\documentclass{amsart}

\title{Understanding $y''+by'+cy$}

\begin{document}

\maketitle

\section{Introduction}
We begin this lab with a theorem:
\begin{quote}
Any $n^{\rm th}$ order differential equation
$$
y^{(n)}=f(t, y, y', \ldots, y^{n-1})
$$
can be converted into
and analyzed as a system of $n$ first order equations.
\end{quote}
We will illustrate this theorem by converting the second
order linear equation into a system of first order equations,
then we will analyze the behavior of the solutions in
these terms.

First, we convert\footnote{Your textbook writes this equation
as $ay''+by'+cy=0$.  But, since we assume $a\neq 0$, we divided
both sides by $a$ to get the simplified equation, $y''+By'+Cy=0$}
$y''+by'+cy=0$ into a system of first order
equations.  We do this by defining new variables.  Let
$$
u=y \qquad v=y'
$$
Then we write the system of differential equations for $u, v$:
$$
\begin{array}{rl}
\dot{u}=&y'=v\\
\dot{v}=&y''=-by'-cy=-bv-cu
\end{array}
$$
Therefore, the solutions to $y''+by'+cy=0$ are equivalent to
the solutions to the system:
$$
\begin{array}{rl}
\dot{u}&=v\\
\dot{v}&=-bv-cu
\end{array}
$$
Some notes:
\begin{enumerate}
\item If we want $y(t)$, all we really need out of the
system is $u(t)$.
\item The phase plane will be a plot of $(u,v)$, or position
{\bf and velocity}.
\item We could get analytic solutions from the linear system,
but that would require some linear algebra.
\item In problems from physics, $b\geq 0$ and $c\geq 0$.  However,
we can examine the solutions for all $b$ and all $c$.
\end{enumerate}

\section{The Lab}

The purpose of this lab is to analyze the behavior of solutions
to $y''+by'+cy=0$ in a different way than you've seen in Differential
Equations class- we will study this using a system of equations,
and analyze the corresponding phase planes.


We know that to solve $y''+by'+cy=0$, we solve the characteristic
equation,
$$
r^2+br+c=0 \Rightarrow r=\frac{-b\pm \sqrt{b^2-4c}}{2}
$$
Therefore, the critical quantity to analyze is $b^2-4c$, which
we will call the discriminant, $\Delta$.  If $\Delta>0$, there
are two real, distinct roots to the characteristic equation.  If
$\Delta=0$ there is a single root, and if $\Delta<0$, there are
complex solutions to the characteristic equation.  In ODE class,
you learned how to write the solution in each case.  In Calc
Lab, we will learn how to perform a ``qualitative analysis''.

To summarize the material we will discuss, you should draw
by hand, on one page, the parabola $b^2-4c=0$, where the
``$x-$axis'' is $b$, and the ``$y-$axis'' is $c$.  So,
in addition to the four quadrants, we have subdivided the
plane into points inside the parabola, on the parabola,
and outside the parabola.  To keep these regions more
concrete, let's focus on some particular values:
\begin{itemize}
\item Along the $c-$axis:  $b=0, c=1$. (In this case, $\Delta<0$)
\item Inside the parabola:  $b=1, c=1$. (In this case, $\Delta<0$)
\item On the parabola:  $b=2, c=1$. (In this case, $\Delta=0$)
\item Outside the parabola:  $b=3, c=1$. (In this case, $\Delta>0$)
\item Along the $b-$axis:  $b=3, c=0$. (In this case, $\Delta>0$)
\end{itemize}
On your piece of paper, plot these points.

\subsection{Lab Questions}

\begin{enumerate}
\item For each point listed above,
\begin{enumerate}
\item Find the equilibria
\item Plot the corresponding phase plane in Maple,
together with some solutions (you may choose the initial conditions).
\item Draw a ``typical'' phase plot on your paper in the
appropriate place.
\end{enumerate}
\item Continue this exploration at some other values of $(b,c)$:
What happens if $b=0$ and $c=0$?
\item Why did we say that for ``physical'' problems, $b>0$ and
$c>0$?  (Consider what the solutions do in the other quadrants).
\item The form $y''+by'+cy$ is especially nice if $y$ is an
exponential function, a polynomial, sines and cosines, or a
product of exponentials with polynomials or exponentials with
sines and cosines.  Write down the result of using this
differential operation on:
\begin{enumerate}
\item $y(t)={\rm e}^{-2t}\left(\sin(3t)+2\cos(3t) \right)$
\item $y(t)=3t^2+2$
\item $y(t)=\cos(2t)$
\end{enumerate}
Given this experience, what is the form of the particular part of
the solution if $y''+by'+cy=3t^2+t$?

\item In the next question we set $c=1$, and add a periodic forcing
function:
$$
y''+by'+y=\sin(t)
$$
What happens to the solution as $b\rightarrow 0$?  Use any
visualization technique- it might be useful to set $y(0)=1$ and
$y'(0)=1$.  You should consider the homogeneous part of the
solution separately from the particular part of the solution
(until $b=0$).
\end{enumerate}
\end{document}
