Solutions - Lecture 5

Transcription

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Solutions - Lecture 5
(Revised 15 February, 2009 @ 14:15)
Professor Stephen H Saperstone
Department of Mathematical Sciences
George Mason University
Fairfax, VA 22030
email: [email protected]
Copyright © 2009 by Stephen H Saperstone
All rights reserved
5.1.
(a) Differentiate
Divide by
implicitly with respect to
We get
to get
and multiply out so that
Finally solve for
which is the original ODE. [Note that the ODE can be inverted by the method of Lecture 4
to get a linear ODE with as the independent variable and as the dependent variable,
file://F:\OnLineODE\Solutions\Lec05\S05-S09.htm
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namely
(b) First we must calculate from the initial condition
the solution to obtain
which implies
Substitute these values into
Thus we have the particular
solution
we must have that
Because
for any value of
.
This is the implicit solution we seek. Use the GCalc-3 Implicit Function Plugin to plot
Note that GCalc-3 uses -variables so that you must change all
instances of to Finally, GCalc-3 requires that the implicit function be of the form
. (This is similar to our notation
.) Type x*y^2-1in the Function box
of the Implicit Function Plugin of GCalc-3 and (left) click on Graph! The resulting image
is a plot over the default window
Because there any vertical line
intersects the graph in two points, the plot below cannot be the graph of a function. The
initial condition is represented by the point with coordinates
a point on the upper
curve. Consequently, the graph of the solution we seek must be given by the upper branch.
(c) It also appears that the graph through
has a vertical asymptote at x=0 (
in the
notation of the exercise) and that the graph extends to the right beyond the default window
We zoom out to see how far the solution extends to the right. Modify
the Ranges under the "View" tab with the following parameters
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to get the following plot.
Without further mathematical analysis it appears that the solution is defined for all
Thus we take the maximum interval of definition to be
5.2.
(a) Note that is the independent variable and is the dependent variable. The ODE is
separable, so write:
Integrate to get
Then
Rearrange to get the implicit solution
(b)
implies
Thus
so the particular (implicit) solution is
(c) Use the GCalc-3 Implicit Function Plugin to plot the solution from part (b). Note that
the variables are already -variables. Finally, GCalc-3 requires that the implicit function
be of the form
. (This is similar to our notation of
.) Type y^5/5-
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2*y-x^3/3 in the Function box of the Implicit Function Plugin of GCalc-3 and (left)
click on Graph! The resulting image is a plot over the default window
Note that the initial condition is represented by the point with coordinates
It appears
that the solution through
extends to about
on the left and about
on the right.
(These values are substantiated by the image in part (d) next.)
(d) We need to zoom in on the plot in part (c) to examine more closely solutions through
and
Zoom in on the region of interest by modifying the Ranges under the
"View" tab. The following view parameters yield the plot below
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It appears that the solution through
has maximal interval of definition
solution through
has maximal interval of definition
.
5.3.
(a) The ODE is separable; rewrite as
; the
Integrate the equation with respect to
As
we have
Integrate
where we have lumped all constants of integration into
the implicit solution
(b)
implies
Combine and transpose to get
We get
so the particular
(implicit) solution is
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(c) Use the GCalc-3 Implicit Function Plugin to plot the solution from part (b). Note that
the -variables must be transformed to -variables. Finally, GCalc-3 requires that the
implicit function be of the form
.) Type
x+sin(x)-y^2/2+(1-y)*exp(y)+1/2-pi in the Function box of the Implicit
Function Plugin of GCalc-3 and (left) click on Graph! (Warning: Be sure to type in as a
lower case pi.) The resulting image is a plot over the default window
Note that the initial condition is represented by the point with coordinates
It appears
that the solution through
extends to about
on the left and indefinitely on the right.
We zoom in on this image to see how the solution behaves as (alias increases. Modify
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It appears that the solution through
extends to about
on the left and indefinitely on
the right. We zoom in on this image again to see how the solution behaves on a larger
scale. Modify the Ranges under the "View" tab as listed below
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It appears from this last plot that the interval of definition of this solution is bounded below
by approximately
There does not appear to be any limit on how large can be. We can
conjecture that the maximal interval of definition is
5.4.
Integrate to obtain
(b)
implies
solution is
We get
so the particular (implicit)
(c) (i) Use the GCalc-3 Implicit Function Plugin to plot the solution from part (b). Note that
.) Type
cos(x)+x*sin(x)+x-x^6-1 in the Function box of the Implicit Function Plugin of
GCalc-3 and (left) click on Graph! The resulting image is a plot over the default window
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We zoom in on this image to see how the solution behaves as (alias
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increases. Modify
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(c) (ii). Note that the initial condition is represented by the point with coordinates
It
appears that the solution through
extends to about
on the left and to about
on
the right. The last figure suggests that the maximal interval of definition is approximately
5.5..
Integrate to get
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(b)
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implies
Thus the implicit solution is given by
(c) (i) Use the GCalc-3 Implicit Function Plugin to plot the solution from part (b). Note that
.) Type
csc(y)-cot(y)-exp(x) in the Function box of the Implicit Function Plugin of
GCalc-3 and (left) click on Graph! The resulting image is a plot over the default window
(c) (i).
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We zoom in on this image to see how the solution behaves as (alias
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increases. Modify
(c) (ii). It appears from this last plot that the interval of definition of this solution is
Take note of the equilibrium solution
that seems to issue from the "S"
shaped solution through
Although
is an acutal solution, the presence of the
graph of
above is actually an artifact of the numerical procedure underlying the
implicit Function Plugin. In particular, the graph of
was not produced by design; it
was an acident of the plotting software.
5.6.
(a)
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(b)
(c)
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5.7.
Parts (a), (b), & (c) are all plotted on the following figure.
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Exercise 5.7
5.8.
Parts (a), (b), & (c) are all plotted on the following figure.
Exercise 5.8
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5.10. Parts (a), (b), & (c) are all plotted on the following figure.
5.12. First we rewrite the ODE in the form
where
or more simply
Now plot the slope field by hand.. Parts (a), (b), & (c) are all plotted on the following
figure.
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Exercise 5.12
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Exercise 5.12: Software Solution
5.15.
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Exercise 5.15
5.16.
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Exercise 5.16
5.17.
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Exercise 5.17
5.18.
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Exercise 5.18
5.19
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The figure below (from Rychlik's Slope Field applet) confirms the hand drawn integral
curve above.
Slope field for
and integral curve through
5.20
(a)
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Isoclines for
(b) If
for
represents the solution through
This means that
then becomes asymptotic to
(c) Sketch of proof. Let
represent the solution through
must have slope equal to zero at
since
Immediately the graph of
enters the (blue) region
as
Then the solution
depicted in the figure below.
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Throughout
the slope of lies between and that is
. Now the graph of
cannot cross the bottom of this region: the isocline
If the graph did cross this
isocline, it's slope at any crossing point would have to be zero. This is impossible, so the
graph of must remain above the graph of
or equivalently,
Thus
for all
By a similar argument the graph of cannot cross into the (green) region
at any point
This is because at
the isocline between
and
is
Thus the solution would have to cross into
with slope
But this is impossible as the
slope of this isocline is strictly less that at all points
We see this by first
representing the isocline
by
, so that
When
we have
so that
Consequently there is no way for the graph of to cross this isocline at any time
It is possible for the graph of to cross the isocline prior to
In this event the graph
of enters
where the slopes of all solutions must lie between and
This will force
the graph of to remain below the upper border of
the isocline
or
equivalently
We conclude that the graph of must remain trapped in the union of the regions
and
THis means that
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Next we must show that the separation of the two isoclines tends to zero as
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that is
In fact
This proves our conjecture stated in part (b).
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Solutions - Lecture 5

Transcription

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