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\begin{document}

\newpage
% You only need to change the NAME OF SCENARIO in the next line and delete the footnote. See above for instructions on toggling from Teacher Version to Student Version.
\title{\ifnum\TeacherEdition=1 TEACHER VERSION  \else STUDENT VERSION \fi\\CONJECTURING  SOLUTIONS FOR LINEAR SYSTEM OF DIFFERENTIAL EQUATIONS}




\markboth{Conjecturing Solutions Linear Systems}{Conjecturing Solutions Linear Systems}





\author{Brian Winkel\\ Director
SIMIODE\\
Chardon OH 44024 USA\\ BrianWinkel@SIMIODE.org}

\makeStitlePDFLaTex



\begin{abstract}
We lead students from  the solution for $y(t) = k\, y(t)$ to a natural extension for the solution conjecture for a system of two constant coefficient, homogeneous, linear differential equations introducing eigenvalues and eigenvectors through student discovery.
\end{abstract}


\section*{INTRODUCTION}





We consider a system of linear differential equations


\begin{eqnarray}\label{eq:system}
x'(t) &=& a \, x(t) + b \,y(t) \\ \nonumber
y'(t) &=& c \,x(t) + d \,y(t) 
\end{eqnarray}     
with  $x(0) = x_0$   and   $y(0) = y_0$.
 

We can put this in matrix form:

\begin{equation}\label{eq:matrix}
\left[
\begin{array}{c}
x'(t) \\
y'(t)
\end{array} 
\right]  = \left[
\begin{array}{c}
a \, x(t) + b \,y(t) \\
c \,x(t) + d \,y(t)
\end{array} 
\right] = \left[
\begin{array}{cc}
a&b \\
c&d
\end{array} 
\right] \cdot \left[
\begin{array}{c}
x(t) \\
y(t)
\end{array} 
\right] 
\end{equation}
     with    $x(0) = x_0$   and   $y(0) = y_0$. 

  
OR
 
$$\displaystyle{X'(t) = A\cdot X(t)} \, , $$
where

\begin{equation}\label{eq:Xarray}
X'(t) = \left[ \begin{array}{c}
x'(t) \\
y'(t)
\end{array}\right] \quad , \quad A = \left[
\begin{array}{cc}
a&b \\
c&d
\end{array} 
\right] \quad ,\quad  X(t) = \left[ \begin{array}{c}
x(t) \\
y(t)
\end{array}\right] 
\end{equation}

$X'(t) = A\cdot X(t)$  is very suggestive for a solution of the form  $X(t) = c e^{A t}$. Thus one could conjecture a solution and try it out to see what $c$ would have to be, but we have to remember $A$ is a matrix, not a number, and while there are theories of what $e^A$ would mean for a matrix $A$ we choose not to go there now. 

Instead, we conjecture  a vector solution for  $X(t)$ which has the best of both worlds (1) nice exponential form with no matrices in the exponent and (2) vector status where we can see the pieces of the solution:

\begin{equation}\label{eq:vecsol}        
X(t) = \left[ \begin{array}{c}
x(t) \\
y(t)
\end{array}\right]    =  \left[ \begin{array}{c}
u \\
v
\end{array}\right] \cdot e^{\lambda t}\, .                                                                                     
\end{equation}
So here we are suggesting  $x(t) = u e^{\lambda t}$ and $y(t) = v e^{\lambda t}$.  

\subsection*{Assignment}

\begin{enumerate}
\item  Now put the conjecture (\ref{eq:vecsol}) in (\ref{eq:Xarray}) and see what has to happen to $a$, $b$, $c$, $d$ and $u$, $v$ in order to obtain solutions - interesting solutions, i.e. other than  $u = 0$  and  $v=0$.  Go for it!!!   Summarize your conclusions in a nice write-up 

\item Illustrate the solution strategy from assignment (1) using the following system
\begin{eqnarray}
x'(t) &=&-4 \, x(t) -2\,y(t) \\ \nonumber
y'(t) &=& -1 \,x(t) -3 \,y(t) 
\end{eqnarray}     
with  $x(0) = 2$   and   $y(0) = 1$. 
\end{enumerate}

\teacher{
\section*{NOTES FOR TEACHERS}

We have used this approach successfully for years.  When asked, students substitute conjecture (\ref{eq:vecsol}) into the system of differential equations (\ref{eq:system}) and produce necessary conditions on  $a$, $b$, $c$, $d$ and $u$, $v$ for a solution to exist from the resulting equations with the $e^{\lambda t}$ term canceled out from every term:

 \begin{equation}\label{eq:cancel}
 \left[
\begin{array}{cc}
a -\lambda & b \\
c  & d -\lambda 
\end{array} 
\right] \cdot \left[
\begin{array}{c}
x(t) \\
y(t)
\end{array} 
\right] = \left[
\begin{array}{c}
0 \\
0
\end{array} 
\right] 
\end{equation}


Students then see the only non-trivial or non-zero solutions which can occur is when in non-matrix form we have the following:

\begin{eqnarray}\label{eq:nonmat}
(a - \lambda)\,x(t) + b \,y(t) &=& 0\\ \nonumber
c \,x(t) + (d - \lambda) \,y(t) &=& 0\, .
\end{eqnarray}
 This means plot of both equations (being lines) in the $x(t)$-$y(t)$ plane have to be the same, i.e. they go through the origin and have identical slopes, 

\begin{equation}\label{eq:slopes}
\frac{-(a - \lambda)}{b} = \frac{-c}{(d - \lambda)}\, .
\end{equation}
 
This implies that solutions to the original differential equation, (\ref{eq:system}),  must satisfy the following quadratic equation:


\begin{equation}\label{eq:char}
\lambda^2 - (a+d)\lambda+ a d - b c = 0 \, .     
\end{equation}

This equation characterizes the solutions of (\ref{eq:system}) and so is called the {\it characteristic equation\/} for the system (\ref{eq:system}) and the roots of this equation, $\lambda_1$ and $\lambda_2$, are called {\it eigenvalues} and their corresponding vectors $\left[\begin{array}{c}u_i\\ v_i\end{array}\right]$, $i = 1, 2$,  are called {\it eigenvectors}.

\subsection*{Practical Issues}

When teaching this material using  by-hand analysis we do not dwell on all the cases of the roots of a quadratic (i.e. real and unequal, real and equal, and complex), but rather we confine this discussion  to  real and unequal roots, always with negative eigenvalues as no real system modeling in this way can have positive eigenvalue. We study the cases,   real and equal  and  complex eigenvalues, shortly thereafter in context of a modeling experience and use technology to obtain the corresponding eigenvalues and eigenvectors as well as handle the complexities of the computation.  We  engage in appropriate algebra and graphical feedback to relate the behavior of the solution to the nature of the eigenvalues.

With the littlest of prompting students can actually derive eigenvalues and eigenvectors without ever having seen a definition or application. Indeed, this approach is one of discovery in which students actually find these objects, but may not have a name for them. However, they sure do have a reason for knowing about them as they see them (they actually found them!) to be the building blocks for at least a 2 by 2 system of linear differential equations.


This approach NEVER FAILS to engage students and draw them in to discovery of something very useful - eigenvalues and eigenvectors.
}
\end{document}

