
# 4.2: Linear Systems of Differential Equations

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A  first order system of  differential equations that can be written in
the form
\label{eq:10.2.1}
\begin{array}{ccl}
y'_1&=&a_{11}(t)y_1+a_{12}(t)y_2+\cdots+a_{1n}(t)y_n+f_1(t)\\
y'_2&=&a_{21}(t)y_1+a_{22}(t)y_2+\cdots+a_{2n}(t)y_n+f_2(t)\\
&\vdots\\
y'_n&
=&a_{n1}(t)y_1+a_{n2}(t)y_2+\cdots+a_{nn}(t)y_n+f_n(t)\end{array}

is called a {\color{blue}\it linear system\/}.

The linear system \eqref{eq:10.2.1}  can be written in matrix form as
$$\col{y'}n=\matfunc ann\col yn+\colfunc fn,$$
or more briefly as
\label{eq:10.2.2}
{\bf y}'=A(t){\bf y}+{\bf f}(t),

where
$${\bf y}=\col yn,\quad A(t)=\matfunc ann,\mbox{\quad and \quad}{\bf f}(t)=\colfunc fn.$$
We call $A$  the {\color{blue}\it coefficient matrix\/} of \eqref{eq:10.2.2} and
${\bf f}$  the {\color{blue}\it forcing function\/}.   We'll say that $A$
and ${\bf f}$ are {\color{blue}\it continuous\/} if their entries are continuous.
If ${\bf f}={\bf 0}$, then \eqref{eq:10.2.2} is {\color{blue}\it
homogeneous\/};
otherwise, \eqref{eq:10.2.2} is {\color{blue}\it nonhomogeneous\/}.

An  initial value problem  for \eqref{eq:10.2.2} consists of
finding a solution of \eqref{eq:10.2.2} that equals a given constant
vector
$${\bf k} =\col kn.$$
at some initial point $t_0$. We write this initial value problem as
$${\bf y}'=A(t){\bf y}+{\bf f}(t), \quad {\bf y}(t_0)={\bf k}.$$

The next theorem gives sufficient conditions for the existence
of solutions of initial value problems for \eqref{eq:10.2.2}. We omit the
proof.

\begin{theorem}\color{blue} \label{thmtype:10.2.1}
Suppose the coefficient matrix $A$ and the forcing function ${\bf f}$ are continuous on $(a,b)$, let $t_0$ be in $(a,b)$, and let ${\bf k}$ be an arbitrary constant $n$-vector. Then the initial value
problem
$${\bf y}'=A(t){\bf y}+{\bf f}(t), \quad {\bf y}(t_0)={\bf k}$$
has a unique solution on $(a,b)$.
\end{theorem}

\begin{example}\label{example:10.2.1}
\rm \mbox{}\newline
\begin{alist}
\item % (a)
Write the system
\label{eq:10.2.3}
\begin{array}{rcl}
y_1'&=&\phantom{2}y_1+2y_2+2e^{4t} \\[1\jot]
y_2'&=&2y_1+\phantom{2}y_2+\phantom{2}e^{4t}
\end{array}

in matrix form and conclude from Theorem~\ref{thmtype:10.2.1} that every
initial value problem for \eqref{eq:10.2.3} has a unique solution on
$(-\infty,\infty)$.
\item % (b)
Verify that
\label{eq:10.2.4}
{\bf y}=
{1\over5}\twocol87e^{4t}+c_1\twocol11e^{3t}+c_2\twocol1{-1}e^{-t}

is a solution of \eqref{eq:10.2.3} for all values of the constants $c_1$
and $c_2$.
\item % (c)
Find the  solution of the initial value problem
\label{eq:10.2.5}
y}(0)={1\over5}\twocol3{22}.

\end{alist}
\end{example}

\solutionpart{a}
The system  \eqref{eq:10.2.3} can be written in matrix form as
$${\bf y}'=\twobytwo1221{\bf y}+\twocol21e^{4t}.$$
An initial value problem for \eqref{eq:10.2.3} can be written as
$${\bf y}'=\twobytwo1221{\bf y}+\twocol21e^{4t}, \quad y(t_0)=\twocol{k_1}{k_2}.$$
Since the coefficient matrix and the forcing function are both
continuous on $(-\infty,\infty)$, Theorem~\ref{thmtype:10.2.1} implies that
this problem has a unique solution on $(-\infty,\infty)$.

\solutionpart{b}
If ${\bf y}$ is given by \eqref{eq:10.2.4}, then
\begin{eqnarray*}
A{\bf y}+{\bf f}&=&
{1\over5}\twobytwo1221\twocol87e^{4t}+
c_1\twobytwo1221\twocol11e^{3t}\\[2\jot]
&&+c_2\twobytwo1221\twocol1{-1}e^{-t}
+\twocol21e^{4t}\\[2\jot]
&=&{1\over5}\twocol{22}{23}e^{4t}+c_1\twocol33e^{3t}+c_2\twocol{-1}1e^{-t}
+\twocol21e^{4t}\\[2\jot]
&=&{1\over5}\twocol{32}{28}e^{4t}+3c_1\twocol11e^{3t}-c_2\twocol1{-1}e^{-t}
={\bf y}'.
\end{eqnarray*}

\solutionpart{c}
We must choose $c_1$ and $c_2$ in \eqref{eq:10.2.4} so that
$${1\over5}\twocol87+c_1\twocol11+c_2\twocol1{-1}={1\over5}\twocol3{22},$$
which is equivalent to
$$\twobytwo111{-1}\twocol{c_1}{c_2}=\twocol{-1}3.$$
Solving this system yields $c_1=1$, $c_2=-2$, so
$${\bf y}={1\over5}\twocol87e^{4t}+\twocol11e^{3t}-2\twocol1{-1}e^{-t}$$
is the solution of  \eqref{eq:10.2.5}.

\color{blue}
\remark{The theory of $n\times n$ linear systems of differential
equations is analogous to the theory of the scalar $n$-th order
equation
\label{eq:10.2.6}
P_0(t)y^{(n)}+P_1(t)y^{(n-1)}+\cdots+P_n(t)y=F(t),

as
developed in Sections~9.1. For example, by rewriting
\eqref{eq:10.2.6} as an equivalent linear system it can be shown that
Theorem~\ref{thmtype:10.2.1} implies Theorem~\ref{thmtype:9.1.1}
(Exercise~\ref{exer:10.2.12}).}
\color{black}

\newpage
\exercises

\begin{exerciselist}

\item\label{exer:10.2.1}
Rewrite the system in matrix form and
verify that the given vector function satisfies the  system for
any choice of the constants $c_1$ and $c_2$.

\begin{alist}
\item % (a)
$\begin{array}{ccl}y'_1&=&2y_1 + 4y_2\\ y_2'&=&4y_1+2y_2;\end{array} \quad {\bf y}=c_1\twocol11e^{6t}+c_2\twocol1{-1}e^{-2t}$

\item % (b)
$\begin{array}{ccl}y'_1&=&-2y_1 - 2y_2\\ y_2'&=&-5y_1 + \phantom{2}y_2;\end{array} \quad {\bf y}=c_1\twocol11e^{-4t}+c_2\twocol{-2}5e^{3t}$

\item % (c)
$\begin{array}{ccr}y'_1&=&-4y_1 -10y_2\\ y_2'&=&3y_1 + \phantom{1}7y_2;\end{array} \quad {\bf y}=c_1\twocol{-5}3e^{2t}+c_2\twocol2{-1}e^t$

\item % (d)
$\begin{array}{ccl}y'_1&=&2y_1 +\phantom{2}y_2 \\ y_2'&=&\phantom{2}y_1 + 2y_2;\end{array} \quad {\bf y}=c_1\twocol11e^{3t}+c_2\twocol1{-1}e^t$
\end{alist}

\item\label{exer:10.2.2}
Rewrite the system in matrix form and
verify that the given vector function satisfies the  system for
any choice of the constants $c_1$, $c_2$, and $c_3$.

\begin{alist}
\item % (a)
$\begin{array}{ccr}y'_1&=&- y_1+2y_2 + 3y_3 \\ y_2'&=&y_2 + 6y_3\\y_3'&=&- 2y_3;\end{array}$

${\bf y}=c_1\threecol110e^t+c_2\threecol100e^{-t}+c_3\threecol1{-2}1e^{-2t}$

\item % (b)
$\begin{array}{ccc}y'_1&=&\phantom{2y_1+}2y_2 + 2y_3 \\ y_2'&=&2y_1\phantom{+2y_2} + 2y_3\\y_3'&=&2y_1 + 2y_2;\phantom{+2y_3}\end{array}$

${\bf y}=c_1\threecol{-1}01e^{-2t}+c_2\threecol0{-1}1e^{-2t}+c_3\threecol111e^{4t}$

\item % (c)
$\begin{array}{ccr}y'_1&=&-y_1 +2y_2 + 2y_3\\ y_2'&=&2y_1 -\phantom{2}y_2 +2y_3\\y_3'&=&2y_1 + 2y_2 -\phantom{2}y_3;\end{array}$

${\bf y}=c_1\threecol{-1}01e^{-3t}+c_2\threecol0{-1}1e^{-3t}+c_3\threecol111e^{3t}$

\item % (d)
$\begin{array}{ccr}y'_1&=&3y_1 - \phantom{2}y_2 -\phantom{2}y_3 \\ y_2'&=&-2y_1 + 3y_2 + 2y_3\\y_3'&=&\phantom{-}4y_1 -\phantom{3}y_2 - 2y_3;\end{array}$

${\bf y}=c_1\threecol101e^{2t}+c_2\threecol1{-1}1e^{3t}+c_3\threecol1{-3}7e^{-t}$
\end{alist}

\item\label{exer:10.2.3}
Rewrite the initial value problem in matrix form and
verify that the given vector function is a solution.

\begin{alist}
\item % (a)
$\begin{array}{ccl}y'_1 &=&\phantom{-2}y_1+\phantom{4}y_2\\ y_2'&=&-2y_1 + 4y_2,\end{array} \begin{array}{ccr}y_1(0)&=&1\\y_2(0)&=&0;\end{array}$  \quad
${\bf y}=2\twocol11e^{2t}-\twocol12e^{3t}$

\item % (b)
$\begin{array}{ccl}y'_1 &=&5y_1 + 3y_2 \\ y_2'&=&- y_1 + y_2,\end{array} \begin{array}{ccr}y_1(0)&=&12\\y_2(0)&=&-6;\end{array}$  \quad
${\bf y}=3\twocol1{-1}e^{2t}+3\twocol3{-1}e^{4t}$
\end{alist}

\item\label{exer:10.2.4}
Rewrite the initial value problem in matrix form and
verify that the given vector function is a solution.

\begin{alist}
\item % (a)
$\begin{array}{ccr}y'_1&=&6y_1 + 4y_2 + 4y_3 \\ y_2'&=&-7y_1 -2y_2 - y_3,\\y_3'&=&7y_1 + 4y_2 + 3y_3\end{array},\; \begin{array}{ccr}y_1(0)&=&3\\ y_2(0)&=&-6\\ y_3(0)&=&4\end{array}$

${\bf y}=\threecol1{-1}1e^{6t}+2\threecol1{-2}1e^{2t}+\threecol0{-1}1e^{-t}$

\item % (b)
$\begin{array}{ccr}y'_1&=& \phantom{-}8y_1 + 7y_2 +\phantom{1}7y_3 \\ y_2'&=&-5y_1 -6y_2 -\phantom{1}9y_3,\\y_3'&=& \phantom{-}5y_1 + 7y_2 +10y_3,\end{array}\ \begin{array}{ccr}y_1(0)&=&2\\ y_2(0)&=&-4\\ y_3(0)&=&3\end{array}$

${\bf y}=\threecol1{-1}1e^{8t}+\threecol0{-1}1e^{3t}+\threecol1{-2}1e^t$
\end{alist}

\item\label{exer:10.2.5}
Rewrite the system in matrix form and
verify that the given vector function satisfies the  system for
any choice of the constants $c_1$ and $c_2$.

\begin{alist}
\item % (a)
$\begin{array}{ccc}y'_1&=&-3y_1+2y_2+3-2t \\ y_2'&=&-5y_1+3y_2+6-3t\end{array}$

${\bf y}=c_1\left[\begin{array}{c}2\cos t\\3\cos t-\sin t\end{array}\right]+c_2\left[\begin{array}{c}2\sin t\\3\sin t+\cos t \end{array}\right]+\twocol1t$

\item % (b)
$\begin{array}{ccc}y'_1&=&3y_1+y_2-5e^t \\ y_2'&=&-y_1+y_2+e^t\end{array}$

${\bf y}=c_1\twocol{-1}1e^{2t}+c_2\left[\begin{array}{c}1+t\\-t\end{array} \right]e^{2t}+\twocol13e^t$

\item % (c)
$\begin{array}{ccl}y'_1&=&-y_1-4y_2+4e^t+8te^t \\ y_2'&=&-y_1-\phantom{4}y_2+e^{3t}+(4t+2)e^t\end{array}$

${\bf y}=c_1\twocol21e^{-3t}+c_2\twocol{-2}1e^t+\left[\begin{array}{c} e^{3t}\\2te^t\end{array}\right]$

\item % (d)
$\begin{array}{ccc}y'_1&=&-6y_1-3y_2+14e^{2t}+12e^t \\ y_2'&=&\phantom{6}y_1-2y_2+7e^{2t}-12e^t\end{array}$

${\bf y}=c_1\twocol{-3}1e^{-5t}+c_2\twocol{-1}1e^{-3t}+ \left[\begin{array}{c}e^{2t}+3e^t\\2e^{2t}-3e^t\end{array}\right]$

\end{alist}

\item\label{exer:10.2.6}
Convert the linear scalar equation
$$P_0(t)y^{(n)}+P_1(t)y^{(n-1)}+\cdots+P_n(t)y(t)=F(t) \eqno{\rm (A)}$$
into an equivalent $n\times n$ system
$${\bf y'}=A(t){\bf y}+{\bf f}(t),$$
and show that $A$ and ${\bf f}$ are continuous on an interval
$(a,b)$ if and only if (A) is normal on $(a,b)$.

\item\label{exer:10.2.7}
A matrix function
$$Q(t)=\matfunc qrs$$
is  said to be  {\color{blue}\it differentiable\/}
if its entries $\{q_{ij}\}$ are  differentiable.  Then the {\color{blue}\it
derivative\/}  $Q'$ is defined by
$$Q'(t)=\matfunc {q'}rs.$$
\begin{alist}
\item % (a)
Prove: If $P$ and $Q$ are differentiable matrices such that $P+Q$ is
defined and if  $c_1$ and $c_2$ are constants, then
$$(c_1P+c_2Q)'=c_1P'+c_2Q'.$$
\item % (b)
Prove: If $P$ and $Q$ are differentiable matrices such that $PQ$ is
defined, then
$$(PQ)'=P'Q+PQ'.$$
\end{alist}

\item\label{exer:10.2.8}
Verify that $Y' = AY$.

\begin{alist}
\item  % (a)
$\dst{Y = \twobytwo {e^{6t}}{e^{-2t}} {e^{6t}}{-e^{-2t}}, \quad A = \twobytwo 2 4 4 2}$

\item % (b)
$\dst{Y = \twobytwo {e^{-4t}} {-2e^{3t}} {e^{-4t}} {5e^{3t}}, \quad A = \twobytwo {-2} {-2} {-5} {1}}$

\item % (c)
$\dst{Y = \twobytwo {-5e^{2t}} {2e^t} {3e^{2t}} {-e^t}, \quad A = \twobytwo {-4} {-10} 3 7}$

\item % (d)
$\dst{Y = \twobytwo {e^{3t}} {e^t} {e^{3t}} {-e^t}, \quad A = \twobytwo 2 1 1 2}$

\item % (e)
$Y = \left[\begin{array}{crr} e^t&e^{-t}& e^{-2t}\\ e^t&0&-2e^{-2t}\\ 0&0&e^{-2t}\end{array}\right], \quad A = \threebythree {-1} 2 {3} {0} 1 6 0 0 {-2}$

\item % (f)
$\dst{Y = \cthreebythree {-e^{-2t}} {-e^{-2t}} {e^{4t}} 0 {\phantom{-} e^{-2t}} {e^{4t}} {e^{-2t}} 0 {e^{4t}}, \quad A = \threebythree 0 2 2 2 0 2 2 2 0}$

\item % (g)
$\dst{Y = \cthreebythree {e^{3t}} {e^{-3t}} 0 {e^{3t}} 0 {-e^{-3t}} {e^{3t}} {e^{-3t}} {\phantom{-}e^{-3t}}, \quad A = \threebythree {-9}66{-6}36{-6}63}$

\item % (h)
$Y = \left[\begin{array}{crr} e^{2t}&e^{3t}& e^{-t}\\ 0&-e^{3t}&-3e^{-t}\\ e^{2t}&e^{3t}&7e^{-t}\end{array}\right] , \quad A = \threebythree 3 {-1} {-1}{-2} 3 2 4 {-1} {-2}$
\end{alist}

\item\label{exer:10.2.9}
Suppose
$${\bf y}_1=\twocol{y_{11}}{y_{21}}\mbox{\quad and \quad}{\bf y}_2=\twocol{y_{12}}{y_{22}}$$ are solutions of the homogeneous system
$${\bf y}'=A(t){\bf y}, \eqno{\rm (A)}$$
and define
\enlargethispage{1in}
$$Y= \twobytwo{y_{11}}{y_{12}}{y_{21}}{y_{22}}.$$
\begin{alist}
\item % (a)
Show that $Y'=AY$.
\item % (b)
Show that if ${\bf c}$ is a constant vector then ${\bf y}= Y{\bf c}$
is a solution of  (A).
\item % (c)
State generalizations of  \part{a} and \part{b} for $n\times n$
systems.
\end{alist}
\newpage

\item\label{exer:10.2.10}  Suppose $Y$ is a differentiable square
matrix.
\begin{alist}
\item % (a)
Find a formula for the derivative of $Y^2$.
\item % (b)
Find a formula for the derivative of $Y^n$, where $n$ is any
positive integer.
\item % (c)
State how the results obtained in \part{a} and \part{b} are analogous to
results from  calculus  concerning scalar functions.
\end{alist}

\item\label{exer:10.2.11}
It can be shown that if $Y$ is a differentiable and  invertible
square matrix function, then $Y^{-1}$ is differentiable.

\begin{alist}
\item % (a)
Show that ($Y^{-1})'= -Y^{-1}Y'Y^{-1}$.
(Hint: Differentiate the identity $Y^{-1}Y=I$.)
\item % (b)
Find the derivative of $Y^{-n}=\left(Y^{-1}\right)^n$, where $n$
is a positive integer.
\item % (c)
State how the results obtained in \part{a} and \part{b} are analogous to
results from  calculus  concerning scalar functions.
\end{alist}

\item\label{exer:10.2.12} Show that Theorem~\ref{thmtype:10.2.1} implies
Theorem~\ref{thmtype:9.1.1}.  \hint{Write the scalar equation
$$P_0(x)y^{(n)}+P_1(x)y^{(n-1)}+\cdots+P_n(x)y=F(x)$$
as an $n\times n$ system of linear equations.}

\item\label{exer:10.2.13}
Suppose ${\bf y}$ is a solution of the $n\times n$ system  ${\bf y}'=A(t){\bf y}$ on $(a,b)$, and that the $n\times n$ matrix $P$
is invertible and differentiable on $(a,b)$. Find a matrix $B$
such that the function ${\bf x}=P{\bf y}$ is a solution of ${\bf x}'=B{\bf x}$ on $(a,b)$.
\end{exerciselist}