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Mathematics LibreTexts

4.6: Constant Coefficient Homogeneous Systems III

  • Page ID
    17439
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    We now consider the system ${\bf y}'=A{\bf y}$, where $A$ has a complex
    eigenvalue $\lambda=\alpha+i\beta$ with $\beta\ne0$. We continue to
    assume that $A$ has real entries, so the characteristic polynomial of
    $A$ has real coefficients. This implies that
    $\overline\lambda=\alpha-i\beta$ is also an eigenvalue of $A$.

    An eigenvector ${\bf x}$ of $A$ associated with
    $\lambda=\alpha+i\beta$ will have complex entries, so we'll write
    $$
    {\bf x}={\bf u}+i{\bf v}
    $$
    where ${\bf u}$ and ${\bf v}$ have real entries; that is, ${\bf u}$
    and ${\bf v}$ are the real and imaginary parts of ${\bf x}$. Since
    $A{\bf x}=\lambda {\bf x}$,
    \begin{equation} \label{eq:10.6.1}
    A({\bf u}+i{\bf v})=(\alpha+i\beta)({\bf u}+i{\bf v}).
    \end{equation}
    Taking complex conjugates here and recalling that $A$ has real entries
    yields
    $$
    A({\bf u}-i{\bf v})=(\alpha-i\beta)({\bf u}-i{\bf v}),
    $$
    which shows that ${\bf x}={\bf u}-i{\bf v}$ is an eigenvector
    associated with $\overline\lambda=\alpha-i\beta$. The complex
    conjugate eigenvalues $\lambda$ and $\overline\lambda$ can be
    separately associated with linearly independent solutions ${\bf
    y}'=A{\bf y}$;     however, we won't pursue this approach, since solutions
    obtained in this way turn out to be complex--valued. Instead, we'll
    obtain solutions of ${\bf y}'=A{\bf y}$ in the form
    \begin{equation} \label{eq:10.6.2}
    {\bf y}=f_1{\bf u}+f_2{\bf v}
    \end{equation}
    where $f_1$ and $f_2$ are real--valued scalar functions.
    The next theorem shows how to do this.

    \begin{theorem}\color{blue} \label{thmtype:10.6.1}
    Let $A$ be an $n\times n$ matrix with real entries$.$ Let
    $\lambda=\alpha+i\beta$ {\rm(}$\beta\ne0${\rm}) be a complex
    eigenvalue of $A$ and let ${\bf x}={\bf u}+i{\bf v}$ be an associated
    eigenvector$,$ where ${\bf u}$ and ${\bf v}$ have real components$.$ Then
    ${\bf u}$ and ${\bf v}$ are both nonzero and
    $$
    {\bf y}_1=e^{\alpha t}({\bf u}\cos\beta t-{\bf v}\sin\beta t)
    \mbox{\quad and\quad}
    {\bf y}_2=e^{\alpha t}({\bf u}\sin\beta t+{\bf v}\cos\beta t),
    $$
    which are the real and imaginary parts of
    \begin{equation} \label{eq:10.6.3}
    e^{\alpha t}(\cos\beta t+i\sin\beta t)({\bf u}+i{\bf v}),
    \end{equation}
    are linearly independent solutions of  ${\bf y}'=A{\bf y}$.
    \end{theorem}

    \proof
    A function of the form \eqref{eq:10.6.2} is a solution of ${\bf y}'=A{\bf
    y}$ if and only if
    \begin{equation} \label{eq:10.6.4}
    f_1'{\bf u}+f_2'{\bf
    v}=f_1A{\bf u}+f_2A{\bf v}.
    \end{equation}
    Carrying out the multiplication indicated on the  right side
    of \eqref{eq:10.6.1} and collecting the  real and imaginary parts of the
    result yields
    $$
    A({\bf u}+i{\bf v})=(\alpha{\bf u}-\beta{\bf v})+i(\alpha{\bf v}+\beta{\bf
    u}).
    $$
    Equating real and imaginary parts on the two sides of this equation yields
    $$
    \begin{array}{rcl}
    A{\bf u}&=&\alpha{\bf u}-\beta{\bf v}\\
    A{\bf v}&=&\alpha{\bf v}+\beta{\bf u}.
    \end{array}
    $$
    We leave it to you (Exercise~\ref{exer:10.6.25}) to show from this that
    ${\bf u}$ and
    ${\bf v}$ are both nonzero.
    Substituting from these equations into \eqref{eq:10.6.4} yields
    \begin{eqnarray*}
    f_1'{\bf u}+f_2'{\bf v}
    &=&f_1(\alpha{\bf u}-\beta{\bf v})+f_2(\alpha{\bf v}+\beta{\bf u})\\
    &=&(\alpha f_1+\beta f_2){\bf u}+(-\beta f_1+\alpha f_2){\bf v}.
    \end{eqnarray*}
    This is true if
    $$
    \begin{array}{rcr}
    f_1'&=&\alpha f_1+\beta f_2\phantom{,}\\
    f_2'&=&-\beta f_1+\alpha f_2,
    \end{array}
    \mbox{\quad or, equivalently,\quad}
    \begin{array}{rcr}
    f_1'-\alpha f_1&=&\phantom{-}\beta f_2\phantom{.}\\
    f_2'-\alpha f_2&=&-\beta f_1.
    \end{array}
    $$
    If we let $f_1=g_1e^{\alpha t}$ and $f_2=g_2e^{\alpha t}$, where
    $g_1$ and $g_2$ are to be determined, then the last two equations
    become
    $$
    \begin{array}{rcr}
    g_1'&=&\beta g_2\phantom{.}\\
    g_2'&=&-\beta g_1,
    \end{array}
    $$
    which implies that
    $$
    g_1''=\beta g_2'=-\beta^2 g_1,
    $$
    so
    $$
    g_1''+\beta^2 g_1=0.
    $$
    The general solution of this equation is
    $$
    g_1=c_1\cos\beta t+c_2\sin\beta t.
    $$
    Moreover, since $g_2=g_1'/\beta$,
    $$
    g_2=-c_1\sin\beta t+c_2\cos\beta t.
    $$
    Multiplying  $g_1$  and $g_2$  by $e^{\alpha t}$ shows that
    \begin{eqnarray*}
    f_1&=&e^{\alpha t}(\phantom{-}c_1\cos\beta t+c_2\sin\beta t ),\\
    f_2&=&e^{\alpha t}(-c_1\sin\beta t+c_2\cos\beta t).
    \end{eqnarray*}
    Substituting these into \eqref{eq:10.6.2} shows that
    \begin{equation} \label{eq:10.6.5}
    \begin{array}{rcl}
    {\bf y}&=&e^{\alpha t}\left[(c_1\cos\beta t+c_2\sin\beta t){\bf u}
    +(-c_1\sin\beta t+c_2\cos\beta t){\bf v}\right]\\
    &=&c_1e^{\alpha t}({\bf u}\cos\beta t-{\bf v}\sin\beta t)
    +c_2e^{\alpha t}({\bf u}\sin\beta t+{\bf v}\cos\beta t)
    \end{array}
    \end{equation}
    is a solution of ${\bf y}'=A{\bf y}$ for any choice of the constants
    $c_1$ and $c_2$. In particular, by first taking $c_1=1$ and $c_2=0$
    and then taking $c_1=0$ and $c_2=1$, we see that ${\bf y}_1$ and ${\bf
    y}_2$ are solutions of $ {\bf y}'=A{\bf y}$. We leave it to you to
    verify that they are, respectively, the real and imaginary parts of
    \eqref{eq:10.6.3} (Exercise~\ref{exer:10.6.26}), and that they are linearly
    independent (Exercise~\ref{exer:10.6.27}).

    \begin{example}\label{example:10.6.1} \rm
     Find the general solution of
    \begin{equation} \label{eq:10.6.6}
    {\bf y}'=\twobytwo4{-5}5{-2}{\bf y}.
    \end{equation}
    \end{example}

    \solution The characteristic polynomial of the coefficient matrix $A$
    in \eqref{eq:10.6.6} is
    $$
    \left|\begin{array}{cc} 4-\lambda&-5\\ 5&-2-\lambda
    \end{array}\right|=(\lambda-1)^2+16.
    $$
    Hence, $\lambda=1+4i$ is an eigenvalue of $A$. The associated
    eigenvectors satisfy $\left(A-\left(1+4i\right)I\right){\bf x}={\bf
    0}$. The augmented matrix of this system is
    $$
    \left[\begin{array}{cccr} 3-4i&-5&\vdots&0\\
    5&-3-4i&\vdots&0  \end{array}\right],
     $$
    which is row equivalent to
    $$
    \left[\begin{array}{cccr} 1&-{3+4i\over5}&\vdots&0\\ 0&0&\vdots&0
    \end{array}\right].
    $$
    Therefore $x_1=(3+4i)x_2/5$. Taking $x_2=5$ yields $x_1=3+4i$, so
    $$
    {\bf x}=\left[\begin{array}{c}3+4i\\5\end{array}\right]
    $$
    is an eigenvector.  The real and imaginary parts of
    $$
    e^t(\cos4t+i\sin4t)\left[\begin{array}{c}3+4i\\5\end{array}\right]
    $$
     are
    $$
    {\bf y}_1=e^t\left[\begin{array}{c}3\cos4t-4\sin
    4t\\5\cos4t\end{array}\right]\quad\mbox{ and }\quad
    {\bf y}_2=e^t\left[\begin{array}{c}3\sin4t+4\cos4t\\5\sin
    4t\end{array}\right],
    $$
    which are linearly independent solutions of  \eqref{eq:10.6.6}.
    The general solution of \eqref{eq:10.6.6} is
    $$
    {\bf y}=
    c_1e^t\left[\begin{array}{c}3\cos4t-4\sin
    4t\\5\cos4t\end{array}\right]+
    c_2e^t\left[\begin{array}{c}3\sin4t+4\cos4t\\5\sin
    4t\end{array}\right].
    $$

    \begin{example}\label{example:10.6.2} \rm
     Find  the general solution of
    \begin{equation} \label{eq:10.6.7}
    {\bf y}'=\twobytwo{-14}{39}{-6}{16}{\bf y}.
    \end{equation}
    \end{example}

    \solution The characteristic polynomial of the coefficient matrix $A$
    in \eqref{eq:10.6.7} is
    $$
    \left|\begin{array}{cc}-14-\lambda&39\\-6&16-\lambda
    \end{array}\right|=(\lambda-1)^2+9.
    $$
    Hence, $\lambda=1+3i$ is an eigenvalue of $A$. The associated
    eigenvectors satisfy $\left(A-(1+3i)I\right){\bf x}={\bf 0}$. The
    augmented augmented matrix of this system is
    $$
    \left[\begin{array}{cccr}-15-3i&39&\vdots&0\\
    -6&15-3i&\vdots&0  \end{array}\right],
     $$
    which is row equivalent to
    $$
    \left[\begin{array}{cccr} 1&{-5+i\over2}&\vdots&0\\ 0&0&\vdots&0
    \end{array}\right].
    $$
    Therefore $x_1=(5-i)/2$. Taking $x_2=2$ yields $x_1=5-i$, so
    $$
    {\bf x}=\left[\begin{array}{c}5-i\\2\end{array}\right]
    $$
    is an eigenvector. The real and imaginary parts of
    $$
    e^t(\cos3t+i\sin3t)\left[\begin{array}{c}5-i\\2\end{array}\right]
    $$
     are
    $$
    {\bf y}_1=e^t\left[\begin{array}{c}\sin3t+5\cos3t\\2\cos
    3t\end{array}\right]\quad\mbox{ and }\quad
    {\bf y}_2=e^t\left[\begin{array}{c}-\cos3t+5\sin3t\\2\sin
    3t\end{array}\right],
     $$
    which are  linearly independent solutions of  \eqref{eq:10.6.7}.
    The general solution of \eqref{eq:10.6.7} is
    $$
    {\bf y}=c_1e^t\left[\begin{array}{c}\sin3t+5\cos3t\\2\cos
    3t\end{array}\right]+
    c_2e^t\left[\begin{array}{c}-\cos3t+5\sin3t\\2\sin
    3t\end{array}\right].
    $$

    \begin{example}\label{example:10.6.3}  \rm
     Find the general solution of
    \begin{equation} \label{eq:10.6.8}
    {\bf y}'=\threebythree{-5}54{-8}76100{\bf y}.
    \end{equation}
    \end{example}

    \solution  The characteristic
    polynomial of the  coefficient matrix $A$ in  \eqref{eq:10.6.8} is
    $$
    \left|\begin{array}{ccc}-5-\lambda&5&4\\-8&7-\lambda&
    6\\ \phantom{-}1
    &0&-\lambda\end{array}\right|=-(\lambda-2)(\lambda^2+1).
    $$
    Hence, the eigenvalues of $A$ are $\lambda_1=2$, $\lambda_2=i$, and
    $\lambda_3=-i$.
    The augmented matrix of $(A-2I){\bf x=0}$ is
    $$
    \left[\begin{array}{rrrcr}-7&5&4&\vdots&0\\-8&
    5&6&\vdots&0\\ 1&0&-2&\vdots&0
    \end{array}\right],
    $$
    which is row equivalent to
    $$
    \left[\begin{array}{rrrcr} 1&0&-2&\vdots&0\\ 0&1&-2&
    \vdots&0\\ 0&0&0&\vdots&0\end{array}\right].
    $$
    Therefore $x_1=x_2=2x_3$.  Taking $x_3=1$ yields
    $$
    {\bf x}_1=\threecol221,
    $$
    so
    $$
    {\bf y}_1=\threecol221e^{2t}
    $$
    is a solution of  \eqref{eq:10.6.8}.

    The augmented  matrix of $(A-iI){\bf x=0}$ is
    $$
    \left[\begin{array}{ccrccc}-5-i&5&4&\vdots&0\\-8&
    7-i&6&\vdots&0\\ \phantom{-}1&0&-i&\vdots&0
    \end{array}\right],
    $$
    which is row equivalent to
    $$
    \left[\begin{array}{ccccc} 1&0&-i&\vdots&0\\ 0&1&1-i&
    \vdots&0\\ 0&0&0&\vdots&0\end{array}\right].
    $$
    Therefore $x_1=ix_3$ and $x_2=-(1-i)x_3$. Taking $x_3=1$ yields
    the eigenvector
    $$
    {\bf x}_2=\left[\begin{array}{c} i\\-1+i\\ 1\end{array}
    \right].
    $$
    The real and imaginary parts of
    $$
    (\cos t+i\sin t)\left[\begin{array}{c}i\\-1+i\\1\end{array}\right]
    $$
     are
    $$
    {\bf y}_2=
    \left[\begin{array}{c}-\sin t\\-\cos t-\sin t\\\cos t\end{array}\right]
    \quad\mbox{ and }\quad
    {\bf y}_3=\left[\begin{array}{c}\cos t\\\cos t-\sin t\\\sin
    t\end{array}\right],
     $$
    which are solutions of  \eqref{eq:10.6.8}.
    Since the  Wronskian of $\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}$
    at $t=0$  is
    $$
    \left|\begin{array}{rrr}
    2&0&1\\2&-1&1\\1&1&0\end{array}\right|=1,
    $$
     $\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}$ is a
    fundamental set of solutions of \eqref{eq:10.6.8}. The general solution of
    \eqref{eq:10.6.8} is
    $$
    {\bf y}=c_1 \threecol221e^{2t}
    +c_2\left[\begin{array}{c}-\sin t\\-\cos t-\sin t\\\cos t\end{array}\right]
    +c_3\left[\begin{array}{c}\cos t\\\cos t-\sin t\\\sin
    t\end{array}\right].
    $$

    \begin{example}\label{example:10.6.4}  \rm
     Find the general solution of
    \begin{equation} \label{eq:10.6.9}
    {\bf y}'=\threebythree1{-1}{-2}1321{-1}2{\bf y}.
    \end{equation}
    \end{example}

    \solution The characteristic
    polynomial of the  coefficient matrix $A$ in  \eqref{eq:10.6.9} is
    $$
    \left|\begin{array}{ccc} 1-\lambda&-1&-2\\ 1&3-\lambda&
    \phantom{-}2\\ 1
    &-1&2-\lambda\end{array}\right|=
    -(\lambda-2)\left((\lambda-2)^2+4\right).
     $$
    Hence, the eigenvalues of $A$ are $\lambda_1=2$, $\lambda_2=2+2i$,
    and $\lambda_3=2-2i$.
    The augmented matrix of $(A-2I){\bf x=0}$ is
    $$
    \left[\begin{array}{rrrcr}-1&-1&-2&\vdots&0\\1&
    1&2&\vdots&0\\ 1&-1&0&\vdots&0
    \end{array}\right],
    $$
    which is row equivalent to
    $$
    \left[\begin{array}{rrrcr} 1&0&1&\vdots&0\\ 0&1&1&
    \vdots&0\\ 0&0&0&\vdots&0\end{array}\right].
    $$
    Therefore $x_1=x_2=-x_3$.  Taking $x_3=1$ yields
    $$
    {\bf x}_1=\threecol{-1}{-1}1,
    $$
    so
    $$
    {\bf y}_1=\threecol{-1}{-1}1e^{2t}
    $$
    is a solution of  \eqref{eq:10.6.9}.

    The augmented  matrix of $\left(A-(2+2i)I\right){\bf x=0}$ is
    $$
    \left[\begin{array}{ccrcc}-1-2i&-1&-2&\vdots&0\\ 1&
    1-2i&\phantom{-}2&\vdots&0\\ 1&-1&-2i&\vdots&0
    \end{array}\right],
    $$
    which is row equivalent to
    $$
    \left[\begin{array}{rrrcr} 1&0&-i&\vdots&0\\ 0&1&i&
    \vdots&0\\ 0&0&0&\vdots&0\end{array}\right].
    $$
    Therefore $x_1=ix_3$ and $x_2=-ix_3$. Taking $x_3=1$ yields
    the eigenvector
    $$
    {\bf x}_2=\threecol i{-i}1
    $$
    The real and imaginary parts of
    $$
    e^{2t}(\cos2t+i\sin2t)\threecol i{-i}1
    $$
    are
    $$
    {\bf y}_2=e^{2t}\left[\begin{array}{r}-\sin2t\\\sin2t\\\cos
    2t\end{array}\right]\quad\mbox{ and }\quad
    {\bf y}_2=e^{2t}\left[\begin{array}{r}\cos2t\\-\cos2t\\\sin
    2t\end{array}\right],
    $$
    which are solutions of  \eqref{eq:10.6.9}.
    Since  the Wronskian of $\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}$
    at $t=0$ is
    $$
    \left|\begin{array}{rrr}
    -1&0&1\\-1&0&-1\\1&1&0\end{array}\right|=-2,
    $$
    $\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}$ is a fundamental set of solutions
    of \eqref{eq:10.6.9}.  The general solution of \eqref{eq:10.6.9} is
    $$
    {\bf y}=c_1\threecol{-1}{-1}1e^{2t}+
    c_2e^{2t}\left[\begin{array}{r}-\sin2t\\\sin2t\\\cos
    2t\end{array}\right]+
    c_3e^{2t}\left[\begin{array}{r}\cos2t\\-\cos2t\\\sin
    2t\end{array}\right].
    $$

    \boxit{Geometric Properties of Solutions when  $n=2$}


    \noindent
    We'll now  consider the geometric properties of solutions of a
    $2\times2$ constant coefficient system
    \begin{equation} \label{eq:10.6.10}
    \twocol{y_1'}{y_2'}=\left[\begin{array}{cc}a_{11}&a_{12}\\a_{21}&a_{22}
    \end{array}\right]\twocol{y_1}{y_2}
    \end{equation}
    under the assumptions of this section; that is, when the matrix
    $$
    A=\left[\begin{array}{cc}a_{11}&a_{12}\\a_{21}&a_{22}
    \end{array}\right]
    $$
    has a complex eigenvalue $\lambda=\alpha+i\beta$
    {\rm(}$\beta\ne0${\rm}) and ${\bf x}={\bf u}+i{\bf v}$ is an
    associated eigenvector, where ${\bf u}$ and ${\bf v}$ have real
    components. To describe the trajectories accurately it's necessary to
    introduce a new rectangular coordinate system in the $y_1$-$y_2$
    plane. This raises a point that hasn't come up before: It is always
    possible to choose ${\bf x}$ so that $({\bf u},{\bf v})=0$. A special
    effort is required to do this, since not every eigenvector has this
    property. However, if we know an eigenvector that doesn't, we can
    multiply it by a suitable complex constant to obtain one that does. To
    see this, note that if ${\bf x}$ is a $\lambda$-eigenvector of $A$ and
    $k$ is an arbitrary real number, then
    $$
    {\bf x}_1=(1+ik){\bf x}=(1+ik)({\bf u}+i{\bf v})
    =({\bf u}-k{\bf v})+i({\bf v}+k{\bf u})
    $$
    is also a $\lambda$-eigenvector of $A$, since
    $$
    A{\bf x}_1= A((1+ik){\bf x})=(1+ik)A{\bf x}=(1+ik)\lambda{\bf x}=
    \lambda((1+ik){\bf x})=\lambda{\bf x}_1.
    $$
    The real and imaginary parts of ${\bf x}_1$ are
    \begin{equation} \label{eq:10.6.11}
    {\bf u}_1={\bf u}-k{\bf v} \mbox{\quad and \quad}
    {\bf v}_1={\bf v}+k{\bf u},
    \end{equation}
    so
    $$
    ({\bf u}_1,{\bf v}_1)=({\bf u}-k{\bf v},{\bf v}+k{\bf u})
    =-\left[({\bf u},{\bf v})k^2+(\|{\bf v}\|^2-\|{\bf u}\|^2)k
    -({\bf u},{\bf v})\right].
    $$
    Therefore $({\bf u}_1,{\bf v}_1)=0$ if
    \begin{equation} \label{eq:10.6.12}
    ({\bf u},{\bf v})k^2+(\|{\bf v}\|^2-\|{\bf u}\|^2)k-({\bf u},{\bf
    v})=0.
    \end{equation}
    If $({\bf u},{\bf v})\ne0$ we can use the quadratic formula to find
    two real values of $k$ such that $({\bf u}_1,{\bf v}_1)=0$
    (Exercise~\ref{exer:10.6.28}).

    \begin{example}\label{example:10.6.5} \rm  In Example~\ref{example:10.6.1}
    we found the eigenvector
    $$
    {\bf x}=\ctwocol{3+4i}5=\twocol35+i\twocol40
    $$
    for the matrix of the system \eqref{eq:10.6.6}. Here ${\bf u}=\dst{\twocol35}$
    and ${\bf v}=\twocol40$ are not orthogonal, since $({\bf u},{\bf
    v})=12$. Since $\|{\bf v}\|^2-\|{\bf u}\|^2=-18$, \eqref{eq:10.6.12} is
    equivalent to
    $$
    2k^2-3k-2=0.
    $$
    The zeros of this equation are $k_1=2$ and $k_2=-1/2$. Letting $k=2$
    in \eqref{eq:10.6.11} yields
    $$
    {\bf u}_1={\bf u}-2{\bf v}=\twocol{-5}{\phantom{-}5}\mbox{\quad and \quad}
    {\bf v}_1={\bf v}+2{\bf u}=\twocol{10}{10},
    $$
    and $({\bf u}_1,{\bf v}_1)=0$.
     Letting $k=-1/2$
    in \eqref{eq:10.6.11} yields
    $$
    {\bf u}_1={\bf u}+{{\bf v}\over2}=\twocol{5}5\mbox{\quad and \quad}
    {\bf v}_1={\bf v}-{{\bf u}\over2}={1\over2}\twocol{-5}{\phantom{-}5},
    $$
    and again $({\bf u}_1,{\bf v}_1)=0$. \bbox
    \end{example}

    (The numbers don't always work out as nicely as in this example.
    You'll need a calculator or computer to do
    Exercises~\ref{exer:10.6.29}-\ref{exer:10.6.40}.)

    Henceforth, we'll assume that $({\bf u},{\bf v})=0$. Let ${\bf U}$ and
    ${\bf V}$ be unit vectors in the directions of ${\bf u}$ and ${\bf
    v}$, respectively; that is, ${\bf U}={\bf u}/\|{\bf u}\|$ and ${\bf
    V}={\bf v}/\|{\bf v}\|$. The new rectangular coordinate system will
    have the same origin as the $y_1$-$y_2$ system. The coordinates of a
    point in this system will be denoted by $(z_1,z_2)$, where $z_1$ and
    $z_2$ are the displacements in the directions of ${\bf U}$ and ${\bf
    V}$, respectively.

    From \eqref{eq:10.6.5}, the solutions of \eqref{eq:10.6.10} are given by
    \begin{equation} \label{eq:10.6.13} {\bf y}=e^{\alpha t}\left[(c_1\cos\beta
    t+c_2\sin\beta t){\bf u} +(-c_1\sin\beta t+c_2\cos\beta t){\bf
    v}\right]. \end{equation} For convenience, let's call the curve
    traversed by $e^{-\alpha t}{\bf y}(t)$ a {\color{blue}\it shadow trajectory\/} of
    \eqref{eq:10.6.10}. Multiplying \eqref{eq:10.6.13} by $e^{-\alpha t}$ yields
    $$
    e^{-\alpha t}{\bf y}(t)=z_1(t){\bf U}+z_2(t){\bf V},
    $$
    where
    \begin{eqnarray*}
    z_1(t)&=&\|{\bf u}\|(c_1\cos\beta t+c_2\sin\beta
    t)\\ z_2(t)&=&\|{\bf v}\|(-c_1\sin\beta t+c_2\cos\beta t).
    \end{eqnarray*}
    Therefore
    $$ {(z_1(t))^2\over\|{\bf
    u}\|^2}+{(z_2(t))^2\over\|{\bf v}\|^2} =c_1^2+c_2^2
    $$
    (verify!), which means that the shadow trajectories of \eqref{eq:10.6.10}
    are ellipses centered at the origin, with axes of symmetry parallel to
    ${\bf U}$ and ${\bf V}$. Since
    $$
    z_1'={\beta\|{\bf u}\|\over\|{\bf v}\|} z_2\mbox{\quad and \quad}
    z_2'=-{\beta\|{\bf v}\|\over\|{\bf u}\|} z_1,
    $$
     the vector from the
    origin to a point on the shadow ellipse rotates in the same direction
    that ${\bf V}$ would have to be rotated by $\pi/2$ radians to bring it
    into coincidence with ${\bf U}$ (Figures~\ref{figure:10.6.1} and
    \ref{figure:10.6.2}).


    \enlargethispage{3in}
    \begin{figure}[H]
    \color{blue}
      \begin{minipage}[b]{0.5\linewidth}
        \centering
       \scalebox{.65}{
      \includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig100601}}
    \caption{ Shadow trajectories traversed clockwise}
      \label{figure:10.6.1}
    \end{minipage}
      \begin{minipage}[b]{0.5\linewidth}
        \centering
       \scalebox{.65}{
      \includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig100602}}
    \caption{ Shadow trajectories traversed counterclockwise}
      \label{figure:10.6.2}
    \end{minipage}
    \end{figure}

    If $\alpha=0$, then any trajectory of \eqref{eq:10.6.10} is a shadow
    trajectory of \eqref{eq:10.6.10};  therefore, if $\lambda$ is purely
    imaginary, then the trajectories of \eqref{eq:10.6.10} are ellipses
    traversed
    periodically as indicated in Figures~\ref{figure:10.6.1} and
    \ref{figure:10.6.2}.


    If $\alpha>0$, then
    $$
    \lim_{t\to\infty}\|{\bf y}(t)\|=\infty\mbox{\quad and \quad}
    \lim_{t\to-\infty}{\bf y}(t)=0,
    $$
    so the trajectory spirals away from the origin as $t$ varies from
    $-\infty$ to $\infty$. The direction of the spiral depends upon the
    relative orientation of ${\bf U}$ and ${\bf V}$, as shown in
    Figures~\ref{figure:10.6.3} and \ref{figure:10.6.4}.


    If $\alpha<0$,  then
    $$
    \lim_{t\to-\infty}\|{\bf y}(t)\|=\infty\mbox{\quad and \quad}
    \lim_{t\to\infty}{\bf y}(t)=0,
    $$
    \enlargethispage{2in}
    so the trajectory spirals toward the origin as $t$ varies from
    $-\infty$ to $\infty$. Again, the direction of the spiral depends upon
    the relative orientation of ${\bf U}$ and ${\bf V}$, as shown in
    Figures~\ref{figure:10.6.5} and \ref{figure:10.6.6}.
    \newpage


    \begin{figure}[H]
    \color{blue}
      \begin{minipage}[b]{0.5\linewidth}
        \centering
       \scalebox{.65}{
      \includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig010603}}
    \caption{ $\alpha>0$;   shadow trajectory spiraling outward}
      \label{figure:10.6.3}
    \end{minipage}
      \begin{minipage}[b]{0.5\linewidth}
        \centering
       \scalebox{.65}{
      \includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig010604}}
    \caption{ $\alpha>0$;   shadow trajectory spiraling outward}
      \label{figure:10.6.4}
    \end{minipage}
    \end{figure}

    \begin{figure}[H]
    \color{blue}
      \begin{minipage}[b]{0.5\linewidth}
        \centering
       \scalebox{.65}{
      \includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig010605}}
    \caption{ $\alpha<0$;   shadow trajectory spiraling inward}
      \label{figure:10.6.5}
    \end{minipage}
      \begin{minipage}[b]{0.5\linewidth}
        \centering
       \scalebox{.65}{
      \includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig100606}}
    \caption{ $\alpha<0$;   shadow trajectory spiraling inward}
      \label{figure:10.6.6}
    \end{minipage}
    \end{figure}

    \exercises
    In Exercises~\ref{exer:10.6.1}--\ref{exer:10.6.16}  find the general
    solution.

    \begin{exerciselist}
    \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
    \item\label{exer:10.6.1}  $\dst{{\bf y}'=\twobytwo{-1}2{-5}5{\bf y}
    }$&
    \item\label{exer:10.6.2}  $\dst{{\bf y}'=\twobytwo{-11}4{-26}9{\bf y}
    }$
    \end{tabular}

    \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
    \item\label{exer:10.6.3}  $\dst

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    $
    \end{tabular}

    \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
    \item\label{exer:10.6.5}  $\dst

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    $
    \end{tabular}

    \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
    \item\label{exer:10.6.7} $\dst

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    $
    \end{tabular}

    \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
    \item\label{exer:10.6.9}  $\dst

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    $
    \end{tabular}

    \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
    \item\label{exer:10.6.11}  $\dst

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    $
    \end{tabular}

    \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
    \item\label{exer:10.6.13}  $\dst

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    $
    \end{tabular}

    \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
    \item\label{exer:10.6.15}  $\dst{{\bf
    y}'=\threebythree60{-3}{-3}331{-2}6{\bf y}'}$&
    \item\label{exer:10.6.16}  $\dst{{\bf
    y}'=\threebythree12{-2}02{-1}100{\bf y}'}$
    \end{tabular}

    \exercisetext{In Exercises~\ref{exer:10.6.17}--\ref{exer:10.6.24}
    solve the initial value problem.}

    \item\label{exer:10.6.17}
    $\dst{{\bf y}'=\twobytwo4{-6}3{-2}{\bf y},\quad {\bf
    y}(0)=\twocol52}$

    \item\label{exer:10.6.18}
    $\dst{{\bf y}'=\twobytwo7{15}{-3}1{\bf y},\quad {\bf
    y}(0)=\twocol51}$

    \item\label{exer:10.6.19}
    $\dst{{\bf y}'=\twobytwo7{-15}3{-5}{\bf y},\quad {\bf
    y}(0)=\twocol{17}7}$

    \item\label{exer:10.6.20}
    $\dst{{\bf y}'={1\over6}\twobytwo4{-2}52{\bf y},\quad {\bf
    y}(0)=\twocol1{-1}}$

    \item\label{exer:10.6.21}
    $\dst{{\bf y}'=\threebythree52{-1}{-3}22132{\bf y},\quad {\bf
    y}(0)=\threecol406}$

    \item\label{exer:10.6.22}
    $\dst{{\bf y}'=\threebythree4408{10}{-20}23{-2}{\bf y},\quad {\bf
    y}(0)=\threecol865}$

    \item\label{exer:10.6.23}
    $\dst{{\bf
    y}'=\threebythree1{15}{-15}{-6}{18}{-22}{-3}{11}{-15}{\bf
    y},\quad {\bf y}(0)=\threecol{15}{17}{10}}$

    \item\label{exer:10.6.24}
    $\dst{{\bf y}'=\threebythree4{-4}4{-10}3{15}2{-3}1{\bf y},\quad
    {\bf y}(0)=\threecol{16}{14}6}$

    \item\label{exer:10.6.25}
    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.

    \item\label{exer:10.6.26}
    Verify that
    $$
    {\bf y}_1=e^{\alpha t}({\bf u}\cos\beta t-{\bf v}\sin\beta t)
    \mbox{\quad and\quad}
    {\bf y}_2=e^{\alpha t}({\bf u}\sin\beta t+{\bf v}\cos\beta t),
    $$
    are the real and imaginary parts of
    $$
    e^{\alpha t}(\cos\beta t+i\sin\beta t)({\bf u}+i{\bf v}).
    $$

    \item\label{exer:10.6.27}
     Show that if the vectors ${\bf u}$ and ${\bf v}$ are not both ${\bf
    0}$ and $\beta\ne0$ then the vector functions
    $$
    {\bf y}_1=e^{\alpha t}({\bf u}\cos\beta t-{\bf v}\sin\beta t)\quad
    \mbox{ and }\quad
    {\bf y}_2=e^{\alpha t}({\bf u}\sin\beta t+{\bf v}\cos\beta t)
    $$
    are linearly independent on every interval. \hint{There are
    two
    cases to consider: {\rm(i)} $\{{\bf u},{\bf v}\}$ linearly
    independent, and {\rm(ii)}
     $\{{\bf u},{\bf v}\}$ linearly dependent. In either case, exploit the
    the linear independence of $\{\cos\beta t,\sin\beta t\}$
    on every interval.}

    \item\label{exer:10.6.28}
    Suppose ${\bf u}=\dst{\twocol{u_1}{u_2}}$ and ${\bf
    v}=\dst{\twocol{v_1}{v_2}}$ are not orthogonal; that is, $({\bf
    u},{\bf v})\ne0$.
    \begin{alist}
    \item % (a)
    Show that  the quadratic equation
    $$
    ({\bf u},{\bf v})k^2+(\|{\bf v}\|^2-\|{\bf u}\|^2)k-({\bf u},{\bf
    v})=0
    $$
    has a positive root $k_1$ and a negative root $k_2=-1/k_1$.
    \item % (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
    $$
    {\bf u}_1^{(2)}={{\bf v}_1^{(1)}\over k_1}
    \mbox{\quad and \quad}
    {\bf v}_1^{(2)}=-{{\bf u}_1^{(1)}\over k_1}.
    $$
    \item % (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
    $-\pi/2$.
    \end{alist}

    \exercisetext{In Exercises~\ref{exer:10.6.29}-\ref{exer:10.6.32} find vectors
    ${\bf U}$ and ${\bf V}$ parallel to the axes of symmetry of the
    trajectories, and plot some typical trajectories.}

    \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
    \item\label{exer:10.6.29} \CGex\, $\dst

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    $
    \end{tabular}

    \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
    \item\label{exer:10.6.31} \CGex\,  $\dst

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    $
    \end{tabular}

    \exercisetext{In Exercises~\ref{exer:10.6.33}-\ref{exer:10.6.40} find vectors
    ${\bf U}$ and ${\bf V}$ parallel to the axes of symmetry of the shadow
    trajectories, and plot a typical trajectory.}

    \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
    \item\label{exer:10.6.33} \CGex\,  $\dst

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    $
    \end{tabular}

    \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
    \item\label{exer:10.6.35} \CGex\,  $\dst

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    $
    \end{tabular}

    \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
    \item\label{exer:10.6.37} \CGex\,  $\dst

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    $
    \end{tabular}

    \enlargethispage{1in}
    \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
    \item\label{exer:10.6.39} \CGex\,  $\dst

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    $
    \end{tabular}

    \end{exerciselist}