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\centerline{Mathematics 304 -- Ordinary Differential Equations}
\centerline{Selected Solutions -- Problem Set 3}
\centerline{October 25, 2000}
\bigskip
\noindent
\S 4.5/ 54.
\item{a)}  The auxiliary equation of $y'' - s^2y = 0$ is
$0 = r^2 - s^2 = (r - s)(r + s)$, so by results proved in 
class, a fundamental set of solutions is $\{y_1,y_2\} = 
\{e^{sx}, e^{-sx}\}$, and the general solution looks like 
$$y = c_1 e^{sx} + c_2 e^{-sx}$$
where $c_1,c_2$ are arbitrary constants.
\item{b)}  Each function $c_1 e^{sx} + c_2 e^{-sx}$ can also be
written as
$$\eqalign{
c_1 e^{sx} + c_2 e^{-sx} 
&= c_1(\cosh(sx) + \sinh(sx)) + c_2(\cosh(sx) -\sinh(sx))\cr
&= (c_1 + c_2)\cosh(sx) + (c_1-c_2)\sinh(sx)\cr}$$
Conversely, every linear combination of $\cosh(sx), \sinh(sx)$
can also be rewritten as:
$$\eqalign{
d_1\cosh(sx) + d_2 \sinh(sx)
&= d_1 {e^{sx} + e^{-sx}\over 2} + d_2 {e^{sx} - e^{-sx}\over 2}\cr
&= \left({d_1 + d_2\over 2}\right) e^{sx} + \left({d_1 - d_2\over 2}\right)e^{-sx}\cr
}$$
Therefore $d_1\cosh(sx) + d_2 \sinh(sx)$ is also a general solution
of $y'' - s^2y = 0$, since this form produces exactly the same collection
of functions as the form $c_1 e^{sx} + c_2 e^{-sx}$ when $c_i$ and $d_i$ 
take on all possible real values.
\item{c)}  To solve inital value problem for $y'' - s^2y = 0$ with
$y(0) = a$, $y'(0) = b$, we can use either form.  The second form of the 
general solution, $d_1\cosh(sx) + d_2 \sinh(sx)$, is somewhat simpler.
Note that $\cosh(0) = 1$ and $\sinh(0) = 0$ by definition.
Hence from $a = y(0) = d_1 \cosh(0) + d_2\sinh(0) = d_1$, we get
$d_1 = a$ immediately.  Then $y'(0) = s d_1 \sinh(0) + s d_2\cosh(0) = sd_2$
so the other initial condition gives $b = y'(0) = sd_2$, so 
$d_2 = b/s$:
$$y(x) = a\cosh(sx) + {b\over s}\sinh(sx)$$
\bigskip
\noindent
\S 4.2/37.  a)  We assume that $L(y) = y'' + py' + qy = 0$ has $p,q$ continuous
on (a,b), and that $\phi$ is a solution of this ODE that {\it is not 
identically zero}, but has a non-isolated zero $x_0$.  In other words,
every deleted neighborhood of $x_0$ contains another zero of $\phi$:
For all $\delta > 0$, there exists some $x\ne x_0$ with 
$|x - x_0| < \delta$ such that $\phi(x) = 0$.  In particular 
for $\delta = 1/n$, we get an $x_n$ with $|x_n - x_0| < 1/n$ such
that $\phi(x_n) = 0$.  By the construction, the $x_n$ form 
a sequence of real numbers converging to $x_0$.  Now, in the definition
of the derivative $\phi'(x_0)$:
$$\phi'(x_0) = \lim_{x \to x_0} {\phi(x) - \phi(x_0)\over x - x_0},$$
we can let $x \to x_0$ through the terms in the $x_n$ sequence.  Since
we require that $\phi'(x_0)$ exists, we should get the same limit
no matter how $x \to x_0$.  But $\phi(x_n) = 0$ for all $n$, so 
$$\phi'(x_0) = \lim_{n \to \infty} {\phi(x_n) - \phi(x_0)\over x_n - x_0} = 
\lim_{n\to\infty} {0\over x_n - x_0} = 0.$$
Therefore $L(\phi) = 0$, $\phi(x_0) = 0$, and $\phi'(x_0) = 0$.

b)  The {\it zero function} also satisfies the IVP
$L(y) = 0$, $y(x_0) = 0$, and $y'(x_0) = 0$.  Therefore, since we 
assume $p,q$ are continuous on an interval containing $x_0$, 
the Existence and Uniqueness Theorem (Theorem 2, page 165 of text) applies.
Hence $\phi$ from part a) must be the zero function.  But this is 
a contradiction, since we assumed that $\phi$ was {\it not identically
zero}.  Therefore, no solution of $L(y) = 0$ can have a non-isolated
zero.
\bigskip
\noindent
\S 4.3/4.  We are given $y_1 = x^2\cos(\ln(x))$, $y_2 = x^2\sin(\ln(x))$.
To decide whether these are linearly independent or not, we can 
consider a linear combination $c_1 y_1(x) + c_2 y_2(x) \equiv 0$ (i.e.
the linear combination adds up to the zero function).  Then we get
another true equation by substituting any $x \in (0,1)$ in to this equation.
For instance with $x = e^{-\pi}$, we get 
$$\eqalign{0&=c_1 y_1(e^{-\pi}) + c_2 y_2(e^{-\pi})\cr 
&=c_1 e^{-2\pi}\cos(-\pi) + c_2 e^{-2\pi}\sin(-\pi)\cr
&= - c_1 e^{-2\pi}\cr}$$
(since $\sin(-\pi) = 0$).  Hence $c_1 = 0$.  Then substituting
$x = e^{-\pi/2}$, we get $c_2 = 0$ as well.  $\{y_1,y_2\}$ is 
{\it linearly independent}.  The Wronskian works out to 
$$W[y_1,y_2](x) = x^3$$
(using $\cos^2(\ln(x)) + \sin^2(\ln(x)) = 1$).
\bigskip
\noindent
23.  If $y = u e^{-{1\over 2}\int p(x)\ dx}$, then by the Chain and 
Product Rules for derivatives
$$\eqalign{
{dy\over dx} &= u e^{-{1\over 2}\int p(x)\ dx} \left({-1\over 2} p(x)\right) + 
                {du\over dx} e^{-{1\over 2}\int p(x)\ dx}\cr
&= e^{-{1\over 2}\int p(x)\ dx} \left({du\over dx} + {-1\over 2} p(x) \right)\cr}$$
Continuing to compute the second derivative, again by Chain and 
Product Rules:
$${d^2y\over dx^2} = e^{-{1\over 2}\int p(x)\ dx}\left(
{d^2u\over dx^2} - p{du\over dx} -{1\over 2} {dp\over dx} u +{1\over 4}p^2u\right)$$
We substitute these into 
$$0 = L(y) = {d^2y\over dx^2} + p(x) {dy\over dx} + q(x) y$$
and factor out the $e^{-{1\over 2}\int p(x)\ dx}$ from all terms.
{\it The $- p{du\over dx}$ from ${d^2y\over dx^2}$ and the $p{du\over dx}$
from $p{dy\over dx}$ cancel\/} leaving
$$0 = e^{-{1\over 2}\int p(x)\ dx}\left(
{d^2u\over dx^2} + \left({-1\over 2}p' - {1\over 4}p^2 + q\right)u\right)$$
Multiplying by $e^{{1\over 2}\int p(x)\ dx}$, we get a second order
ODE of the desired form for the function $u$:
$$0 = {d^2u\over dx^2} + \left({-1\over 2}p' - {1\over 4}p^2 + q\right)u$$
where $f(x) = {-1\over 2}p'(x) - {1\over 4}p(x)^2 + q(x)$.
\bigskip
\noindent
31.  We have $L(y) = x^2 y'' - 3x y' + 3y = 0$ (a Cauchy-Euler equation).
\item{a)}  $y_1 = x$ and $y_2 = x^3$ are solutions on $(-\infty,\infty)$
since
$$\eqalign{
L(x) &= x^2\cdot 0 - 3x\cdot 1 + 3x = 0\cr
L(x^3) &= x^2\cdot 6x - 3x\cdot 3x^2 + 3x^3 = 0\cr}$$
The Wronskian of $y_1$ and $y_2$ is 
$$W[y_1,y_2](x) = \det\pmatrix{x & x^3\cr   
                               1 & 3x^2\cr} = 2x^3$$
This is not zero (e.g.) at $x = 1$.
\item{b)}  $\{y_1,y_2\}$ is linearly independent, since if 
$c_1 y_1 + c_2 y_2 = c_1 x + c_2 x^3 \equiv 0$,  then we
would get true equations by substituting any value for $x$.  E.g., with
$x = 1$ and $x = 2$, we get the equations
$$\eqalign{0 &= c_1 + c_2\cr
           0 &= 2c_1 + 8c_2\cr}$$
The only solution is $c_1 = c_2 = 0$, so these are linearly independent.
\item{c)}  $y_3 = |x|^3$ is also a solution of $L(y) = 0$ on $(-\infty, \infty)$
since 
$$y_3(x) = \cases{x^3 & if $x \ge 0$\cr
                  -x^3 & if $x < 0$\cr}; \qquad
                  y_3'(x) = \cases{3x^2 & if $x \ge 0$\cr
                  -3x^2 & if $x < 0$\cr}; \qquad
                  y_3''(x) = \cases{6x & if $x \ge 0$\cr
                  -6x & if $x < 0$\cr}$$
Therefore 
$$L(y_3) = \cases{x^2\cdot 6x - 3x\cdot 3x^2 + 3x^3 = 0 & if $x \ge 0$\cr
x^2\cdot (-6x) - 3x\cdot (-3x^2) - 3x^3 = 0& if $x < 0$\cr}$$
so $y_3$ is also a solution defined on $(-\infty,\infty)$.
\item{d)}  Is $y_3 = c_1 y_1 + c_2 y_2$ for some $c_1,c_2$?  The 
answer is no:  Let $x = 1$; we get
$1 = c_1 + c_2$.  But with $x = -1$ we get $1 = -c_1 - c_2$.
These equations are inconsistent.  So there are no $c_1,c_2 \in {\bf R}$
such that $y_3 = c_1 y_1 + c_2 y_2$ for all $x \in {\bf R}$.
(Note that there are such constants, though, if we only take $x > 0$
or only $x < 0$.)
\item{e)}  We are {\it not} seeing a contradiction to the uniqueness
of solutions here since the theorem on existence and uniqueness
of solutions {\it does not apply} on the interval $(-\infty, \infty)$.
To write our equation in the standard form $y'' + py' + qy = 0$, we
would have to divide through by $x^2$:
$$y'' + \left({-3\over x}\right)y' + \left({3\over x^2}\right)y = 0$$
But then the $p = {-3\over x}$ and $q = {3\over x^2}$ functions
have discontinuities at $x = 0$.  The fact that there is not a unique
solution with $y(1) = 1, y'(1) = 3$ is not a contradiction!
\bye
