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\newtheorem{thm}{Theorem}

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\newtheorem{prob}{Problem}
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\begin{document}
\begin{center}
{\large \bf The Theory of Second Order Linear Differential Equations}
\footnote{\copyright Michael C. Sullivan, \today \,(except for parts excerpted form {\em Elementary Differential Equation and Boundary
Value Problems}, by Boyce and DiPrima)}

Michael C. Sullivan 

Math Department

Southern Illinois University

\end{center}

\vspace{.25in}

\hrule

\vspace{.25in}

These notes are intended as a supplement to section 3.3
of the textbook {\em Elementary Differential Equation and Boundary
Value Problems}, by Boyce and DiPrima, 9th edition.

\vspace{.25in}

\hrule

\vspace{.25in}

\section{A motivating example}

Consider the equation $y'' + y =0$. We can see by inspection that 
$y_1 = \sin x$ and $y_2 = \cos x$ are solutions.
It is easily checked that $y = C_1 \sin x + C_2 \cos x$ is a solution
for all choices of the constants $C_1$ and $C_2$. Furthermore, {\bf as the reader should check}, 
for all initial conditions of the from  $y(x_0)=\alpha$ and
$y'(x_0)=\beta$, we can find $C_1$ and $C_2$ such that 
$y = C_1 \sin x + C_2 \cos x$ satisfies these conditions. 
We will see later that we do indeed have all
possible solutions to $y'' + y =0$. 

Now consider the figure below. It is a plane, but the axes are labeled
$C_1$ and $C_2$. To each point in this plane we associate the function
$y = C_1 \sin x + C_2 \cos x$. In this way the solution set of 
$y'' + y =0$ becomes a {\em vector space}. We shall not give a formal
definition of a vector space. But we note that, geometrically, adding
pairs of solutions to our differential equation is just like adding
vectors in this plane. Thus, the tools of linear algebra will 
come into play.

	\begin{center}  
	\psfrag{c1}{$C_1$}
	\psfrag{c2}{$C_2$}
	\psfrag{2}{2}
	\psfrag{3}{3}
	\psfrag{()}{$(3,2) \rightarrow y=3 \sin x + 2 \cos x$}
	\includegraphics[height=1.5in]{plane.eps}
	\end{center}
	\label{plane_fig}

\pagebreak

\section{Review of 2$\times$2 systems of equations} 

The {\em determinant} of a $2\times2$ matrix is 
\[ \det \left[ \begin{array}{cc} 
a & b \\
c & d \end{array} \right] = ad-bc.
\]
Determinants are often denoted by replacing the brackets
with straight lines:
\[ \det \left[ \begin{array}{cc} 
a & b \\
c & d \end{array} \right] = 
\left| \begin{array}{cc} 
a & b \\
c & d \end{array} \right| = ad-bc.
\]

Now, consider a $2\times2$ system of linear equations,
\[ \hspace{.5in} \begin{array}{ccc}  
ax+by & = & e \\
cx+dy & = & f 
\end{array} \hspace{.5in} (1) \]
and assume they represent two lines in the $xy$-plane.
This can be rewritten in matrix notation as,
\[ \left[ \begin{array}{cc} 
a & b \\
c & d \end{array} \right]
\left[ \begin{array}{c}
x \\ y \end{array} \right]
=
\left[ \begin{array}{c}
e \\ f \end{array} \right].
\]
We shall separate our discussion into two cases.

\vspace{.14in}

CASE I: When $e=f=0$, we say the system of equations (1) is {\em homogeneous}.
The pair equations can be thought of as representing two lines
passing through the origin. Hence, if they have the same slope there are
infinitely many points where they intersect. But if they have 
different slopes the only intersection point is the origin. It is easy to
show that the slopes are the same if and only if the determinant of 
$\left[ \begin{array}{cc}
a & b \\
c & d 
\end{array} \right]$ is zero. (Show this, and note that this result does 
not depend on $e$ and $f$ being zero.) Thus, we can state the following:

\begin{eqnarray*}
&&\\
\left| \begin{array}{cc}
a & b \\
c & d 
\end{array} \right| = 0 
& \Longleftrightarrow &
\left\{ \begin{array}{c}
\mbox{There are infinitely many} \\
\mbox{solutions to equation (1).}
\end{array} \right\} \\
%
&&\\
& \mbox{and} &  \\
&&\\
%
\left| \begin{array}{cc}
a & b \\
c & d
\end{array} \right| \neq 0 
& \Longleftrightarrow &
\left\{ \begin{array}{c}
\mbox{There is only one solution,} \\
\mbox{$x=y=0$, to equation (1).}
\end{array} \right\} \\
&&
\end{eqnarray*}



Case II: If either $e$ or $f$ is not zero, then we say the system of equations
(1) is {\em nonhomogeneous}. Now the lines given by the two equations are not 
both going
through the origin. If the slopes are different there is still a unique point
of intersection. But if the slopes are the same either the lines coincide 
as before giving infinitely many solutions, or they are disjoint parallel 
lines and have no points in common. Thus, we can write:

\begin{eqnarray*}
&&\\
\left| \begin{array}{cc}
a & b \\
c & d
\end{array} \right| = 0 
& \Longleftrightarrow &
\left\{ \begin{array}{c}
\mbox{There are infinitely many solutions or} \\ 
\mbox{there are no solutions to equation (1).}
\end{array}\right\} \\
&&\\
&\mbox{and}&\\
&&\\
\left| \begin{array}{cc}
a & b \\
c & d
\end{array} \right| \neq 0 
&\Longleftrightarrow&
\left\{ \mbox{There is only one solution to (1).} \right\} \\
&&
\end{eqnarray*}

The diligent reader will check the conclusions we have reached for several 
examples and draw the relevant graphs.


\section{Linear Independence and the Wronskian}

We now define some tools from linear algebra that will
be useful in our study of the solutions of second order
linear differential equations.

\begin{dfn}[Linear Independence]
Two functions, $f(x)$ and $g(x)$, defined on the same open 
interval $I$ are {\em linearly independent} if neither is a 
nonzero multiple of the other. This is the same as saying
it is impossible to find real numbers, $C_1$ and $C_2$,
not both zero, such that 
\[ C_1 f(x) + C_2 g(x) = 0 \]
for all $x \in I$. If it is possible two find two such
numbers, not both zero, then we say $f$ and $g$ are
{\em linearly dependent}. Again, this means $f$ is a multiple of $g$
or vice versa.
\end{dfn}

\begin{prob} Show that $x$ and $x^2$ are linearly 
independent over the real line.
\end{prob}

\begin{prob} Show that $|x|$ and $2x$ are linearly dependent
on $I=(2,3)$ but are linearly independent on $I=(-1,1)$.
\end{prob}

\begin{dfn}[Wronskian]
Let $f(x)$ and $g(x)$ be differentiable functions. 
Then their {\em Wronskian} at $x$ is defined to be the function
\[ W(x) = W(f,g)(x) = f(x)g'(x) - f'(x)g(x) =
\left| 
\begin{array}{cc}
f(x) & g(x) \\
f'(x) & g'(x) 
\end{array} \right|
\] 
\end{dfn}

\begin{prob} Show that $W(\sin x,\cos x) = -1$ for all $x$.
\end{prob}

\begin{prob} Show that $W(e^{r_1x}, e^{r_2x})$ 
is never zero if $r_1 \neq r_2$. 
\end{prob}

\begin{thm}  \label{thm:W}
Let $f$ and $g$ be differentiable functions defined on an
open interval $I$.
If $f$ and $g$ are linearly dependent then $W(f,g)(x) = 0$ for
all $x \in  I$. 
It follows that if the Wronskian is not zero for at least one point
in $I$ then $f$ and $g$ are not linearly dependent, and are thus
linearly independent.
\end{thm}

Notice that Theorem \ref{thm:W} holds true for the pairs of functions
in the examples above.

\begin{proof}
Suppose $f$ and $g$ are linearly dependent on $I$.
Then there exist constants $C_1$ and $C_2$, not both zero,
such that
\[ C_1f(x) + C_2g(x) = 0,  \hspace{.2in} \mbox{for all $x \in I.$}\]
Taking the derivative gives
\[ C_1f'(x) + C_2g'(x) = 0, \hspace{.2in} \mbox{for all $x \in I.$}\]
But, we can view this as a solution to the $2\times2$
system
\[ \left[ \begin{array}{cc}
f(x) & g(x) \\
f'(x) & g'(x) \end{array} \right]
\left[ \begin{array}{c}
C_1 \\ C_2 \end{array} \right] 
=
\left[ \begin{array}{c}
0 \\ 0 \end{array} \right].
\]
Since the system is homogeneous and $C_1$ and $C_2$ are
not both zero, it follows that the determinant of 
$ \left[ \begin{array}{cc}
f(x) & g(x) \\
f'(x) & g'(x) \end{array} \right] $ is zero. Thus, the Wronskian
is always zero in $I$.
\end{proof}

{\bf Remark.\footnote{This was problem 28 in section 3.3 of the 6th edition but is no longer included.}}
Let $f(t) = t^2|t|$ and $g(t)=t^3$. These give us an example of a pair
of differentiable linearly independent functions for which the Wronskian
is always zero. (Check these claims.) 
Why does this not contradict Theorem 1? However, we will
see later (Theorem 5) that if $f$ and $g$ are solutions to $y''+py'+qy=0$ then they 
are linearly independent if and only if their Wronskian is never zero
in the relevant interval.

\section{The Theory}

We now begin our discussion of differential equations.

\begin{thm}[Existence and Uniqueness: 3.2.1 page 146]\label{EandU}
Let  $p(x)$, $q(x)$, 
and $\phi(x)$ be continuous functions on an open interval $I$ and let $x_0\in I$. 
Then the initial value problem
\[
\begin{array}{cc}
y'' + py' + qy = \phi &\mbox{with $y(x_0)=\alpha$ and 
$y'(x_0)=\beta$,}
\end{array}\]
has a unique solution for $y(x)$ on $I$.
\end{thm}

\begin{exam} Solve the initial value problem,
\[ y'' + \frac{t}{t^2-1}\,y' + \frac{1}{t-2}\,y = e^t \hspace{.2in} y(0)=1\, \& \,y'(0)=2.\]
\end{exam}

\begin{proof}[Solution.]
A unique solution is guaranteed to exist on $(-1,1)$. If we had as initial 
conditions $y(1.5)=7$ \& $y'(1.5)=23$ then a unique solution is guaranteed 
to exist on $(1,2)$. If we had as initial conditions $y(-13.5)=6$ \& $y'(-13.5)=10^{23}$ 
then a unique solution is guaranteed to exist on $(-\infty,-1)$. 
\end{proof}

{\bf Remark.} The proof of Theorem \ref{EandU} is very difficult and we will not do it. However, 
you should understand the statement of the theorem and its many implications 
and applications. It is the most important theorem in Chapter 3. We will, 
in what follows, prove some special cases of this theorem and 
we will use it to gain an understanding of the structure of the solution 
set of a second order linear differential equation. The general idea is this. 
For any second order linear homogeneous equation $(y''+py'+qy=0)$ there 
exists a pair of linearly independent solutions, $f$ and $g$. Any initial 
value problem $(y(a)=b, y'(a)=c)$ can be solved by a unique linear 
combination of $f$ and $g$. 

In sections 3.5 and 3.6 we will show that a solution to a nonhomogeneous
problem can be found by adding an extra term to any solution of the corresponding
homogeneous problem.

\begin{thm}[Linear Superposition: 3.2.2 page 147]\label{super}
If $y=f(x)$ and $y=g(x)$ 
are solutions of the {\em homogeneous} differential equation, 
$ y'' + py' + qy = 0$, then so 
is any linear combination of  $f(x)$ and $g(x)$.
\end{thm}

\begin{proof}
The proof is very easy. Do it. You may be tested on this. This concept
will come up over and over again, both in this course and others.
\end{proof}


\begin{thm}
%[A special case of Theorem~\ref{EandU}: Existence only]
\label{special}
The unique solution of the initial value problem
$ay'' + by' + cy = 0$ with $y(x_0)=\alpha$ and $y'(x_0)=\beta$ has a solution
given by some linear combination of one of the these three pairs of functions, $\{ e^{r_1x}, e^{r_2x} \}, \{ e^{rx}, xe^{rx}\}$ and 
$\{ e^{\gamma x} \sin \lambda x , e^{\gamma x} \cos \lambda x\}$.
\end{thm}

\begin{proof}
We have shown that we can always find two linearly independent 
solutions to $ay'' + by' + cy = 0$ and that these will be one of the
three pairs listed in the theorem. 
Call them $f$ and $g$. So, the only question is
can we solve the system
\[ \left[ \begin{array}{cc}
f(x_0) & g(x_0) \\
f'(x_0) & g'(x_0) \end{array} \right]
\left[ \begin{array}{c}
C_1 \\ C_2 \end{array} \right] 
=
\left[ \begin{array}{c}
\alpha \\ \beta \end{array} \right].
\]
The answer depends on the Wronskian. We know that, because of
linear independence, the Wronskian is not always zero. But what
if it is zero at $x=x_0$? That would be a problem. However, the 
reader can check that for the three possible solution pairs
$\{ e^{r_1x}, e^{r_2x} \}, \{ e^{rx}, xe^{rx}\}$ and 
$\{ e^{rx} \sin \gamma x , e^{rx} \cos \gamma x\}$ the Wronskian is in fact 
always nonzero.
\end{proof}

Can we push this idea further? 

\begin{thm}\label{thm_W}
Let $y=f(x)$ and $y=g(x)$ be linearly independent solutions of
$y'' +p(x)y' + q(x)y = 0$, where $p$ and $q$ are continuous on
an open interval $I$. Then the Wronskian $W(f,g)$ is never
zero in $I$. 
\end{thm}

{\bf Remark:} This means that if we can find a pair ($f$ and $g$) 
of linearly
independent solutions to $y'' +p(x)y' + q(x)y = 0$ then we can
solve any initial value problem $y(x_0)=\alpha$ and $y'(x_0)=\beta$ with a
linear combination of $f$ and $g$. We record this as Theorem 6 below.

\begin{proof}[Easy Proof.]
Suppose that there is a point $x_0 \in I$ where $W(f,g)(x_0)=0$.
Then consider the initial value problem with $y(x_0)=y'(x_0)=0$.
This leads to the system of equations
\[
\left[ \begin{array}{cc}
f(x_0) & g(x_0) \\
f'(x_0) & g'(x_0) \end{array} \right]
\left[ \begin{array}{c}
C_1 \\ C_2 \end{array} \right] 
=
\left[ \begin{array}{c}
0 \\ 0 \end{array} \right].
\]
But if the Wronskian at $x_0$, which is just the determinant of the 
matrix above, is zero, then there are infinitely many values
of $C_1$ and $C_2$ that work. But this contradicts the uniqueness
claim of Theorem~\ref{EandU}.
\end{proof}

Your text has a much longer proof. It makes use of 
{\em Abel's Formula} which is useful in its own right. You should
know Abel's Formula, but the proof below is optional reading.

\begin{proof}[Book's Proof.]
STEP 1.
We will show that the Wronskian is given by the equation
\[ W(f,g)(x) = C \exp\left(-\int p(x) \, dx \right), \]
where $C$ is a constant. (This is Abel's Formula: 3.2.6 page 153.) 
It follows that $W$ is either always zero or never zero.

We know that
\[ f'' + pf' + qf = 0, \]
and
\[ g'' + pg' + qg = 0. \]
Multiply the first equation by $g$ and the second one by $f$.
Thus,
\[ gf'' + pgf' + qgf = 0, \]
and
\[f g'' + pfg' + qfg = 0. \]
Subtract first equation from the second and simplify to get
\[ (fg'' - gf'') + p(fg' - f'g) = 0. \]
Notice that $W = fg' - f'g$ and that $W' = fg'' - gf''$.
Thus $W$ must satisfy the differential equation
\[ W' + pW = 0. \]
But this we can solve, showing that 
 \[ W(x) = C \exp\left(-\int p(x) \, dx\right), \]
for some $C$. Next we must show that $C$ is not zero.

STEP 2.
Suppose that $W$ is always zero in $I$. Let $x_0$ be any point in $I$.
Then the system of equations
\[ C_1f(x_0) + C_2g(x_0) = 0,\]
\[ C_1f'(x_0) + C_2g'(x_0) = 0,\]
has a nontrivial solutions for $C_1$ and $C_2$ (i.e., they are not both
zero). Let $h(x) = C_1f(x) + C_2g(x)$. Then $y=h(x)$ is a solution to the
initial value problem $y''+py'+qy=0$ with $y(x_0)=0$ and $y'(x_0)=0$.
However, it is clear that $y(x)=0$ (the zero function) solves this system.
Thus, by the uniqueness  part of Theorem~\ref{EandU}, $h(x)$ is the 
zero function. This, in turn, means that $C_1f(x) + C_2g(x) = 0$ for all
$x$ in $I$. Thus, $f$ and $g$  are linearly dependent on $I$, contradicting
our hypotheses. Thus, $W$ is never zero.
\end{proof}


\begin{thm}\label{hmm}
Suppose $f$ and $g$ are linearly independent
solutions of 
\[ y'' + py' + qy = 0. \]
Then any initial value problem, $y(x_0)=\alpha$ and $y'(x_0)=\beta$, 
is solved by a linear combination of $f$ and $g$.
\end{thm}

\begin{proof}
In light of Theorem~\ref{thm_W} this is just the same as the proof of 
Theorem~\ref{special}.
\end{proof}

\begin{thm}
If $f$ and $g$ are linearly 
independent solutions of 
\[ y'' + py' + qy = 0, \]
then every other solution can be written as a linear combination
of $f$ and $g$.
\end{thm}

\begin{proof}
Let $h$ be a solution. Let $x_0 \in I$. 
Let $\alpha=h(x_0)$ and $\beta=h'(x_0)$.
Consider the system of equations
\[ C_1 f(x_0) + C_2 g(x_0) = \alpha \]
\[ C_1 f'(x_0) + C_2 g'(x_0) = \beta. \]
Since the Wronskian of $f$ and $g$ is never zero it is not zero at $x=x_0$.
This means we can find unique values of $C_1$ and $C_2$ that solve the 
$2\times2$ system. By the uniqueness part of Theorem~\ref{EandU}
it follows that
$h(x)=C_1 f(x) + C_2 g(x)$ for all $x \in I$.
\end{proof}


\begin{thm}
Every differential equation of the form 
$y'' + py' + qy = 0,$
with $p$ and $q$ continuous on $I$, has two linearly independent solutions.
\end{thm}

\begin{proof}
Let $x_0$ be in $I$. Consider the two initial value problems 
\[ 
\begin{array}{ccccc}
y'' + py' + qy = 0 & \mbox{with}& y(x_0)=1 & \mbox{and} & y'(x_0)=0 
\end{array}\]
and
\[
\begin{array}{ccccc} 
y'' + py' + qy = 0 & \mbox{with} & y(x_0)=0 & \mbox{and}& y'(x_0)=1.
\end{array} \]
Let $f$ be a solution of the first and $g$ of the second.
Their Wronskian at $x=x_0$ is 1 (check this). Thus, $f$ and $g$ are
linearly independent.
\end{proof}

{\bf Conclusions:} So, by using suitable initial values we can show
that there exists a pair of linearly independent solutions to any
second order linear homogeneous differential equation. Except for the 
case of constant coefficients and a few other special
cases discussed in the text we do not have a procedure for finding them.
In section 3.4 (page 170, ``Reduction of Order'') we did see that if you know one solution, 
$y_1(x)$, then you can find another of the form $y_2(x) = v(x)y_1(x)$.
This second solution can be shown to be linearly independent of the first.

\section{Implications and applications}

Students often wonder what the point is of learning the theory.
It can seem rather abstract and remote from practical concerns.
In an attempt to mitigate this phenomena, this section shows some
useful applications that we would not have suspected without 
the theoretical insights gleaned above. 


\subsection{The interlacing theorem}

\begin{thm}[\#33 in 3.3.] \label{thm:interlace}
Suppose that $y_1$ and $y_2$ are linearly independent solutions of $y''+py'+qy=0$.
Suppose $p$ and $q$ are continuous on $I$. 
Suppose $a$ and $b$ are in $I$, $a<b$, and $f(a)=f(b)=0$. 
Then there is a point $c$ in between $a$ and $b$, $a<c<b$,
such that $g(c)=0$. Thus, $y_1$ and $y_2$ are ``interlaced''.
\end{thm}

{\bf Examples:} 
Consider $y'' +y =0$. The functions $\sin x$ and $\cos x$ are a linearly
independent pair of solutions. In between each pair of zeros of $\sin x$,
$\cos x$ has a zero, and vice versa. Thus the graphs of these two functions
are said to be {\em interlaced}. This would work for any solution pair
of the form $\{ e^{\gamma x} \sin \lambda x , e^{\gamma x} \cos \lambda x\}$. 
[Pick numbers for $\gamma$ and $\lambda$, 
and graph these two functions to see this.]

We give two proofs. The might be called an {\em engineer's proof}.
It relies on the graphical intuition.
The second, a {\em mathematicians proof}, is technically correct
but does not give one a sense of {\em why} Theorem \ref{thm:interlace} 
should be true.

\begin{proof}[Pictorial Proof]

Let $a$ and $b$ be consecutive zeros of $y_1$.
It is easy to show $y_1'(a) \neq 0$ since otherwise the 
Wronskian would be
zero at $a$. Likewise $y_1'(b) \neq 0$. Thus, our situation 
is one of the two cases depicted below.

% @@@@@@@@@@@@@@@@@@@@@@@@@@@@ FIGURE @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
	\begin{center}  \psfrag{a}{$a$}\psfrag{b}{$b$}
        \includegraphics[width=3in]{interlace.eps}
	\end{center}
% @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

Notice that in both cases $y_1'(a)$ and $y_1'(b)$ 
have opposite signs.
We know $y_2(a)$ and $y_2(b)$ cannot be zero 
(otherwise the Wronskian 
would have a zero). Suppose $y_2(x)$ is never zero in $[a,b]$. 
Then $y_2(x)$ is either always positive or always negative on
$[a,b]$. In particular $y_2(a)$ and  $y_2(b)$ have the same sign.
Consider $W(y_1,y_2)(t)$ at $a$ and $b$.
\[
W(y_1,y_2)(a) = y_1(a)y_2'(a) - y_1'(a)y_2(a) = -y_1'(a)y_2(a)
\]
\[
W(y_1,y_2)(b) = y_1(b)y_2'(b) - y_1'(b)y_2(b) = -y_1'(b)y_2(b)
\]
Since $y_1'(a)$ and $y_1'(b)$ have opposite signs and $y_2(a)$ and $y_2(b)$ 
have the same signs, $W(y_1,y_2)(a)$ and $W(y_1,y_2)(b)$ must have opposite signs.
Since $W(y_1,y_2)$ is continuous (why?) it must have a zero somewhere in 
between $a$ and $b$. But by Theorem 5 the Wronskian is never zero!
This contradiction proves that $y_2$ must have a zero in $(a,b)$.
\end{proof}

\begin{proof}[Formal Proof] Let $a$ and $b$ be consecutive zeros of $y_1$.
We know the Wronskian is never zero on $[a,b]$. Therefore $y_2(a)$ and $y_2(b)$ are not zero.
Assume $y_2(t)$ is never zero for $t$ in $(a,b)$. Let $f(t) = y_1(t)/y_2(t)$. It is differentiable 
for all $t \in (a,b)$ and continuous on $[a,b]$. Now
\[
f'(t) = \frac{y'_1(t) y_2(t) - y_1(t)y'_2(t)}{y^2_2} = \frac{W(t)}{y_2^2(t)}.
\]
Therefore $f'(t)$ is never zero in $(a,b)$. But $f(a) = f(b) = 0$. By the Mean Value Theorem
there exists a $c\in(a,b)$ such that $f'(c)=0$. But this is a contradiction. Therefore the
assumption that $y_2(t)$ is never zero in $(a,b)$ is false 
\end{proof}

\subsection{Impossible pairs Theorems}



\begin{exam} Suppose $t$ and $t^2$ solve $y''+p(t)y'+q(t)y=0$.
Show that $p$ is discontinuous at zero.
$W(t,t^2) = t^2 = 0$ at $t=0$. {\em Solution.}
This is impossible if $p(t)$ is continuous
at $t=0$, by Theorem 5. 
%If fact we can ``back solve'' and find $p=-1/t$ and $q=1/t^2$.
\end{exam}

\begin{prob}
Suppose $y_1(t)=t$ and $y_2(t)=t^2$ solve $y''+p(t)y'+q(t)y=0$.
Substitute these for $y$ in $y''+p(t)y'+q(t)y=0$ to yield
two equations with $p$ and $q$ as unknowns. 
Solve for $p(t)$ and $q(t)$.
\end{prob}

\vspace{.2in}

In the theorems below assume $p$ and $q$ are continuous on an interval 
$I$ and that
$y_1$ and $y_2$ are linearly independent solutions to $y'' + py'+qy=0$.

\begin{thm}[\#38, section 3.2]
The functions $y_1$ and $y_2$ cannot both be zero at a point in $I$.
\end{thm}

\begin{proof}
Try to prove this! (We already used this in the proof of 
Theorem \ref{thm:interlace}.)
\end{proof}


\begin{thm}[\#39, section 3.2]
The functions $y_1$ and $y_2$ cannot have a max or min at the same point in $I$.
\end{thm}

\begin{proof}
Try to prove this!
\end{proof}


\begin{thm}[\#40, section 3.2]
The functions $y_1$ and $y_2$ cannot have  an inflection point at  
the same point $a\in I$, unless $p$ and $q$ are both zero at that
point.
\end{thm}

\begin{proof} 
Suppose $a$ is an inflection point of $y_1$ and $y_2$. Then
$y_1''(a) = y_2''(a) =0$. Thus $y_1'(a)p(a)+ y_1(a)q(a)=0$
and $y_2'(a)p(a) +y_2(a)q(a)=0$. We can rewrite this as,
\[
\mat{ y_1'(a) & y_1(a)\\
y_2'(a) & y_2(a)} 
\mat{p(a)\\q(a)} = \mat{0\\0}
\]
The determinant is equal to $W(y_1,y_2)(a)$ even though the matrix 
is a little different. Thus the determinant is not zero at $a$ since
$W(y_1,y_2)(t)$ is never zero on $I$. Thus, our matrix equation has 
the unique solution $p(a) = q(a) =0$.
\end{proof}


\end{document}
