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\centerline{Mathematics 304, section 2 -- Ordinary Differential Equations}
\centerline{Review Sheet -- Exam 1, Selected Answers}
\centerline{September 30, 2004}
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\noindent
C) Sketch the phase line for each equation, identify equilibria
as sinks, sources, or neither.  Then for the given initial conditions,
sketch qualitative graphs of the corresponding solutions:
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\item{} $y' = y \cos y$, $y(0) = \pi, 0, 5\pi/3$
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{\it Answers:}  Equilbrium points at $y = 0$ and 
all solutions of $\cos(y) = 0$, that is all odd integer
multiples of $\pi/2$:
$$y = 0, {(2k+1)\pi/2}, {\rm all} k \in {\bf Z}$$
By considering the sign of $\cos(y)$, we can see that the 
sources and sinks {\it alternate}:
$$\ldots, -3\pi/2, 0, 3\pi/2, \ldots$$
 are sources, and 
$$\ldots,-5\pi/2, -\pi/2, \pi/2, 5\pi/2, \ldots $$
are sinks.
(These can also be seen by using our derivative 
criterion: For instance, if $f(y) = y \cos(y)$, then
$f'(y) = \cos(y) - y\sin(y)$.  So for instance
$f'(\pi/2) = \cos(\pi/2) - \pi/2 \sin(\pi/2) = -\pi/2 < 0$.
Hence $y = \pi/2$ is a sink.)

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The solution with $y(0) = \pi$ decreases toward the sink at $\pi/2$,
the solution with $y(0) = 0$ is the equilibrium solution $y(t) = 0$
all $t$.  The solution with $y(0) = 5\pi/3$ increases toward
the sink at $5\pi/2$ since the initial point lies above
the source at $y = 3\pi/2$.  

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\noindent
D)  Sketch the phase line for the ODE 
$$y' = {1\over (y-2)(y+1)}$$  
and discuss the behavior of the solution with $y(0) = 1/2$.
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{\it Answer:}  The right side has vertical asymptotes at $y = 2, -1$.
Between the two asymptotes $1/((y-2)(y+1)) < 0$, so the 
solution with $y(0) = 1/2$ is decreasing.
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\noindent
E)  Sketch the bifurcation diagram for the family of equations
$$y' = y(1-y)^2 + a$$
and describe the bifurcations that occur as $a$ increases
from $-\infty$ to $+\infty$.
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\noindent
With $a = 0$, the graph $z = y(1-y)^2$ crosses the $y$-axis at
$y = 0$ and it is tangent to the $y$-axis at the double root
$y = 1$ (a local min).  It is a standard cubic polynomial graph
with a local max and local min.  The coefficient
of $y^3$ is positive, so it starts negative for $y << 0$
and ends positive for $y >> 0$.
The effect of the $a$ if $a \ne 0$ is to shift this graph
up or down.  If $a > 0$, then the double root at $y = 1$
disappears and there is just one negative root.  If
$a < 0$, then the graph shifts down.  For a range of negative
values $a_0 < a < 0$, there are three equilibrium points.
We get a negative double root eventually at some $a_0 < 0$.  
Then if $a$ is very negative ($a < a_0$), the graph shifts down so far that
the local max is below the $y$-axis.  From this description
you should be able to generate the bifurcation diagram.
\bye