SOLUTIONS TO SAMPLE
PROBLEMS EXAM 1
If it is not already on your hard drive, you will need to download
the free DPGraph Viewer to view some of
the pictures linked to on this page. |
QuickTime
6 free download. |
Maple
Worksheet for Examples 1 - 9
Maple
Worksheet--Slope Fields
PowerPoint
Presentation of Solutions to Separable First Order Differential Equations
Example 1: Solving a
Separable Differential Equation
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The graph of the solution is shown above. Click on the graph to see an animation of the direction field
vectors moving across the screen for increasing values of x along with an
animated solution point. |
Example 2: Solving a Separable DE--Circles
yo = 1 yo
= 2 yo = 3
yo = 4 |
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Example 3: Solving a Separable
DE--Logistics
Equation
Above is the graph of the
solution to the logistics equation.
QT animation
as y(0) varies from 0 to 40.
dy/dx = .01y(100-y), y(0) = 10.
The solution is y = 100ex / (9 + ex).
Here is an
Animation (Quicktime
version) of the changing graph of the solution as y(0) varies from
-50 to 150 with x between 0 and 10 and here is an Animation
(Quicktime version) with x between -10 and 10. Notice the significance of
the blue horizontal lines and their relationship to the zeroes of
.01y(100-y). Look at examples 3 and 4 in Section 2.1 discussing
autonomous first order differential equations (DE's of the form F(y,y') =
0 or in normal form dy/dx = f(y)). If the DE was modeling a
population then y(0) would have to be positive. If y(0) = yo
then
y = 100 is a singular solution (see pages 7-8 in
your text). |
Solution
Details
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Example
4: Section 2.2#36 (It is separable)
Note: y = 0.9 is a singular solution that
could not be obtained from the general solution shown above. Click
here to see animated solution graphs as yo varies from -2.1
to 3.9 demonstrating the "missing" singular solution. Quicktime
version
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Solution graphs for yo = 1
Click
here to see animated solution graphs as yo varies from -3
to 3. Quicktime version |
Example
5: Another variables separable example
The
picture on the right shows the graphs of particular solutions with y0
= 1/4 (blue) and y0
= -1/4 (red).
Click here or on the picture to
see an animation of solutions as y0 varies from -3 to 3.
In the case of the blue solution on the right, does y continue to increase
as x continues to increase beyond 8? Click
here for a pictorial answer.
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Quicktime
Version
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PowerPoint
Presentation of Solutions to Non-Separable First Order Differential Equations
Example
6: Homogeneous Example
The pictures below show solutions to the
equation at the right for values of y1 (-3
red, -1 green,
1 blue, 2
orange, 3 purple).
Click on the picture to see an animation of the solutions as y1
goes from -3 to 3. Quicktime
Version
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Solution
Graph with Slope Field for y(1) = 1
The purple and orange solution graphs would also
not be meaningful to the right of their vertical asymptotes.
Investigate solutions to this problem using the First
Order DE Solution Grapher.
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Example
7: Exact Example
The picture at the right shows a particular
solution corresponding to y(0) = yo = 1. Click
here or click on the picture at the right to see an animation with yo
varying from -1/2 to -5 (blue) and yo varying from -1/2 to 4 (red).
Click here to see a similar
animation that includes negative values for x.
Here is a
DPGraph picture of z = -ye-x
+ cos(x) - y2
and z =
C. You can use the scrollbar to vary C. The intersection of
the two surfaces would correspond to the solution for each value of
C. You can also use the z-slice feature to look at various solutions
corresponding to z = C.
Here is a
DPGraph picture of z
= -ye-x + cos(x) - y2 - 1 + a + a2
where a = yo. C
= 1 - a - a2 The
view is from the top with z between -0.0001 and +0.0001 so you have the
effect of looking at the graph of -ye-x
+ cos(x) - y2 - 1 + a + a2 = 0 You
can use the scrollbar to vary a (yo) from -4 to 5. The
default value for a is 1. |
Quicktime
version of the animation
Try using the First
Order DE Solution Grapher to solve the DE in standard form and observe what
happens when you reach a value for x where the analytical solution is
undefined.
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Integrating Factor Formulas
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Example
8: Not Exact But You Can Use An Integrating Factor
The
three pictures on the right show particular solutions to the equation above
with y(0) = yo = -3, 1, and 3. EC: What is not clearly
shown in the pictures where yo = -3 and 3? (Hint:
Your graphing calculator does the same thing.) Click
here to see an animation of particular solutions with yo
varying from -5 to 5. Quicktime
version
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Example
9: Linear Equation Example
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Below is the graph of the particular solution to
the example above with y1 = 1. Click
here or on the graph below to see an animation of particular solutions with
y1 varying from -3 to 3. The graph on the left shows a
particular solution with y1 = 2 along with direction
vectors.
QT |
Maple
Worksheet for Examples 1 - 9
Example
10: Section 2.3 #31 Another Linear Equation
Example
Maple
Worksheet for Examples 11 - 14 and 16
Examples 11
and 12: Two equations that are not linear but can be turned into linear equations by making an appropriate substitution.
Click here
to see animated solutions as yo varies from -3 to 3. Quicktime
version Notice that y = 0 is a singular solution.
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If y(1) = y1 then C = ln(y1)
- 2
Click here
to see animated solutions as y1 varies from 1/100 to 501/100.
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Example
13: Here is another example of an equation that is not
separable but can be made separable using an appropriate substitution.
Example
14: Another Homogeneous Equation (2.5 #14)
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This one is not so easy to graph even
by point plotting but you could approximate the graph of the solution by looking at
and using the First
Order DE Solution Grapher. We can also look
at this DPGraph Picture of the
solution. The view is from the top with z between -0.0001 and
+0.0001 so you have the effect of looking at the graph of or
we can look at this DPGraph
Picture of the graph of and
look at the z-slice corresponding to z = 0. |
Example
15: Exponential
Population Growth Example
Exponential Population Growth
The population of a country is growing at a rate
that is proportional to the population of the country. The
population in 1990 was 20 million and in 2000 the population was 22
million. Estimate the population in 2020. |
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Solution
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Example
16: Solution
to a mixture problem: We start with a tank containing 50
gallons of salt water with the salt concentration being 2 lb/gal. Salt
water with a salt concentration of 3 lb/gal is then poured into the top of the
tank at the rate of 3 gal/min and salt water is at the same time drained from
the bottom of the tank at the rate of 3 gal/min. We will consider the
water and salt mixture in the tank to be well-stirred and at all times to have a
uniform concentration of salt. Find the function S(t) that gives the
amount of salt in the tank as a function of time (t) since we began pouring in
salt water at the top and simultaneously draining salt water from the bottom of
the tank. How long before there will be 120 pounds of salt in the tank?
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Graph
of the solution function
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Maple
Worksheet for Examples 11 - 14 and 16
Example
17: Solution to
a Newton's Law of Cooling problem:
A pot of
liquid is put on the stove to boil. The temperature of the liquid reaches
170oF and then the pot is taken off the burner and placed on a
counter in the kitchen. The temperature of the air in the kitchen is 76oF.
After two minutes the temperature of the liquid in the pot is 123oF.
How long before the temperature of the liquid in the pot will be 84oF?
Example
18: Differential
Equations Sailing Application Example (Equation similar to that
governing Newton's Law of Cooling)
Example
19: An
Application Involving a Snowplow
On
a Tuesday morning in February before noon in rural Minnesota it started to
snow. There had been no snow on the ground before it started to
snow. Literally immediately it was snowing at a steady, constant rate so
that the thickness of the snow on the ground was increasing at a constant
rate. A snowplow began clearing the snow from the streets at noon.
The speed of the snowplow in clearing the snow is inversely proportional to the
thickness of the snow. The snowplow traveled two miles during the first
hour after noon and traveled one mile during the second hour after noon.
At what time did it begin snowing?
a
= The length of time before noon when it started snowing. t
= The length of time the snowplow had been traveling (and plowing). v
= The speed of the snowplow. s
= The distance traveled by the snowplow. T
= The thickness of the snow. |
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Click on the picture to see a snowplow
in action (with sound). Click
here for more action (takes longer to load). |
s(0)
= 0, s(1) = 2, s(2) = 3 v
= k/(t + a) since the speed of the snowplow in clearing the
snow is inversely proportional to the thickness of the snow. Thus Here
is an animation showing the snowplow
plowing (and slowing down). Quicktime
version Solution
Euler's Method
Check out 10 steps of Euler's
Method applied to
y' = 1 - t + 4y y(0) = 1
EC: Find the analytical solution to the
equation above. The graph of the
analytical solution is given below.
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Example
Here is an example of the first step
(finding y1) in applying Euler's Method.
xo = 1, yo = e1/2,
h = 0.1
y1 = yo + (xoyo)(0.1)
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The
analytical solution is shown in
red above and the line
segment in
green above is
representing applying Euler's Method once to find y1.
Below is a graph of part of the
analytical solution of the differential equation above along with the tangent to the graph at the point where
x = 1.
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A Nice Euler's Method Example For The TI (The
Analytical Solution Is Shown Below)
The arrow above means press the STORE key.
The second ENTER
generates y1. Pressing ENTER again will produce y2.
Do you know why? Press the ENTER key repeatedly to generate
additional Euler solution values (y3, y4, . . .).
With yo = 10 you should get approximately y1 = 19, y2
= 34.39, y3 = 56.95, y4 = 81.47, y5 = 96.57, y6
= 99.88. The points along with the analytical solution are plotted in the
graph on the right below.
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To the left is the graph of the
solution to the logistics equation
dy/dx = .01y(100-y), y(0) = 10.
The solution is y = 100ex / (ex
+ 9). |
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Euler's Method Extra Credit
Generate the data points for the example above
with x going from 0 to 250 and a step size of 2.5. The picture at
the right shows the analytical solution in red and the data points in blue
for x going from 0 to 40 with a step size of 2.5. Ideally you will
use a computer for this and generate a graph of the data points using your
computer and/or graphing calculator. What appears to be happening? |
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Another Euler's
Method Example Compared to an Analytical Solution |
Improved Euler's Method Compared To Euler's
Method--One Step
Here is an Excel
graph comparing the Euler solution, Improved Euler solution and the
analytical solution to
y'(x) = cos(x),
y(0) = 0 over the interval from 0 to two pi with a step size of
pi/8.
Formulas for Applying the Improved Euler's Method and
a Classical Fourth Order Runge-Kutta Method
return
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