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\centerline{Mathematics 304, Section 2 -- Ordinary Differential Equations}
\centerline{Solutions Midterm Exam 1 -- October 1, 2004}
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\noindent
{\it Directions:}
Do all work in the blue exam booklet.  You may use a calculator and
the table of integrals at any point.  There are 100 total regular
points and 10 Extra Credit points.
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\noindent
I.  Consider the first order ODE 
$$y' = 2t(y-2)^{2/3}\leqno(1)$$
\item{A)}  (15) Find a solution of this equation that satisfies
the initial condition $y(0) = 3$.
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\noindent
{\it Solution:}  This is a separable equation, so we solve as usual:
$$\eqalign{\int (y-2)^{-2/3}\ dy &= \int 2t\ dt\cr
3(y-2)^{1/3} &= t^2 + c\cr
y &= \left({t^2\over 3} + c\right)^3 + 2\cr}$$
From the initial condition,
$$3 = c^3 + 2 \Rightarrow c = 1.$$
So the solution is 
$$y = \left({t^2\over 3} + 1\right)^3 + 2$$
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\item{B)}  (10) What does the Existence and Uniqueness Theorem say
about solutions of the initial value problem for (1) with
$y(0) = 2$?  Show that the constant function $y(t) = 2$ for all $t$ is 
one solution.  Is it the only one?
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\noindent
{\it Solution:}  The right-hand side function in this equation is
$$f(t,y) = 2t(y-2)^{2/3} \Rightarrow {\partial f\over \partial y} = 
{4t\over 3(y-2)^{1/3}}$$
The partial derivative function is not continuous on any rectangle
in the $(t,y)$-plane containing points with $y = 2$.  Hence,
the hypotheses of the Existence and Uniqueness Theorem are
{\it not satisfied}.  However, the most we can say in this case
is that we are not guaranteed that there exists a unique solution
to the initial value problem with $y(0) = 2$.  (There could
be more than one, but that's not guaranteed either.)  In this case there 
is more than one solution.  First, $y = 2$ for all $t$ satisfies
the differential equation since
$$y' = 0 = 2t(2 - 2)^{2/3}$$
is true for all $t$.  Hence this constant function is a solution.
But then, we can also get a solution of this ODE with $y(0) = 2$
by using the solutions found in part A.  If $y(0) = 2$, we get $c = 0$ 
and a different solution 
$$y = \left({t^2\over 3}\right)^3 + 2 = {t^6\over 27} + 2.$$ 
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\noindent
II. (25) A tank with capacity 200 gal. originally contains 50 gal. water
with 30 lb of salt in solution.  Water containing .5 lb of 
salt per gallon is entering at 3 gal./minute and the 
tank is thoroughly mixed so the salt concentration is 
uniform at all times.   Contents of the tank are also flowing out of an 
exit pipe at 2 gal/minute.  Find the amount of salt in the tank at all 
times prior to the time the tank overflows.  What is the amount of
salt at the instant the tank starts to overflow?
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{\it Solution:}  From the statement of the problem, letting
$S = $ the total amount of salt in the tank (not salt per gallon,
the total), we have $50 + t$ gallons of liquid after $t$ minutes,
and
$$S' = 1.5 - {2 \over 50 + t}S$$
This is a first order linear equation with $g(t) = {-2\over 50+t}$
and $r(t) = 1.5$.  Solving as usual,
$$e^{\int g(t)\ dt} = e^{-2\int {dt\over 50+t}} = e^{-2\ln|50+t|} = 
(50+t)^{-2}$$
Then 
$$\eqalign{y(t) = c(50+t)^{-2} + {3\over 2}(50+t)^{-2}\int (50+t)^2 \ dt\cr
&=c(50+t)^{-2} + {1\over 2}(50+t)^{-2} (50+t)^3\cr
&=c(50+t)^{-2} + {1\over 2}(50+t)\cr}$$
The initial condition is $S(0) = 30$ (total salt in tank at $t = 0$).
Then 
$${c\over 2500} + 25 = 30 \Rightarrow c = 12500$$
Finally, since the capacity of the tank is $200$ gallons, the tank
starts to overflow when $50 + t =200$ or $t = 150$ minutes.
At that time, the total amount of salt is
$$S(150) = {12500\over 200^2} + {1\over 2}(200) = 100.31$$
pounds of salt.
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\noindent
III.  Let $T,M$ be positive constants with $T < M$, and consider the 
autonomous 1st order ODE
$$P' = -P\left(1 - {P\over T}\right)\left(1 - {P\over M}\right)\leqno(2)$$
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\item{A)} (15) What are the equilibrium solutions of this equation?
Which of the equilibria are sinks and 
which are sources?
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{\it Solution:}  Setting 
$$-P\left(1 - {P\over T}\right)\left(1 - {P\over M}\right) = 0$$, 
we see
$P = 0$, $P = T$, or $P = M$.  Since this is a cubic polynomial
in $P$ with a negative leading coefficient and three horizontal
axis intercepts, $P' > 0$ if $P < 0$, or $T < P < M$.
$P' < 0$ if $0 < P < T$ or $P > M$.  Hence $P = 0$ and $P = M$
are sinks, while $P = T$ is a source.
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\noindent
\item{B)} (10)  Suppose the equation (2) models growth of a population
of a species of organisms in a particular environment.
Based on your answer to part A, what happens to solutions of initial value 
problems for (1) if $P(0) < T$? if $T < P(0) < M$? if $P(0) > M$?  
What are the meanings of the $P$ values $P = T$
and $P = M$? 
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\noindent
{\it Solution:}  If $0 < P(0) < T$, then the solution decreases and tends to 
$P = 0$ as $t \to +\infty$; if $T < P(0) < M$, then the solution
increases and tends to $P = M$ as $t \to +\infty$;  if
$P(0) > M$, then the solution decreases and tends to $P = M$ as
$t \to +\infty$.  In biological terms, $M$ represents the 
maximum sustainable population, while $T$ represents the 
minimum sustainable population (or population threshold).  If
$P(0) < T$, then the population would eventually die out.
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\noindent
IV. (25) Sketch the bifurcation diagram for the family of ODE 
$$y' = y^3 - 3ay$$
containing the parameter $a$.  Describe all bifurcations that occur.
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{\it Partial Solution:}
This is a ``pitchfork bifurcation''.  There are equilibria
at $y = 0$ for all $a$, and at $y = \pm \sqrt{3a}$ for $a > 0$.
For $a \le 0$, $y = 0$ is a source.  For $a \ge 0$, the ``new''
equilibria $y = \pm \sqrt{3a}$ are sources, while $y = 0$
changes to a sink.
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