
# 4.5: Constant Coefficient Homogeneous Systems II

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We saw in Section~10.4 that if an $n\times n$
constant matrix
$A$ has $n$ real eigenvalues $\lambda_1$, $\lambda_2$, \dots, $\lambda_n$
(which need not be distinct) with associated linearly independent
eigenvectors ${\bf x}_1$, ${\bf x}_2$, \dots, ${\bf x}_n$, then the general
solution of ${\bf y}'=A{\bf y}$ is
$${\bf y}=c_1{\bf x}_1e^{\lambda_1t}+c_2{\bf x}_2e^{\lambda_2 t} +\cdots+c_n{\bf x}_ne^{\lambda_n t}.$$
In this section we consider the case where $A$ has $n$ real
eigenvalues, but does not have $n$ linearly independent eigenvectors.
It is shown in linear algebra that this occurs if and only if $A$ has
at least one eigenvalue of multiplicity $r>1$ such that the associated
eigenspace has dimension less than $r$. In this case $A$ is said to be
{\color{blue}\it defective\/}. Since it's beyond the scope of this book to give a
complete analysis of systems with defective coefficient matrices, we
will restrict our attention to some commonly occurring special cases.

\begin{example}\label{example:10.5.1}\rm
\space  Show that the system
\label{eq:10.5.1}
{\bf y}'=\twobytwo{11}{-25}4{-9}{\bf y}

does not have a fundamental set of solutions of the form $\{{\bf x}_1e^{\lambda_1t},{\bf x}_2e^{\lambda_2t}\}$, where $\lambda_1$ and
$\lambda_2$ are eigenvalues of the coefficient matrix $A$ of
\eqref{eq:10.5.1} and ${\bf x}_1$, and ${\bf x}_2$ are associated
linearly independent eigenvectors.
\end{example}

\solution   The characteristic polynomial of $A$ is
\begin{eqnarray*}
\twochar{11}{-25}4{-9}
&=&(\lambda-11)(\lambda+9)+100\\
&=&\lambda^2-2\lambda+1=(\lambda-1)^2.
\end{eqnarray*}
Hence, $\lambda=1$ is the only eigenvalue of $A$. The augmented
matrix of the system $(A-I){\bf x}={\bf 0}$ is
$$\left[\begin{array}{rrcr}10&-25&\vdots&0\\4& -10&\vdots&0\end{array}\right],$$
which is row equivalent to
$$\left[\begin{array}{rrcr}1&-\dst{5\over2}&\vdots&0\\[10pt]0& 0&\vdots&0\end{array}\right].$$
Hence, $x_1=5x_2/2$ where $x_2$ is arbitrary. Therefore all
eigenvectors of $A$ are scalar multiples of ${\bf x}_1=\dst{\twocol52}$,
so $A$ does not have a set of two linearly independent eigenvectors.
\bbox

From Example~\ref{example:10.5.1}, we know that all scalar multiples of
${\bf y}_1=\dst{\twocol52}e^t$ are solutions of \eqref{eq:10.5.1};
however,
to find the general solution we must find a second solution ${\bf y}_2$ such that $\{{\bf y}_1,{\bf y}_2\}$ is linearly independent.
Based on your recollection of the procedure for solving a constant
coefficient scalar equation
$$ay''+by'+cy=0$$
in the case where the characteristic polynomial has a repeated root,
you might expect to obtain a second solution of \eqref{eq:10.5.1} by
multiplying the first solution by $t$. However, this yields ${\bf y}_2=\dst{\twocol52}te^t$, which doesn't work, since
$${\bf y}_2'=\twocol52(te^t+e^t),\mbox{\quad while \quad } \twobytwo{11}{-25}4{-9}{\bf y}_2=\twocol52te^t.$$

The next theorem shows what to do in this situation.

\begin{theorem}\color{blue}\label{thmtype:10.5.1}
Suppose the $n\times n$ matrix $A$ has an eigenvalue $\lambda_1$
of multiplicity $\ge2$ and the associated eigenspace has dimension
$1;$ that is$,$ all $\lambda_1$-eigenvectors of $A$ are scalar
multiples
of  an eigenvector ${\bf x}.$  Then there are infinitely many vectors
${\bf u}$ such that
\label{eq:10.5.2}
(A-\lambda_1I){\bf u}={\bf x}.

Moreover$,$ if ${\bf u}$ is any such vector  then
\label{eq:10.5.3}
{\bf y}_2={\bf u}e^{\lambda_1t}+{\bf x}te^{\lambda_1t}

are linearly independent  solutions of ${\bf y}'=A{\bf y}.$
\end{theorem}

A complete proof of this theorem is beyond the scope of this book. The
difficulty is in proving that there's a vector ${\bf u}$ satisfying
\eqref{eq:10.5.2}, since $\det(A-\lambda_1I)=0$. We'll take this without
proof and verify the other assertions of the theorem.

We already know that ${\bf y}_1$ in \eqref{eq:10.5.3} is a solution of
${\bf y}'=A{\bf y}$. To see that ${\bf y}_2$ is also a solution, we
compute
\begin{eqnarray*}
{\bf y}_2'-A{\bf y}_2&=&\lambda_1{\bf u}e^{\lambda_1t}+{\bf
x} e^{\lambda_1t}
+\lambda_1{\bf x} te^{\lambda_1t}-A{\bf
u}e^{\lambda_1t}-A{\bf x} te^{\lambda_1t}\\
&=&(\lambda_1{\bf u}+{\bf x} -A{\bf
u})e^{\lambda_1t}+(\lambda_1{\bf x} -A{\bf x} )te^{\lambda_1t}.
\end{eqnarray*}
Since $A{\bf x}=\lambda_1{\bf x}$, this can be written as
$${\bf y}_2'-A{\bf y}_2=- \left((A-\lambda_1I){\bf u}-{\bf x}\right)e^{\lambda_1t},$$
and now \eqref{eq:10.5.2} implies that
${\bf y}_2'=A{\bf y}_2$.

To see that ${\bf y}_1$ and ${\bf y}_2$  are linearly independent,
suppose   $c_1$ and $c_2$ are constants such that
\label{eq:10.5.4}
c_1{\bf y}_1+c_2{\bf y}_2=c_1{\bf x}e^{\lambda_1t}+c_2({\bf
u}e^{\lambda_1t} +{\bf x}te^{\lambda_1t})={\bf 0}.

We must show that $c_1=c_2=0$.  Multiplying \eqref{eq:10.5.4}
by $e^{-\lambda_1t}$ shows that
\label{eq:10.5.5}
c_1{\bf x}+c_2({\bf u} +{\bf x}t)={\bf 0}.

By differentiating this with respect to $t$, we see that $c_2{\bf x}={\bf 0}$, which implies $c_2=0$, because ${\bf x}\ne{\bf 0}$.
Substituting  $c_2=0$ into \eqref{eq:10.5.5}  yields $c_1{\bf x}={\bf 0}$,
which implies that $c_1=0$, again because ${\bf x}\ne{\bf 0}$

\begin{example}\label{example:10.5.2}\rm
Use Theorem~\ref{thmtype:10.5.1} to find the general solution of the system
\label{eq:10.5.6}
{\bf y}'=\twobytwo{11}{-25}4{-9}{\bf y}

considered in Example~\ref{example:10.5.1}.
\end{example}

\solution In Example~\ref{example:10.5.1} we saw that $\lambda_1=1$ is an
eigenvalue of multiplicity $2$ of the coefficient matrix $A$ in
\eqref{eq:10.5.6}, and that all of the eigenvectors of $A$ are multiples of
$${\bf x}=\twocol52.$$
Therefore
$${\bf y}_1=\twocol52e^t$$
is a solution of \eqref{eq:10.5.6}. From Theorem~\ref{thmtype:10.5.1}, a second
solution is given by ${\bf y}_2={\bf u}e^t+{\bf x}te^t$, where
$(A-I){\bf u}={\bf x}$. The augmented matrix of this system is
$$\left[\begin{array}{rrcr}10&-25&\vdots&5\\4&-10&\vdots&2\end{array}\right],$$
which is row equivalent to
$$\dst{\left[\begin{array}{rrcr}1&-{5\over2}&\vdots&1\over2\\ 0&0&\vdots&0\end{array}\right]}.$$
Therefore the components of ${\bf u}$ must satisfy
$$u_1-{5\over2}u_2={1\over2},$$
where  $u_2$ is arbitrary. We choose $u_2=0$, so that $u_1=1/2$ and
$${\bf u}=\twocol{1\over2}0.$$
Thus,
$${\bf y}_2=\twocol10{e^t\over2}+\twocol52te^t.$$
Since ${\bf y}_1$ and ${\bf y}_2$ are linearly independent by
Theorem~\ref{thmtype:10.5.1}, they form a fundamental set of solutions of
\eqref{eq:10.5.6}. Therefore the general solution of \eqref{eq:10.5.6} is
$${\bf y}=c_1\twocol52e^t+c_2\left(\twocol10{e^t\over2}+\twocol52te^t\right).\bbox$$

Note that choosing the arbitrary constant $u_2$ to be nonzero is
equivalent to adding a scalar multiple of ${\bf y}_1$ to the second
solution ${\bf y}_2$ (Exercise~\ref{exer:10.5.33}).

\begin{example}\label{example:10.5.3}\rm
Find the general solution of
\label{eq:10.5.7}
{\bf y}'=\threebythree34{-10}21{-2}22{-5} {\bf y}.

\end{example}

\solution  The characteristic polynomial of
the coefficient matrix $A$ in  \eqref{eq:10.5.7} is
$$\left|\begin{array}{ccc} 3-\lambda & 4 & -10\\ 2 & 1-\lambda & -2\\ 2 & 2 &-5-\lambda\end{array}\right| =- (\lambda-1)(\lambda+1)^2.$$
Hence, the eigenvalues are $\lambda_1=1$ with multiplicity~$1$ and
$\lambda_2=-1$ with  multiplicity~$2$.

Eigenvectors associated with $\lambda_1=1$ must satisfy $(A-I){\bf x}={\bf 0}$. The augmented matrix of this system is
$$\left[\begin{array}{rrrcr} 2 & 4 & -10 &\vdots & 0\\ 2& 0 & -2 &\vdots & 0\\ 2 & 2 & -6 & \vdots & 0\end{array}\right],$$
which is row equivalent to
$$\left[\begin{array}{rrrcr} 1 & 0 & -1 &\vdots& 0\\ 0 & 1 & -2 &\vdots& 0\\ 0 & 0 & 0 &\vdots&0\end{array}\right].$$
Hence, $x_1 =x_3$ and  $x_2 =2 x_3$, where $x_3$ is arbitrary.
Choosing $x_3=1$ yields the eigenvector
$${\bf x}_1=\threecol121.$$
Therefore
$${\bf y}_1 =\threecol121e^t$$
is a solution of  \eqref{eq:10.5.7}.

Eigenvectors associated with $\lambda_2 =-1$ satisfy $(A+I){\bf x}={\bf 0}$. The  augmented matrix of this system is
$$\left[\begin{array}{rrrcr} 4 & 4 & -10 &\vdots & 0\\ 2 & 2 & -2 & \vdots & 0\\2 & 2 & -4 &\vdots & 0\end{array}\right],$$
which is row equivalent to
$$\left[\begin{array}{rrrcr} 1 & 1 & 0 &\vdots& 0\\ 0 & 0 & 1 &\vdots& 0 \\ 0 & 0 & 0 &\vdots&0\end{array}\right].$$
Hence, $x_3=0$ and $x_1 =-x_2$, where $x_2$ is
arbitrary. Choosing $x_2=1$  yields the eigenvector
$${\bf x}_2=\threecol{-1}10,$$
so
$${\bf y}_2 =\threecol{-1}10e^{-t}$$
is a solution of  \eqref{eq:10.5.7}.

Since all the eigenvectors of $A$ associated with $\lambda_2=-1$ are
multiples of ${\bf x}_2$, we must now use Theorem~\ref{thmtype:10.5.1} to
find a third solution of \eqref{eq:10.5.7} in the form
\label{eq:10.5.8}
{\bf y}_3={\bf u}e^{-t}+\threecol{-1}10te^{-t},

where ${\bf u}$ is a solution of $(A+I){\bf u=x}_2$.
The  augmented matrix  of this system is
$$\left[\begin{array}{rrrcr} 4 & 4 & -10 &\vdots & -1\\ 2 & 2 & -2 & \vdots & 1\\ 2 & 2 & -4 &\vdots & 0\end{array}\right],$$
which is  row equivalent to
$$\left[\begin{array}{rrrcr} 1 & 1 & 0 &\vdots& 1\\ 0 & 0 & 1 &\vdots& {1\over2} \\ 0 & 0 & 0 &\vdots&0\end{array}\right].$$
Hence, $u_3=1/2$ and $u_1 =1-u_2$, where $u_2$  is
arbitrary. Choosing $u_2=0$ yields
$${\bf u} =\threecol10{1\over2},$$
and substituting this into  \eqref{eq:10.5.8}
yields the solution
$${\bf y}_3=\threecol201{e^{-t}\over2}+\threecol{-1}10te^{-t}$$
of  \eqref{eq:10.5.7}.

Since the Wronskian of $\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}$
at $t=0$ is
$$\left|\begin{array}{rrr} 1&-1&1\\2&1&0\\1&0&1\over2\end{array}\right|={1\over2},$$
$\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}$ is a fundamental set of solutions
of \eqref{eq:10.5.7}. Therefore the general solution of \eqref{eq:10.5.7}
is
$${\bf y}=c_1\threecol121e^t+c_2\threecol{-1}10e^{-t}+c_3\left (\threecol201{e^{-t}\over2}+\threecol{-1}10te^{-t}\right).$$

\begin{theorem}\color{blue}\label{thmtype:10.5.2}
Suppose the $n\times n$ matrix $A$ has an eigenvalue $\lambda_1$ of
multiplicity $\ge 3$ and the associated eigenspace is
one--dimensional$;$ that is$,$ all eigenvectors associated with
$\lambda_1$
are scalar multiples of the eigenvector ${\bf x}.$ Then there are
infinitely many vectors ${\bf u}$ such that
\label{eq:10.5.9}
(A-\lambda_1I){\bf u}={\bf x},

and, if ${\bf u}$ is any such vector$,$  there are infinitely many
vectors ${\bf v}$ such that
\label{eq:10.5.10}
(A-\lambda_1I){\bf v}={\bf u}.

If ${\bf u}$ satisfies {\rm\eqref{eq:10.5.9}}  and ${\bf v}$ satisfies
{\rm\eqref{eq:10.5.10}},  then
\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*}
are linearly independent solutions of  ${\bf y}'=A{\bf y}$.
\end{theorem}

Again, it's beyond the scope of this book to prove that there are
vectors ${\bf u}$ and ${\bf v}$ that satisfy \eqref{eq:10.5.9} and
\eqref{eq:10.5.10}. Theorem~\ref{thmtype:10.5.1} implies that ${\bf y}_1$ and
${\bf y}_2$ are solutions of ${\bf y}'=A{\bf y}$. We leave the rest of
the proof to you (Exercise~\ref{exer:10.5.34}).

\begin{example}\label{example:10.5.4}\rm
Use Theorem~\ref{thmtype:10.5.2} to find the general solution of
\label{eq:10.5.11}
{\bf y}'=\threebythree11113{-1}022{\bf y}.

\end{example}

\solution  The characteristic polynomial of
the coefficient matrix $A$ in  \eqref{eq:10.5.11} is
$$\left|\begin{array}{ccc} 1-\lambda & 1 & \phantom{-}1\\ 1 & 3-\lambda & -1\\ 0 & 2 & 2-\lambda\end{array}\right| =-(\lambda-2)^3.$$
Hence, $\lambda_1=2$ is an eigenvalue of multiplicity $3$. The
associated eigenvectors satisfy $(A-2I){\bf x=0}$. The augmented
matrix of this system is
$$\left[\begin{array}{rrrcr} -1 & 1 & 1 &\vdots & 0\\ 1& 1 & -1 &\vdots & 0\\ 0 & 2 & 0 & \vdots & 0\end{array}\right],$$
which is row equivalent to
$$\left[\begin{array}{rrrcr} 1 & 0 &- 1 &\vdots& 0\\ 0 & 1 & 0 &\vdots& 0 \\ 0 & 0 & 0 &\vdots&0\end{array}\right].$$
Hence, $x_1 =x_3$ and  $x_2 = 0$, so the eigenvectors are all scalar
multiples of
$${\bf x}_1=\threecol101.$$
Therefore
$${\bf y}_1=\threecol101e^{2t}$$
is a solution of  \eqref{eq:10.5.11}.

We  now find a second solution of  \eqref{eq:10.5.11}  in the form
$${\bf y}_2={\bf u}e^{2t}+\threecol101te^{2t},$$
where ${\bf u}$ satisfies $(A-2I){\bf u=x}_1$.
The  augmented matrix  of this system is
$$\left[\begin{array}{rrrcr} -1 & 1 & 1 &\vdots & 1\\ 1& 1 & -1 &\vdots & 0\\ 0 & 2 & 0 & \vdots & 1\end{array}\right],$$
which is row equivalent to
$$\left[\begin{array}{rrrcr} 1 & 0 &- 1 &\vdots& -{1\over2}\\ 0 & 1 & 0 &\vdots& {1\over2}\\ 0 & 0 & 0 &\vdots&0\end{array}\right].$$
Letting $u_3=0$ yields $u_1=-1/2$ and $u_2=1/2$; hence,
$${\bf u}={1\over2}\threecol{-1}10$$
and
$${\bf y}_2=\threecol{-1}10{e^{2t}\over2}+\threecol101te^{2t}$$
is a solution of  \eqref{eq:10.5.11}.

We  now find a third solution of  \eqref{eq:10.5.11}  in the form
$${\bf y}_3={\bf v}e^{2t}+\threecol{-1}10{te^{2t}\over2}+\threecol101{t^2e^{2t}\over2}$$
where ${\bf v}$ satisfies $(A-2I){\bf v}={\bf u}$.
The  augmented matrix  of this system is
$$\left[\begin{array}{rrrcr} -1 & 1 & 1 &\vdots &-{1\over2}\\ 1& 1 & -1 &\vdots & {1\over2}\\ 0 & 2 & 0 & \vdots & 0\end{array}\right],$$
which is row equivalent to
$$\left[\begin{array}{rrrcr} 1 & 0 &- 1 &\vdots& {1\over2}\\ 0 & 1 & 0 &\vdots& 0\\ 0 & 0 & 0 &\vdots&0\end{array}\right].$$
Letting $v_3=0$ yields $v_1=1/2$ and $v_2=0$; hence,
$${\bf v}={1\over2}\threecol100.$$
Therefore
$${\bf y}_3=\threecol100{e^{2t}\over2}+ \threecol{-1}10{te^{2t}\over2}+\threecol101{t^2e^{2t}\over2}$$
is a solution of  \eqref{eq:10.5.11}. Since ${\bf y}_1$, ${\bf y}_2$, and
${\bf y}_3$ are linearly independent by Theorem~\ref{thmtype:10.5.2}, they
form a fundamental set of solutions of \eqref{eq:10.5.11}. Therefore the
general solution of \eqref{eq:10.5.11} is
\begin{eqnarray*}
{\bf y} &=&\dst{c_1\threecol101e^{2t}+
c_2\left(\threecol{-1}10{e^{2t}\over2}+\threecol101te^{2t}\right)}\\[2\jot]
&&+c_3\dst{\left(\threecol100{e^{2t}\over2}+
\threecol{-1}10{te^{2t}\over2}+\threecol101{t^2e^{2t}\over2}\right)}.
\end{eqnarray*}

\enlargethispage{1in}

\begin{theorem}\color{blue}\label{thmtype:10.5.3}
Suppose the $n\times n$ matrix $A$ has an eigenvalue $\lambda_1$ of
multiplicity $\ge 3$ and the associated eigenspace is
two--dimensional; that is, all eigenvectors of $A$ associated with
$\lambda_1$ are linear combinations of two linearly independent
eigenvectors ${\bf x}_1$ and ${\bf x}_2$$. Then there are constants \alpha and \beta (not both zero) such that if \label{eq:10.5.12} {\bf x}_3=\alpha{\bf x}_1+\beta{\bf x}_2, then there are infinitely many vectors {\bf u} such that \label{eq:10.5.13} (A-\lambda_1I){\bf u}={\bf x}_3. If {\bf u} satisfies {\rm\eqref{eq:10.5.13}}, then \begin{eqnarray} {\bf y}_1&=&{\bf x}_1 e^{\lambda_1t},\nonumber\\ {\bf y}_2&=&{\bf x}_2e^{\lambda_1t},\mbox{and }\nonumber\\ {\bf y}_3&=&{\bf u}e^{\lambda_1t}+{\bf x}_3te^{\lambda_1t}\label{eq:10.5.14}, \end{eqnarray} are linearly independent solutions of {\bf y}'=A{\bf y}. \end{theorem} We omit the proof of this theorem. \newpage \begin{example}\label{example:10.5.5}\rm Use Theorem~\ref{thmtype:10.5.3} to find the general solution of \label{eq:10.5.15} {\bf y}'=\threebythree001{-1}11{-1}02{\bf y}. \end{example} \solution The characteristic polynomial of the coefficient matrix A in \eqref{eq:10.5.15} is$$ \left|\begin{array}{ccc} -\lambda & 0 & 1\\ -1 & 1-\lambda & 1\\ -1 & 0 & 2-\lambda\end{array}\right| =-(\lambda-1)^3. $$Hence, \lambda_1=1 is an eigenvalue of multiplicity 3. The associated eigenvectors satisfy (A-I){\bf x=0}. The augmented matrix of this system is$$ \left[\begin{array}{rrrcr} -1 & 0 & 1 &\vdots & 0\\ -1& 0 & 1 &\vdots & 0\\ -1 & 0 & 1 & \vdots & 0\end{array}\right], $$which is row equivalent to$$ \left[\begin{array}{rrrcr} 1 & 0 &- 1 &\vdots& 0\\ 0 & 0 & 0 &\vdots& 0 \\ 0 & 0 & 0 &\vdots&0\end{array}\right]. $$Hence, x_1 =x_3 and x_2 is arbitrary, so the eigenvectors are of the form$$ {\bf x}_1=\threecol{x_3}{x_2}{x_3}=x_3\threecol101+x_2\threecol010. $$Therefore the vectors \label{eq:10.5.16} {\bf x}_1 =\threecol101\quad\mbox{and }\quad {\bf x}_2=\threecol010 form a basis for the eigenspace, and$$ {\bf y}_1 =\threecol101e^t\quad\mbox{and }\quad {\bf y}_2=\threecol010e^t $$are linearly independent solutions of \eqref{eq:10.5.15}. To find a third linearly independent solution of \eqref{eq:10.5.15}, we must find constants \alpha and \beta (not both zero) such that the system \label{eq:10.5.17} (A-I){\bf u}=\alpha{\bf x}_1+\beta{\bf x}_2 has a solution {\bf u}. The augmented matrix of this system is$$ \left[\begin{array}{rrrcr} -1 & 0 & 1 &\vdots &\alpha\\ -1& 0 & 1 &\vdots &\beta\\ -1 & 0 & 1 & \vdots &\alpha\end{array}\right], $$which is row equivalent to \label{eq:10.5.18} \left[\begin{array}{rrrcr} 1 & 0 &- 1 &\vdots& -\alpha\\ 0 & 0 & 0 &\vdots&\beta-\alpha\\ 0 & 0 & 0 &\vdots&0\end{array} \right]. Therefore \eqref{eq:10.5.17} has a solution if and only if \beta=\alpha, where \alpha is arbitrary. If \alpha=\beta=1 then \eqref{eq:10.5.12} and \eqref{eq:10.5.16} yield$$ {\bf x}_3={\bf x}_1+{\bf x}_2= \threecol101+\threecol010=\threecol111, $$and the augmented matrix \eqref{eq:10.5.18} becomes$$ \left[\begin{array}{rrrcr} 1 & 0 &- 1 &\vdots& -1\\ 0 & 0 & 0 &\vdots& 0\\ 0 & 0 & 0 &\vdots&0\end{array} \right]. $$This implies that u_1=-1+u_3, while u_2 and u_3 are arbitrary. Choosing u_2=u_3=0 yields$$ {\bf u}=\threecol{-1}00. $$Therefore \eqref{eq:10.5.14} implies that$$ {\bf y}_3={\bf u}e^t+{\bf x}_3te^t=\threecol{-1}00e^t+\threecol111te^t $$is a solution of \eqref{eq:10.5.15}. Since {\bf y}_1, {\bf y}_2, and {\bf y}_3 are linearly independent by Theorem~\ref{thmtype:10.5.3}, they form a fundamental set of solutions for \eqref{eq:10.5.15}. Therefore the general solution of \eqref{eq:10.5.15} is$$ {\bf y}=c_1\threecol101e^t+c_2\threecol010e^t +c_3\left(\threecol{-1}00e^t+\threecol111te^t\right).\bbox $$\boxit{Geometric Properties of Solutions when n=2} \noindent We'll now consider the geometric properties of solutions of a 2\times2 constant coefficient system \label{eq:10.5.19} \twocol{y_1'}{y_2'}=\left[\begin{array}{cc}a_{11}&a_{12}\\a_{21}&a_{22} \end{array}\right]\twocol{y_1}{y_2} under the assumptions of this section; that is, when the matrix$$ A=\left[\begin{array}{cc}a_{11}&a_{12}\\a_{21}&a_{22} \end{array}\right] $$has a repeated eigenvalue \lambda_1 and the associated eigenspace is one-dimensional. In this case we know from Theorem~\ref{thmtype:10.5.1} that the general solution of \eqref{eq:10.5.19} is \label{eq:10.5.20} {\bf y}=c_1{\bf x}e^{\lambda_1t}+c_2({\bf u}e^{\lambda_1t}+{\bf x}te^{\lambda_1t}), where {\bf x} is an eigenvector of A and {\bf u} is any one of the infinitely many solutions of \label{eq:10.5.21} (A-\lambda_1I){\bf u}={\bf x}. We assume that \lambda_1\ne0. \begin{figure}[H] \centering \scalebox{.8}{ \includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig100501}} \caption{Positive and negative half-planes} \label{figure:10.5.1} \end{figure} Let L denote the line through the origin parallel to {\bf x}. By a {\color{blue}\it half-line\/} of L we mean either of the rays obtained by removing the origin from L. Eqn.~\eqref{eq:10.5.20} is a parametric equation of the half-line of L in the direction of {\bf x} if c_1>0, or of the half-line of L in the direction of -{\bf x} if c_1<0. The origin is the trajectory of the trivial solution {\bf y}\equiv{\bf 0}. Henceforth, we assume that c_2\ne0. In this case, the trajectory of \eqref{eq:10.5.20} can't intersect L, since every point of L is on a trajectory obtained by setting c_2=0. Therefore the trajectory of \eqref{eq:10.5.20} must lie entirely in one of the open half-planes bounded by L, but does not contain any point on L. Since the initial point (y_1(0),y_2(0)) defined by {\bf y}(0)=c_1{\bf x}_1+c_2{\bf u} is on the trajectory, we can determine which half-plane contains the trajectory from the sign of c_2, as shown in Figure~\pageref{figure:10.5.1}. For convenience we'll call the half-plane where c_2>0 the {\color{blue}\it positive half-plane\/}. Similarly, the-half plane where c_2<0 is the {\color{blue}\it negative half-plane\/}. You should convince yourself (Exercise~\ref{exer:10.5.35}) that even though there are infinitely many vectors {\bf u} that satisfy \eqref{eq:10.5.21}, they all define the same positive and negative half-planes. In the figures simply regard {\bf u} as an arrow pointing to the positive half-plane, since wen't attempted to give {\bf u} its proper length or direction in comparison with {\bf x}. For our purposes here, only the relative orientation of {\bf x} and {\bf u} is important; that is, whether the positive half-plane is to the right of an observer facing the direction of {\bf x} (as in Figures~\ref{figure:10.5.2} and \ref{figure:10.5.5}), or to the left of the observer (as in Figures~\ref{figure:10.5.3} and \ref{figure:10.5.4}). Multiplying \eqref{eq:10.5.20} by e^{-\lambda_1t} yields$$ e^{-\lambda_1t}{\bf y}(t)=c_1{\bf x}+c_2{\bf u}+c_2t {\bf x}. $$Since the last term on the right is dominant when |t| is large, this provides the following information on the direction of {\bf y}(t): \begin{alist} \item % (a) Along trajectories in the positive half-plane (c_2>0), the direction of {\bf y}(t) approaches the direction of {\bf x} as t\to\infty and the direction of -{\bf x} as t\to-\infty. \item % (b) Along trajectories in the negative half-plane (c_2<0), the direction of {\bf y}(t) approaches the direction of -{\bf x} as t\to\infty and the direction of {\bf x} as t\to-\infty. \end{alist} Since$$ \lim_{t\to\infty}\|{\bf y}(t)\|=\infty\mbox{\quad and \quad} \lim_{t\to-\infty}{\bf y}(t)={\bf 0}\mbox{\quad if \quad}\lambda_1>0, $$or$$ \lim_{t-\to\infty}\|{\bf y}(t)\|=\infty\mbox{\quad and \quad} \lim_{t\to\infty}{\bf y}(t)={\bf 0}\mbox{\quad if \quad}\lambda_1<0, $$there are four possible patterns for the trajectories of \eqref{eq:10.5.19}, depending upon the signs of c_2 and \lambda_1. Figures~\ref{figure:10.5.2}-\ref{figure:10.5.5} illustrate these patterns, and reveal the following principle: {\color{blue}\it If \lambda_1 and c_2 have the same sign then the direction of the traectory approaches the direction of -{\bf x} as \|{\bf y} \|\to0 and the direction of {\bf x} as \|{\bf y}\|\to\infty. If \lambda_1 and c_2 have opposite signs then the direction of the trajectory approaches the direction of {\bf x} as \|{\bf y} \|\to0 and the direction of -{\bf x} as \|{\bf y}\|\to\infty.} \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]{fig100502}} \caption{ Positive eigenvalue; motion away from the origin} \label{figure:10.5.2} \end{minipage} \begin{minipage}[b]{0.5\linewidth} \centering \scalebox{.65}{ \includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig100503}} \caption{ Positive eigenvalue; motion away from the origin} \label{figure:10.5.3} \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]{fig100504} } \caption{ Negative eigenvalue; motion toward the origin} \label{figure:10.5.4} \end{minipage} \begin{minipage}[b]{0.5\linewidth} \centering \scalebox{.65}{ \includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig100505} } \caption{ Negative eigenvalue; motion toward the origin} \label{figure:10.5.5} \end{minipage} \end{figure} \newpage \exercises In Exercises~\ref{exer:10.5.1}--\ref{exer:10.5.12} find the general solution. \begin{exerciselist} \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}} \item\label{exer:10.5.1} \dst ParseError: invalid DekiScript (click for details) Callstack: at (Courses/Mount_Royal_University/MATH_3200:_Mathematical_Methods/4:_Linear_Systems_of_Ordinary_Differential_Equations_(LSODE)/4.5:__Constant_Coefficient_Homogeneous_Systems_II), /content/body/p[44]/span, line 1, column 1  \end{tabular} \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}} \item\label{exer:10.5.3} \dst ParseError: invalid DekiScript (click for details) Callstack: at (Courses/Mount_Royal_University/MATH_3200:_Mathematical_Methods/4:_Linear_Systems_of_Ordinary_Differential_Equations_(LSODE)/4.5:__Constant_Coefficient_Homogeneous_Systems_II), /content/body/p[45]/span, line 1, column 1  \end{tabular} \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}} \item\label{exer:10.5.5} \dst ParseError: invalid DekiScript (click for details) Callstack: at (Courses/Mount_Royal_University/MATH_3200:_Mathematical_Methods/4:_Linear_Systems_of_Ordinary_Differential_Equations_(LSODE)/4.5:__Constant_Coefficient_Homogeneous_Systems_II), /content/body/p[46]/span, line 1, column 1  \end{tabular} \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}} \item\label{exer:10.5.7} \dst ParseError: invalid DekiScript (click for details) Callstack: at (Courses/Mount_Royal_University/MATH_3200:_Mathematical_Methods/4:_Linear_Systems_of_Ordinary_Differential_Equations_(LSODE)/4.5:__Constant_Coefficient_Homogeneous_Systems_II), /content/body/p[47]/span, line 1, column 1  \end{tabular} \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}} \item\label{exer:10.5.9} \dst ParseError: invalid DekiScript (click for details) Callstack: at (Courses/Mount_Royal_University/MATH_3200:_Mathematical_Methods/4:_Linear_Systems_of_Ordinary_Differential_Equations_(LSODE)/4.5:__Constant_Coefficient_Homogeneous_Systems_II), /content/body/p[48]/span, line 1, column 1  \end{tabular} \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}} \item\label{exer:10.5.11} \dst ParseError: invalid DekiScript (click for details) Callstack: at (Courses/Mount_Royal_University/MATH_3200:_Mathematical_Methods/4:_Linear_Systems_of_Ordinary_Differential_Equations_(LSODE)/4.5:__Constant_Coefficient_Homogeneous_Systems_II), /content/body/p[49]/span, line 1, column 1  \end{tabular} \exercisetext{In Exercises~\ref{exer:10.5.13}--\ref{exer:10.5.23} solve the initial value problem.} \item\label{exer:10.5.13} \dst{{\bf y}'=\twobytwo{-11}8{-2}{-3}{\bf y} ,\quad{\bf y}(0)=\twocol62} \item\label{exer:10.5.14} \dst{{\bf y}'=\twobytwo{15}{-9}{16}{-9}{\bf y} ,\quad{\bf y}(0)=\twocol58} \item\label{exer:10.5.15} \dst{{\bf y}'=\twobytwo{-3}{-4}1{-7}{\bf y},\quad{\bf y}(0)=\twocol23} \item\label{exer:10.5.16} \dst{{\bf y}'=\twobytwo{-7}{24}{-6}{17}{\bf y} ,\quad{\bf y}(0)=\twocol31} \item\label{exer:10.5.17} \dst{{\bf y}'=\twobytwo{-7}3{-3}{-1}{\bf y} ,\quad{\bf y}(0)=\twocol02} \item\label{exer:10.5.18} \dst{{\bf y}' =\threebythree{-1}101{-1}{-2}{-1}{-1}{-1}{\bf y},\quad {\bf y}(0)=\threecol65{-7}} \item\label{exer:10.5.19} \dst{{\bf y}' =\threebythree{-2}21{-2}21{-3}32{\bf y},\quad {\bf y}(0)=\threecol{-6}{-2}0} \item\label{exer:10.5.20} \dst{{\bf y}' =\threebythree{-7}{-4}4{-1}01{-9}{-5}6{\bf y},\quad\bf {\bf y}(0)=\threecol{-6}9{-1}} \item\label{exer:10.5.21} \dst{{\bf y}' =\threebythree{-1}{-4}{-1}361{-3}{-2}3\bf y,\quad\bf y(0)=\threecol{-2}13} \item\label{exer:10.5.22} \dst{{\bf y}' =\threebythree4{-8}{-4}{-3}{-1}{-3}1{-1}9{\bf y},\quad {\bf y}(0)=\threecol{-4}1{-3}} \item\label{exer:10.5.23} \dst{{\bf y}'= \threebythree{-5}{-1}{11}{-7}1{13}{-4}08{\bf y},\quad {\bf y}(0)=\threecol022} \exercisetext{The coefficient matrices in Exercises~\ref{exer:10.5.24}--\ref{exer:10.5.32} have eigenvalues of multiplicity 3. Find the general solution.} \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}} \item\label{exer:10.5.24} \dst ParseError: invalid DekiScript (click for details) Callstack: at (Courses/Mount_Royal_University/MATH_3200:_Mathematical_Methods/4:_Linear_Systems_of_Ordinary_Differential_Equations_(LSODE)/4.5:__Constant_Coefficient_Homogeneous_Systems_II), /content/body/p[63]/span, line 1, column 1  \end{tabular} \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}} \item\label{exer:10.5.26} \dst ParseError: invalid DekiScript (click for details) Callstack: at (Courses/Mount_Royal_University/MATH_3200:_Mathematical_Methods/4:_Linear_Systems_of_Ordinary_Differential_Equations_(LSODE)/4.5:__Constant_Coefficient_Homogeneous_Systems_II), /content/body/p[64]/span, line 1, column 1  \end{tabular} \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}} \item\label{exer:10.5.28} \dst ParseError: invalid DekiScript (click for details) Callstack: at (Courses/Mount_Royal_University/MATH_3200:_Mathematical_Methods/4:_Linear_Systems_of_Ordinary_Differential_Equations_(LSODE)/4.5:__Constant_Coefficient_Homogeneous_Systems_II), /content/body/p[65]/span, line 1, column 1  \end{tabular} \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}} \item\label{exer:10.5.30} \dst{{\bf y}' =\threebythree{-4}0{-1}{-1}{-3}{-1}10{-2}{\bf y}}& \item\label{exer:10.5.31} \dst{{\bf y}' =\threebythree{-3}{-3}445{-8}23{-5}\bf y} \end{tabular} \item\label{exer:10.5.32} {\bf y}'={\threebythree{-3}{-1}01{-1}0{-1}{-1}{-2}}{\bf y} \item\label{exer:10.5.33} Under the assumptions of Theorem~\ref{thmtype:10.5.1}, suppose {\bf u} and \hat{\bf u} are vectors such that$$ (A-\lambda_1I){\bf u}={\bf x}\quad\mbox{and }\quad (A-\lambda_1I)\hat{\bf u}={\bf x}, $$and let$$ {\bf y}_2={\bf u}e^{\lambda_1t}+{\bf x}te^{\lambda_1t} \quad\mbox{and }\quad \hat{\bf y}_2=\hat{\bf u}e^{\lambda_1t}+{\bf x}te^{\lambda_1t}. $$Show that {\bf y}_2-\hat{\bf y}_2 is a scalar multiple of {\bf y}_1={\bf x}e^{\lambda_1t}. \item\label{exer:10.5.34} Under the assumptions of Theorem~\ref{thmtype:10.5.2}, let \begin{eqnarray*} {\bf y}_1 &=&{\bf x} e^{\lambda_1t},\\ {\bf y}_2&=&{\bf u}e^{\lambda_1t}+{\bf x} te^{\lambda_1t},\mbox{ and }\\ {\bf y}_3&=&{\bf v}e^{\lambda_1t}+{\bf u}te^{\lambda_1t}+{\bf x} {t^2e^{\lambda_1t}\over2}. \end{eqnarray*} Complete the proof of Theorem~\ref{thmtype:10.5.2} by showing that {\bf y}_3 is a solution of {\bf y}'=A{\bf y} and that \{{\bf y}_1,{\bf y}_2,{\bf y}_3\} is linearly independent. \item\label{exer:10.5.35} Suppose the matrix$$ A=\left[\begin{array}{cc}a_{11}&a_{12}\\a_{21}&a_{22} \end{array}\right]$$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${\bf x}$. \exercisetext{In Exercises~\ref{exer:10.5.36}-~\ref{exer:10.5.45} plot trajectories of the given system.} \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}} \item\label{exer:10.5.36} \CGex\,${\bf y}'=\dst{\twobytwo{-3}{-1}41}{\bf
y}$& \item\label{exer:10.5.37} \CGex\,${\bf y}'=\dst{\twobytwo2{-1}10}{\bf y}$\end{tabular} \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}} \item\label{exer:10.5.38} \CGex\,${\bf y}'=\dst{\twobytwo{-1}{-3}35}{\bf
y}$& \item\label{exer:10.5.39} \CGex\,${\bf y}'=\dst{\twobytwo{-5}3{-3}1}{\bf
y}$\end{tabular} \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}} \item\label{exer:10.5.40} \CGex\,${\bf y}'=\dst{\twobytwo{-2}{-3}34}{\bf
y}$& \item\label{exer:10.5.41} \CGex\,${\bf y}'=\dst{\twobytwo{-4}{-3}32}{\bf
y}$\end{tabular} \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}} \item\label{exer:10.5.42} \CGex\,${\bf y}'=\dst{\twobytwo0{-1}1{-2}}{\bf
y}$& \item\label{exer:10.5.43} \CGex\,${\bf y}'=\dst{\twobytwo01{-1}2}{\bf y}$\end{tabular} \begin{tabular}[t]{@{}p{168pt}@{}p{168pt}} \item\label{exer:10.5.44} \CGex\,${\bf y}'=\dst{\twobytwo{-2}1{-1}0}{\bf
y}$& \item\label{exer:10.5.45} \CGex\,${\bf y}'=\dst{\twobytwo0{-4}1{-4}}{\bf
y}\$
\end{tabular}

\end{exerciselist}