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\centerline{Mathematics 304, section 2 -- Ordinary Differential Equations}
\centerline{Problem Set 2}
\centerline{{\it Due:} Friday, September 17}
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\noindent
A.  From Section 1.8 in the handouts, do problem 27 -- {\bf Hint}:
measure the amount of dioxin in units of ``parts per billion times
gallons.''
\bigskip
\noindent
B.  From Chapter 1 in the main text, do problems 1, 2abe, 3, 4, 10ab.
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\noindent
C.  (Differential equations with discontinuous coefficients).  ``Reasonable''
solutions of first order linear differential equations 
$y' = g(t) y + r(t)$ can still be 
obtained in situations where one or both of $g(t), r(t)$ have 
jump discontinuities.  In this problem, you will see how this works
in the case 
$$y' = \cases{y - 4 & if $t < 5$\cr
              2 - y & if $t \ge 5$\cr}\leqno(1)$$
(that is, 
$$g(t) = \cases{1 & if $t < 5$\cr
                -1 & if $t \ge 5$\cr} \qquad r(t) = \cases{-4 & if $t < 5$\cr
                2 & if $t \ge 5$)\cr}$$
and various initial conditions $y(0) = y_0$.
Here's the process:  First, we find the general solution for $t < 5$ using
the appropriate method.  Determine the constant to satisfy the initial 
condition $y(0) = y_0$.  Then find the general solution for $t \ge 5$ 
and choose the arbitrary constant there so that the resulting function
is {\it continuous} at $t = 5$.  The function obtained this way satisfies the 
equation (1), except at the point $t = 5$ (where the derivative is 
usually undefined).  
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\item{1)}  Carry out this process for the initial condition $y(0) = 4$.
Describe the qualitative behavior of this solution.
\item{2)}  Same question for the initial condition $y(0) = 2$.
\item{3)}  Give a general description of the qualitative behavior 
of solutions of (1), identifying all cases.
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\noindent
D.  (More on solutions of the logistic equation)  Recall that we have
seen that solutions of $y' = ay(1-y)$, $a > 0$ can be written in the form
$$y = {Ke^{at} \over 1 + Ke^{at}}$$
where $K$ is an arbitrary constant, related to the value of $y$ at 
time $t = 0$ by $K = {y(0)\over 1 - y(0)}$.
\medskip
\item{1)}  Show that if $y(0) < 0$, then the solution becomes unbounded
(approaches a vertical asymptote) in a finite amount of time.
\item{2)}  Show that the solutions with $y(0) > 1$ also
have vertical asymptotes if $t$ is extended to negative values.
\bye