&1\end{array} \right]
\left[\begin{array}{c}{u_1}\\{u_2}\\{u_3}\end{array} \right]}=\displaystyle{\left[\begin{array}{c}{-1}\\0\\1\end{array} \right]}\).
The augmented matrix of this system is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&0&1&\vdots&{3\over4}\\0&1&0&
\vdots&-{1\over4}\\0&0&0&\vdots&0\end{array}\right]}\).
Let \( u_3=0\), \( u_1=\displaystyle{3\over4}\), and \( u_2=-\displaystyle{1\over4}\). Then \( {\bf
y}_3=\displaystyle{\left[\begin{array}{c}3\\{-1}\\0\end{array} \right]}{e^{8t}\over4}+\displaystyle{\left[\begin{array}{c}{-1}\\0\\1 \end{array} \right]}te^{8t}\).
The general solution is
\(c_1\displaystyle{\left[\begin{array}{c}1\\1\\0\end{array} \right]}e^{t}+
c_2\displaystyle{\left[\begin{array}{c}{-1}\\0\\1\end{array} \right]}e^{8t}+
c_3\displaystyle\left({\left[\begin{array}{c}3\\{-1}\\0\end{array} \right]}{e^{8t}\over4}+
\displaystyle{\left[\begin{array}{c}{-1}\\0\\1\end{array} \right]}te^{8t}\right)\).
Now \( {\bf y}(0)=\displaystyle{\left[\begin{array}{c}{-4}\\1\\{-3}\end{array} \right]}\Rightarrow
c_1\displaystyle{\left[\begin{array}{c}1\\1\\0\end{array} \right]}+
c_2\displaystyle{\left[\begin{array}{c}{-1}\\0\\1\end{array} \right]}+
c_3\displaystyle{\left[\begin{array}{c}{3\over4}\\{-{1\over4}}\\0\end{array} \right]}=\displaystyle{\left[\begin{array}{c}{-4}\\1\\{-3}\end{array} \right]}\),
so \(c_1=-1\), \(c_2=-3\), and \(c_3=-8\). Therefore,
\( {\bf y}=\displaystyle{-\left[\begin{array}{c}1\\1\\0\end{array} \right]e^{-4t}+\left[\begin{array}{c}{-3}\\2\\{-3}\end{array} \right]e^{8t}+
\left[\begin{array}{c}8\\0\\{-8}\end{array} \right]te^{8t}}\).
Exercise \(\PageIndex{23}\)
\(\displaystyle{{\bf y}'= \left[ \begin{array}{c}{-5} & {-1} & {11} \\ {-7} & 1 & {13} \\ {-4} & 0 & 8 \end{array} \right] {\bf y} , \quad {\bf y}(0) = \left[ \begin{array}{c}0 \\ 2 \\ 2 \end{array} \right] }\)
The coefficient matrices in Exercises \((4.5E.24)\) to \((4.5E.32)\) have eigenvalues of multiplicity \(3\). Find the general solution.
Exercise \(\PageIndex{24}\)
\(\displaystyle{{\bf y}' = \left[ \begin{array}{c}5 & {-1} & 1 \\ {-1} & 9 & {-3} \\ {-2} & 2 & 4 \end{array} \right] {\bf y}}\)
- Answer
-
\(\left|\begin{array}{ccc}5-\lambda&{-1}&1\\{-1}&9-\lambda&{-3}\\{-2}&2&4-\lambda\end{array} \right|
=-(\lambda-6)^3\).
Hence \( \lambda_1=6\). The eigenvectors
satisfy the system with augmented matrix
\( \displaystyle{\left[\begin{array}{rrrcr}1&-1&1&\vdots&0\\-1&3&-3&
\vdots&0\\-2&2&-2&\vdots&0\end{array}\right]}\),
which is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&0&0&\vdots&0\\0&1&-1&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \( x_1=0\) and \( x_2=x_3\). Taking \( x_3=1\)
yields
\( {\bf y}_1=\displaystyle{\left[\begin{array}{c}0\\1\\1 \end{array} \right]}e^{6t}\).
For a second solution we need a vector \( {\bf u}\) such that
\( \displaystyle{\left[\begin{array}{ccc}1&{-1}&1\\{-1}&3&{-3}\\{-2}&2&{-2} \end{array} \right]\left[\begin{array}{c}{u_1}\\{u_2}\\{u_3} \end{array} \right]}
=\displaystyle{\left[\begin{array}{c}0\\1\\1 \end{array} \right]}\).
The augmented matrix of this system is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&0&0&\vdots&-{1\over4}\\0&1&-1&
\vdots&{1\over4}\\0&0&0&\vdots&0\end{array}\right]}\).
Let \( u_3=0\), \( u_1=-\displaystyle{1\over4}\), and \( u_2=\displaystyle{1\over4}\). Then \( {\bf
y}_2=\displaystyle{\left[\begin{array}{c}{-1}\\1\\0 \end{array} \right]}{e^{6t}\over4}+\displaystyle{\left[\begin{array}{c}0\\1\\1 \end{array} \right]}te^{6t}\).
For a third solution we need a vector \( {\bf v}\) such that
\( \displaystyle{\left[\begin{array}{ccc}1&{-1}&1\\{-1}&3&{-3}\\{-2}&2&{-2} \end{array} \right]\left[\begin{array}{c}{v_1}\\{v_2}\\{v_3} \end{array} \right]}
=\displaystyle{\left[\begin{array}{c}{-{1\over4}}\\{1\over4}\\0 \end{array} \right]}\).
The augmented matrix of this system is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&0&0&\vdots&{1\over8}\\0&1&-1&
\vdots&{1\over8}\\0&0&0&\vdots&0\end{array}\right]}\).
Let \(v_3=0\), \(v_1=\displaystyle{1\over8}\), and \(v_2=\displaystyle{1\over8}\). Then
\( {\bf y}_3= \displaystyle{\left[\begin{array}{c}1\\1\\0 \end{array} \right]}{e^{6t}\over8}+
\displaystyle{\left[\begin{array}{c}{-1}\\1\\0 \end{array} \right]}{te^{6t}\over4}+\displaystyle{\left[\begin{array}{c}0\\1\\1 \end{array} \right]}{t^2e^{6t}\over2}\).
The general solution is
\( {\bf y}=\displaystyle{c_1\left[\begin{array}{c}0\\1\\1 \end{array} \right]e^{6t}+
c_2\left(\left[\begin{array}{c}{-1}\\1\\0 \end{array} \right]{e^{6t}\over4}+\left[\begin{array}{c}0\\1\\1 \end{array} \right]te^{6t}\right)}+c_3\left(\left[\begin{array}{c}1\\1\\0 \end{array} \right]{e^{6t}\over8}+\left[\begin{array}{c}{-1}\\1\\0 \end{array} \right]{t
e^{6t}\over4}+\left[\begin{array}{c}0\\1\\1 \end{array} \right]{t^2e^{6t}\over2}\right).\)
Exercise \(\PageIndex{25}\)
\(\displaystyle{{\bf y}' = \left[ \begin{array}{c}1 & {10} & {-12} \\ 2 & 2 & 3 \\ 2 & {-1} & 6 \end{array} \right] {\bf y}}\)
Exercise \(\PageIndex{26}\)
\(\displaystyle{{\bf y}' = \left[ \begin{array}{c}{-6} & {-4} & {-4} \\ 2 & {-1} & 1 \\ 2 & 3 & 1 \end{array} \right] {\bf y}}\)
- Answer
-
\(\left|\begin{array}{ccc}{-6-\lambda}&{-4}&{-4}\\2&{-1-\lambda}&1\\2&3&1-\lambda\end{array} \right|
=-(\lambda+2)^3\).
Hence \( \lambda_1=-2\).
The eigenvectors
satisfy the system with an augmented matrix
\( \displaystyle{\left[\begin{array}{rrrcr}-4&-4&-4&\vdots&0\\2&1&1&
\vdots&0\\2&3&3&\vdots&0\end{array}\right]}\),
which is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&0&0&\vdots&0\\0&1&1&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \( x_1=0\) and \( x_2=-x_3\). Taking \( x_3=1\)
yields
\( {\bf y}_1=\displaystyle{\left[\begin{array}{c}0\\{-1}\\1 \end{array} \right]}e^{-2t}\).
For a second solution we need a vector \( {\bf u}\) such that
\( \displaystyle{\left[\begin{array}{ccc}{-4}&{-4}&{-4}\\2&1&1\\2&3&3 \end{array} \right]\left[\begin{array}{c}{u_1}\\{u_2}\\{u_3} \end{array} \right]}
=\displaystyle{\left[\begin{array}{c}0\\{-1}\\1 \end{array} \right]}\).
The augmented matrix of this system is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&0&0&\vdots&-1\\0&1&1&
\vdots&1\\0&0&0&\vdots&0\end{array}\right]}\).
Let \( u_3=0\), \( u_1=-1\), and \( u_2=1\). Then
\( {\bf y}_2=\left[\begin{array}{c}-1\\1\\0 \end{array} \right]e^{-2t}+\left[\begin{array}{c}0\\{-1}\\1 \end{array} \right]te^{-2t}\).
For a third solution we need a vector \( {\bf v}\) such that
\( \displaystyle{\left[\begin{array}{ccc}{-4}&{-4}&{-4}\\2&1&1\\2&3&3 \end{array} \right]\left[\begin{array}{c}{v_1}\\{v_2}\\{v_3} \end{array} \right]}
=\displaystyle{\left[\begin{array}{c}{-1}\\1\\0 \end{array} \right]}\).
The augmented matrix of this system is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&0&0&\vdots&{3\over4}\\0&1&1&
\vdots&-{1\over2}\\0&0&0&\vdots&0\end{array}\right]}\).
Let \(v_3=0\), \(v_1=\displaystyle{3\over4}\), and \(v_2=-\displaystyle{1\over2}\). Then
\( {\bf y}_3= \left[\begin{array}{c}3\\{-2}\\0 \end{array} \right]\displaystyle{e^{-2t}\over4}+\left[\begin{array}{c}{-1}\\1\\0 \end{array} \right]t
e^{-2t}+\left[\begin{array}{c}0\\{-1}\\1 \end{array} \right]\displaystyle{t^2e^{-2t}\over2}\).
The general solution is
\( {\bf y}=\displaystyle{c_1\left[\begin{array}{c}0\\{-1}\\1 \end{array} \right]e^{-2t}+
c_2\left(\left[\begin{array}{c}-1\\1\\0 \end{array} \right]e^{-2t}+\left[\begin{array}{c}0\\{-1}\\1 \end{array} \right]te^{-2t}\right)}+c_3\left(\left[\begin{array}{c}3\\{-2}\\0 \end{array} \right]{e^{-2t}\over4}+\left[\begin{array}{c}{-1}\\1\\0 \end{array} \right]t
e^{-2t}+\left[\begin{array}{c}0\\{-1}\\1 \end{array} \right]{t^2e^{-2t}\over2}\right)\)
Exercise \(\PageIndex{27}\)
\(\displaystyle{{\bf y}' = \left[ \begin{array}{c}0 & 2 & {-2} \\ {-1} & 5 & {-3} \\ 1 & 1 & 1 \end{array} \right] {\bf y}}\)
Exercise \(\PageIndex{28}\)
\(\displaystyle{{\bf y}' = \left[ \begin{array}{c}{-2} & {-12} & {10} \\ 2 & {-24} & {11} \\ 2 & {-24} & 8 \end{array} \right] {\bf y}}\)
- Answer
-
\(\left|\begin{array}{ccc}{-2-\lambda}&{-12}&{10}\\2&{-24-\lambda}&{11}\\2&{-24}&8-\lambda\end{array} \right|
=-(\lambda+6)^3\).
Hence \( \lambda_1=-6\).
The eigenvectors
satisfy the system with augmented matrix
\( \displaystyle{\left[\begin{array}{rrrcr}4&-12&10&\vdots&0\\2&-18&11&
\vdots&0\\2&-24&14&\vdots&0\end{array}\right]}\),
which is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&0&1&\vdots&0\\0&1&-{1\over2}&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \( x_1=-x_3\) and \( x_2=\displaystyle{x_3\over2}\). Taking \( x_3=2\)
yields
\( {\bf y}_1=\displaystyle{\left[\begin{array}{c}{-2}\\1\\2 \end{array} \right]}e^{-6t}\).
For a second solution we need a vector \( {\bf u}\) such that
\( \displaystyle{\left[\begin{array}{ccc}4&{-12}&{10}\\2&{-18}&{11}\\2&{-24}&{14} \end{array} \right] \left[\begin{array}{c}{u_1}\\{u_2}\\{u_3} \end{array} \right]}
=\displaystyle{\left[\begin{array}{c}{-2}\\1\\2 \end{array} \right]}\).
The augmented matrix of this system is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&0&1&\vdots&-1\\0&1&-{1\over2}&
\vdots&-{1\over6}\\0&0&0&\vdots&0\end{array}\right]}\).
Let \( u_3=0\), \( u_1=-1\), and \( u_2=-\displaystyle{1\over6}\). Then
\( {\bf y}_2=-\left[\begin{array}{c}6\\1\\0 \end{array} \right]\displaystyle{e^{-6t}\over 6}+\left[\begin{array}{c}{-2}\\1\\2 \end{array} \right]te^{-6t}\).
For a third solution we need a vector \( {\bf v}\) such that
\( \displaystyle{\left[\begin{array}{ccc}4&{-12}&{10}\\2&{-18}&{11}\\2&{-24}&{14} \end{array} \right]\left[\begin{array}{c}{v_1}\\{v_2}\\{v_3} \end{array} \right]}
=\displaystyle{\left[\begin{array}{c}{-1}\\{-{1\over6}}\\0 \end{array} \right]}\).
The augmented matrix of this system is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&0&1&\vdots&-{1\over3}\\0&1&-{1\over2}&
\vdots&-{1\over36}\\0&0&0&\vdots&0\end{array}\right]}\).
Let \(v_3=0\), \(v_1=-\displaystyle{1\over3}\), and \(v_2=-\displaystyle{1\over36}\). Then
\( {\bf y}_3= -\left[\begin{array}{c}{12}\\1\\0 \end{array} \right]\displaystyle{e^{-6t}\over36}-\left[\begin{array}{c}6\\1\\0 \end{array} \right]{t
e^{-6t}\over6}+\left[\begin{array}{c}{-2}\\1\\2 \end{array} \right]{t^2e^{-6t}\over2}\).
The general solution is
\( {\bf y}=\displaystyle{c_1\left[\begin{array}{c}{-2}\\1\\2 \end{array} \right]e^{-6t}+
c_2\left(-\left[\begin{array}{c}6\\1\\0 \end{array} \right]{e^{-6t}\over6}+\left[\begin{array}{c}{-2}\\1\\2 \end{array} \right]te^{-6t}\right)}+c_3\left(-\left[\begin{array}{c}{12}\\1\\0 \end{array} \right]{e^{-6t}\over36}-\left[\begin{array}{c}6\\1\\0 \end{array} \right]{t
e^{-6t}\over6}+\left[\begin{array}{c}{-2}\\1\\2 \end{array} \right]{t^2e^{-6t}\over2}\right).\)
Exercise \(\PageIndex{29}\)
\(\displaystyle{{\bf y}' = \left[ \begin{array}{c}{-1} & {-12} & 8 \\ 1 & {-9} & 4 \\ 1 & {-6} & 1 \end{array} \right] {\bf y}}\)
Exercise \(\PageIndex{30}\)
\(\displaystyle{{\bf y}' = \left[ \begin{array}{c}{-4} & 0 & {-1} \\ {-1} & {-3} & {-1} \\ 1 & 0 & {-2} \end{array} \right] {\bf y}}\)
- Answer
-
\(\left|\begin{array}{ccc}{-4-\lambda}&0&{-1}\\{-1}&{-3-\lambda}&{-1}\\1&0&{-2-\lambda}\end{array} \right|
=-(\lambda+3)^3\).
Hence \( \lambda_1=3\).
The eigenvectors
satisfy the system with augmented matrix
\( \displaystyle{\left[\begin{array}{rrrcr}-1&0&-1&\vdots&0\\-1&0&-1&
\vdots&0\\1&0&1&\vdots&0\end{array}\right]}\),
which is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&0&1&\vdots&0\\0&0&0&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \( x_1=-x_3\) and \( x_2\) is arbitrary. Taking \( x_2=0\) and \( x_3=1\)
yields
\( {\bf y}_1=\left[\begin{array}{c}{-1}\\0\\1 \end{array} \right]e^{-3t}\).
Taking \( x_2=1\) and \( x_3=0\)
yields \( {\bf y}_2=\left[\begin{array}{c}0\\1\\0 \end{array} \right]e^{-3t}\).
For a third solution we need constants \( \alpha\) and \( \beta\) and a
vector \( {\bf u}\) such that
\( \displaystyle{\left[\begin{array}{ccc}{-1}&0&{-1}\\{-1}&0&{-1}\\1&0&1 \end{array} \right]\left[\begin{array}{c}{u_1}\\{u_2}\\{u_3} \end{array} \right]}
=\alpha\displaystyle{\left[\begin{array}{c}{-1}\\0\\1 \end{array} \right]}+\beta\left[\begin{array}{c}0\\1\\0 \end{array} \right]\).
The augmented matrix of this system is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcc}1&0&1&\vdots&\alpha\\0&0&0&
\vdots&\alpha+\beta\\0&0&0&\vdots&0\end{array}\right]}\);
hence the system
has a solution if \( \alpha=-\beta=1\), which yields the eigenvector
\( {\bf x}_3=\left[\begin{array}{c}{-1}\\{-1}\\1 \end{array} \right]\). Taking \( u_1=1\) and \( u_2=u_3=0\)
yields the solution
\( {\bf y}_3=\left[\begin{array}{c}1\\0\\0 \end{array} \right]e^{-3t}+\left[\begin{array}{c}{-1}\\{-1}\\1 \end{array} \right]te^{-3t}\).
The general solution is
\( {\bf
y}=\displaystyle{c_1\left[\begin{array}{c}{-1}\\0\\1 \end{array} \right]e^{-3t}+c_2\left[\begin{array}{c}0\\1\\0 \end{array} \right]e^{-3t}+c_3\left
(\left[\begin{array}{c}1\\0\\0 \end{array} \right]e^{-3t}+\left[\begin{array}{c}{-1}\\{-1}\\1 \end{array} \right]te^{-3t}\right)}\)
Exercise \(\PageIndex{31}\)
\(\displaystyle{{\bf y}' = \left[ \begin{array}{c}{-3} & {-3} & 4 \\ 4 & 5 & {-8} \\ 2 & 3 & {-5} \end{array} \right] \bf y}\)
Exercise \(\PageIndex{32}\)
\({\bf y}' = { \left[ \begin{array}{c}{-3} & {-1} & 0 \\ 1 & {-1} & 0 \\ {-1} & {-1} & {-2} \end{array} \right]}{\bf y}\)
- Answer
-
\(\left|\begin{array}{ccc}{-3-\lambda}&{-1}&0\\1&{-1-\lambda}&0\\{-1}&{-1}&{-2-\lambda}\end{array} \right|
=-(\lambda+2)^3\).
Hence \( \lambda_1=-2\).
The eigenvectors
satisfy the system with augmented matrix
\( \displaystyle{\left[\begin{array}{rrrcr}-1&-1&0&\vdots&0\\1&1&0&
\vdots&0\\-1&-1&0&\vdots&0\end{array}\right]}\),
which is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcr}1&1&0&\vdots&0\\0&0&0&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \( x_1=-x_2\) and \( x_3\) is arbitrary. Taking \( x_2=1\) and \( x_3=0\)
yields
\( {\bf y}_1=\left[\begin{array}{c}{-1}\\1\\0 \end{array} \right]e^{-2t}\).
Taking \( x_2=0\) and \( x_3=1\) yields
\( {\bf y}_2=\left[\begin{array}{c}0\\0\\1 \end{array} \right]e^{-2t}\).
For a third solution we need constants \( \alpha\) and \( \beta\) and a
vector \( {\bf u}\) such that
\( \displaystyle{\left[\begin{array}{ccc}{-1}&{-1}&0\\1&1&0\\{-1}&{-1}&0 \end{array} \right]\left[\begin{array}{c}{u_1}\\{u_2}\\{u_3} \end{array} \right]}
=\alpha\displaystyle{\left[\begin{array}{c}{-1}\\1\\0 \end{array} \right]}+\beta\left[\begin{array}{c}0\\0\\1 \end{array} \right]\).
The augmented matrix of this system is row equivalent to
\( \displaystyle{\left[\begin{array}{rrrcc}1&1&0&\vdots&\alpha\\0&0&0&
\vdots&\alpha+\beta\\0&0&0&\vdots&0\end{array}\right]}\);
hence the system
has a solution if \( \alpha=-\beta=1\), which yields the eigenvector
\( {\bf x}_3=\left[\begin{array}{c}{-1}\\1\\{-1} \end{array} \right]\). Taking \( u_1=1\) and \( u_2=u_3=0\)
yields the solution
\( {\bf y}_3=
\left[\begin{array}{c}1\\0\\0 \end{array} \right]e^{-2t}+\left[\begin{array}{c}{-1}\\1\\{-1} \end{array} \right]te^{-2t}\).
The general solution is
\( {\bf y}=\displaystyle{
c_1\left[\begin{array}{c}{-1} \\1\\0 \end{array} \right]e^{-2t}+c_2\left[\begin{array}{c}0\\0\\1 \end{array} \right]e^{-2t}+c_3\left
(\left[\begin{array}{c}1\\0\\0 \end{array} \right]e^{-2t}+\left[\begin{array}{c}{-1}\\1\\{-1} \end{array} \right]te^{-2t}\right)}\).
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 \({\bf y}_1={\bf x}e^{\lambda_1t}\).
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
-
\begin{eqnarray*}
{\bf y}_3'-A{\bf y}_3&=&(\lambda_1I-A){\bf v}e^{\lambda_1t}
+(\lambda_1I-A){\bf u}te^{\lambda_1t}+{\bf u}e^{\lambda_1t}\\
&&+(\lambda_1I-A){\bf x}{t^2e^{\lambda_1t}\over2}+{\bf
x}te^{\lambda_1t}\\
&=&-{\bf u}e^{\lambda_1t}-{\bf x}te^{\lambda_1t}+{\bf
u}e^{\lambda_1t}+0+{\bf x}te^{\lambda_1t}=0.
\end{eqnarray*}
Now suppose that \(c_1{\bf y}_1+c_2{\bf y}_2+c_3{\bf y}_3={\bf 0}\).
Then
$$
c_1{\bf x}+c_2({\bf u}+t{\bf x})+c_3\displaystyle\left({\bf
v}+t{\bf u}+{t^2\over2}{\bf x}\right)={\bf 0}.
\eqno{\rm(A)}
$$
Differentiating this twice yields \(c_3{\bf x}=0\), so \(c_3=0\) since
\( {\bf x}\ne{\bf 0}\). Therefore,(A) reduces to
(B) \(c_1{\bf x}+c_2({\bf u}+t{\bf x})={\bf 0}\).
Differentiating this yields \(c_2{\bf x}=0\), so \(c_2=0\) since
\( {\bf x}\ne{\bf 0}\). Therefore,(B) reduces to \(c_3{\bf x}={\bf 0}\),
so \(c_1=0\) since
\( {\bf x}\ne{\bf 0}\). Therefore,\( {\bf y}_1\), \( {\bf y}_2\), and \( {\bf
y}_3\) are linearly independendent.
Exercise \(\PageIndex{35}\)
Suppose the matrix
\begin{eqnarray*}
A = \left[ \begin{array}{c}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 \({\bf x}\).
In Exercises \((4.5E.36)\) to \((4.5E.45)\), plot trajectories of the given system.
Exercise \(\PageIndex{36}\)
\({\bf y}'=\displaystyle{ \left[ \begin{array}{c}{-3} & {-1} \\ 4 & 1 \end{array} \right] }{\bf y}\)
Exercise \(\PageIndex{37}\)
\({\bf y}' = \displaystyle{ \left[ \begin{array}{c}2 & {-1} \\ 1 & 0 \end{array} \right]}{\bf y}\)
Exercise \(\PageIndex{38}\)
\({\bf y}' = \displaystyle{ \left[ \begin{array}{c}{-1} & {-3} \\ 3 & 5 \end{array} \right] }{\bf y}\)
Exercise \(\PageIndex{39}\)
\({\bf y}' = \displaystyle{ \left[ \begin{array}{c}{-5} & 3 \\ {-3} & 1 \end{array} \right] }{\bf y}\)
Exercise \(\PageIndex{40}\)
\({\bf y}' = \displaystyle{ \left[ \begin{array}{c}{-2} & {-3} \\ 3 & 4 \end{array} \right] }{\bf y}\)
Exercise \(\PageIndex{41}\)
\({\bf y}' = \displaystyle { \left[ \begin{array}{c}{-4} & {-3} \\ 3 & 2 \end{array} \right] }{\bf y}\)
Exercise \(\PageIndex{42}\)
\({\bf y}' = \displaystyle{ \left[ \begin{array}{c}0 & {-1} \\ 1 & {-2} \end{array} \right] }{\bf y}\)
Exercise \(\PageIndex{43}\)
\({\bf y}' = \displaystyle{ \left[ \begin{array}{c}0 & 1 \\ {-1} & 2 \end{array} \right] }{\bf y}\)
Exercise \(\PageIndex{44}\)
\({\bf y}' = \displaystyle{ \left[ \begin{array}{c}{-2} & 1 \\ {-1} & 0 \end{array} \right] }{\bf y}\)
Exercise \(\PageIndex{45}\)
\({\bf y}' = \displaystyle{ \left[ \begin{array}{c}0 & {-4} \\ 1 & {-4} \end{array} \right] }{\bf y}\)