-
Verify that
x
=
e
−
t
+
1
is a solution of
x
′
+
x
=
1.
-
Verify that
x
=
a
t
2
+
b
t
+
c
is a solution of
x
′
′
′
=
0.
[Note that
a
,
b
and
c
are arbitrary constants.]
-
Verify that
x
=
c
e
−
r
t
is a solution to Example 1.5.
-
Verify that
x
=
t
is a solution of
x
′
′
−
t
x
′
+
x
=
0.
-
Verify that
x
=
K
1
+
c
e
−
r
t
is a solution to Example 1.6.
-
Verify that
q
=
c
1
e
−
t
cos
t
+
c
2
e
−
t
sin
t
solves
q
′
′
+
2
q
′
+
2
q
=
0
.
[Note that
c
1
and
c
2
are arbitrary constants.]
-
Use the law of reflection (angle of incidence = angle of reflection) and
geometrical analysis to derive Eqn. (1.11) in Example 1.10.
-
Solve Eqn. (1.13) when
w
(
x
)
has the constant value
w
0
.
[Hint: Recognize in this instance that
w
0
is the 4-th derivative of the unknown function
y
(
x
)
.
]
-
Verify that
y
=
(
1
+
t
2
)
−
1
is a solution of
(
1
+
t
2
)
y
′
′
+
4
t
y
′
+
2
y
=
0.
-
Verify that
x
=
c
1
cos
2
t
+
c
2
sin
2
t
is a solution of
x
′
′
+
4
x
=
0.
[Note that
c
1
and
c
2
are arbitrary constants.]
-
Verify that
x
=
[
t
ln
(
t
)
+
c
t
]
−
1
is a solution of
t
2
x
′
(
t
)
+
t
x
(
t
)
+
t
2
x
2
=
0.
-
Verify that
t
x
+
c
=
ln
|
x
|
is an implicit solution of
x
′
(
t
)
=
x
2
/
(
1
−
t
x
)
.
[Note that
c
is an arbitrary constant.]
-
Verify that
x
−
1
+
c
e
−
x
=
t
is an implicit solution of
x
′
=
1
/
(
x
−
t
)
.
[Note that
c
is an arbitrary constant.]
-
Verify that
x
2
−
y
2
+
2
x
y
=
c
is an implicit solution to
ⅆ
y
ⅆ
x
=
y
+
x
y
−
x
where
c
is an arbitrary constant.
-
Verify that
x
2
=
c
1
t
+
c
2
is an implicit solution of
x
x
′
′
+
(
x
′
)
2
=
0.
[Note that
c
1
and
c
2
are arbitrary constants.]
-
Verify that
x
4
−
8
x
2
+
16
y
+
y
4
+
20
=
0
is an implicit solution of
(
4
+
y
3
)
ⅆ
y
ⅆ
x
=
4
x
−
x
3
.
-
Determine all values of the parameter
ω
so that
x
=
2
cos
ω
t
is a solution of
x
′
′
+
8
x
=
0.
-
Determine all values of the parameters
A
and
ω
so that
x
=
A
sin
ω
t
is a solution of
x
′
′
+
9
x
=
0
-
Determine values of the parameters
A
and
B
so that
x
=
e
t
+
A
t
+
B
is a solution of
x
′
′
−
x
=
t
+
2.
-
Determine values of the parameter
r
so that
x
=
t
r
is a solution of
t
2
x
′
′
+
6
t
x
′
+
4
x
=
0
-
What is the only solution of
x
′
2
+
x
2
=
0
?
Why are there no others?
-
Consider the ODE
x
′
+
e
−
t
=
1.
-
Verify that
x
=
t
+
e
−
t
+
c
is a solution.
-
Determine the solution for the initial condition
x
(
0
)
=
−
1.
-
Consider the ODE
x
′
+
ln
t
−
2
t
=
0.
-
Verify that
x
=
t
2
+
t
−
t
ln
t
+
c
is a solution.
-
Determine a solution for the initial condition
x
(
1
)
=
2.
-
Consider the ODE
x
′
+
4
t
3
x
2
=
0.
-
Verify that
x
=
1
/
(
t
4
+
c
)
is a solution.
-
Determine a particular solution for the initial condition
x
(
0
)
=
1.
-
Consider the ODE
t
2
x
′
+
x
=
0.
-
Verify that
x
=
c
e
1
/
t
is a solution.
-
Determine a solution for the initial condition
x
(
−
1
)
=
e
−
1
.
-
Consider the ODE
x
′
+
x
2
cos
t
=
0.
-
Verify that
x
=
1
/
(
sin
t
+
c
)
is a solution.
-
Determine a solution for the initial condition
x
(
0
)
=
1
/
2.
-
Determine a solution for the initial condition
x
(
0
)
=
1.
-
Consider the ODE
t
2
x
′
−
t
x
−
x
2
=
0
.
-
Verify that
x
=
−
t
/
ln
(
c
t
)
is a solution.
-
Determine a solution for the initial condition
x
(
1
)
=
1.
-
Determine a solution for the initial condition
x
(
1
)
=
−
2.
-
x
′
−
e
t
=
1
,
x
(
0
)
=
0.
[See Exercise 1.22.]
-
x
′
+
ln
t
−
2
t
=
0
,
x
(
1
)
=
2.
[See Exercise 1.23.]
-
x
′
+
4
t
3
x
2
=
0
,
x
(
0
)
=
1.
[See Exercise 1.24.]
-
t
2
x
′
+
x
=
0
,
x
(
−
1
)
=
e
−
1
.
[See Exercise 1.25.]
-
x
′
+
x
2
cos
t
=
0
,
x
(
0
)
=
1
/
2.
[See Exercise 1.26.]
-
x
′
+
x
2
cos
t
=
0
,
x
(
0
)
=
1.
[See Exercise 1.26.]
-
t
2
x
′
−
t
x
−
x
2
=
0
,
x
(
1
)
=
−
2.
[See Exercise 1.27. Be careful!]
-
The ODE
(
t
x
−
1
)
x
′
=
2
t
x
−
x
2
has the general (implicit solution)
t
x
−
ln
|
x
|
−
t
2
=
c
.
Some solutions of this equation are plotted below for various values of
c
.
(Note that you cannot solve the implicit solution for
x
in terms of
t
.
)
To answer each of the following questions you will need to print a copy of the
graph below.
-
Mark the solution to the ODE that satisfies the initial data
x
(
0
)
=
1
and "eyeball" the maximum interval of definition.
BE CAREFUL!
-
Mark the solution to the ODE that satisfies the initial data
x
(
−
1
2
)
=
−
1
and "eyeball" the maximum interval of definition.
BE
CAREFUL!
-
The ODE
(
2
sin
x
−
t
)
x
′
=
e
t
+
x
has the general (implicit solution)
e
t
+
t
x
+
2
cos
x
=
c
.
The surface defined by
z
=
e
t
+
t
x
+
2
cos
x
is depicted below. Some of the surface contours are projected onto the
t
x
-plane.
Portions of the branches of these level curves represent particular solutions
to the
ODE.
To
answer each of the following questions you will need to print a copy of the
graph
below.
-
Mark the solution to the ODE that satisfies the initial data
x
(
0
)
=
π
/
2
and "eyeball" the maximum interval of definition.
BE CAREFUL!
-
Mark the solution to the ODE that satisfies the initial data
x
(
0
)
=
−
π
/
2
and "eyeball" the maximum interval of definition.
BE CAREFUL!
Determine all
equilibrium solutions (if any) to the following ODEs.
-
x
′
=
x
(
4
−
x
)
-
t
2
x
′
+
x
=
0
-
x
′
+
x
2
cos
t
=
0
-
x
′
−
e
t
=
1
-
x
′
+
2
t
x
2
=
x
2
-
x
′
=
t
x
−
1
+
x
−
t
-
x
′
=
t
sin
x
Determine
by inspection any solutions to the following ODEs. Be sure to verify that your
answer is a solution.
-
x
′
=
−
(
x
−
t
)
2
+
5
[Hint: Examine how the singular solution to the ODE of
Example
2.8 was arrived at.]
-
x
′
=
x
2
+
1
−
t
2
-
t
2
x
′
=
x
−
t
−
1
−
1
-
y
′
=
e
−
x
y
2
+
y
−
e
x
[Hint: Factor the right side of the ODE appropriately.]
-
t
2
x
′
+
t
x
+
t
2
x
2
=
1
[Hint: Factor the ODE appropriately.]
-
Solve the ODE
x
′
+
ln
t
−
2
t
=
0.
-
Solve the ODE
t
2
x
′
+
x
=
0.
-
Solve the ODE
t
x
′
+
x
=
0.
-
Solve the ODE
x
′
+
4
t
3
x
2
=
0.
-
Solve the ODE
x
′
+
2
t
x
2
=
x
2
.
-
Solve the ODE
x
′
+
x
2
cos
t
=
0.
-
Solve the ODE
x
(
y
+
2
)
y
′
=
y
2
(
x
2
−
1
)
-
Solve the ODE
x
′
=
t
x
−
1
+
x
−
t
.
[Hint: Factor the right side of the ODE appropriately.]
-
Solve the ODE
(
1
−
x
)
x
′
=
e
t
−
x
.
-
Solve the ODE
θ
′
+
4
sin
θ
=
0.
-
Solve the ODE
x
′
=
x
2
−
x
−
2.
-
Solve the ODE
x
2
y
′
+
y
2
+
1
=
0
..
-
Consider the ODE
x
′
=
e
x
.
-
Calculate a general solution.
-
Determine a particular solution for the initial values
x
(
0
)
=
x
0
.
-
Determine the maximal interval of definition for your solution to part (b)
-
Plot a graph of your solution to part (b).
-
Determine all equilibrium solutions (if any) to the ODE. Be sure to explain
why any such solution you obtain cannot be a particular solution.
-
Consider the ODE
x
′
=
(
1
−
x
2
)
/
(
1
−
t
2
)
.
-
Calculate an (implicit) general solution.
-
Determine a particular solution for the initial values
x
(
0
)
=
1
2
.
-
Determine the maximal interval of definition for your solution to part (b).
-
Determine all equilibrium solutions (if any) to the ODE.
-
Consider the ODE
x
′
=
3
t
x
1
/
3
.
-
Calculate a general solution
-
Determine a particular solution for the initial values
x
(
0
)
=
0.
-
Determine the maximal interval of definition of your solution to part (b).
-
Plot a graph of your solution to part (b).
-
Determine all equilibrium solutions.
-
Consider the IVP
x
′
−
2
cos
2
x
=
0
,
x
(
0
)
=
0.
-
Solve for
x
(
t
)
.
-
What is the maximal interval of definition
I
of your solution?
-
Sketch a graph of your solution on
I
.
-
Consider the IVP
x
′
=
sin
(
t
+
x
)
,
x
(
0
)
=
0.
-
Solve for
x
(
t
)
.
[Hint: Make the substitution
u
=
t
+
x
so as to transform the ODE to a separable equation of the form
u
′
=
g
(
t
,
u
)
.
]
-
What is the maximal interval of definition
I
of your solution?
-
Sketch a graph of your solution on
I
.
-
Consider the ODE
x
′
=
t
x
3
.
-
Determine a the general solution to the ODE
-
Calculate the particular solution and the maximal interval of definition for
each of the following initial condition:
-
x
(
0
)
=
1
-
x
(
0
)
=
1
2
-
x
(
0
)
=
2
-
Plot the three solution from part (c) on a single
t
x
-coordinate
system.
-
Now suppose the initial condition has the form
x
(
0
)
=
α
,
with
α
>
0.
-
Show that as
α
approaches zero from the right, the maximal interval of definition approaches
R
.
-
Show that as
α
approaches
∞
,
the maximal interval of definition shrinks to a single point.
-
The ODE for the velocity of a body falling to earth and subject to air
resistance is given by
v
′
=
g
−
k
v
2
,
where
g
is the acceleration of gravity and
k
is a constant that is a measure of the drag or friction resulting from air
resistance.
-
Derive the implicit
solution
v
¯
+
v
v
¯
−
v
=
c
e
2
g
k
t
where
v
¯
=
g
/
k
.
-
Interpret the term
v
¯
=
g
/
k
.
-
If the body has an initial velocity of
v
0
;
i.e.,
v
(
0
)
=
v
0
,
show
that
v
=
(
v
¯
+
v
0
)
−
(
v
¯
−
v
0
)
e
−
2
g
k
t
(
v
¯
+
v
0
)
+
(
v
¯
−
v
0
)
e
−
2
g
k
t
v
¯
-
When
g
=
9.81
,
k
=
0.2
,
sketch plots of
v
vs.
t
for
v
0
=
30
and
v
0
=
0.
-
Suppose
p
(
t
)
is a continuous function on
R
and consider the ODE
x
′
=
p
(
t
)
x
.
Derive the general solution
formula
x
=
c
e
∫
p
(
t
)
ⅆ
t
where
c
is an arbitrary constant.
-
Determine whether or not the ODE
(
t
x
2
+
x
+
e
2
t
)
+
(
t
+
t
2
x
−
x
)
x
′
=
0
is exact.
-
Determine whether or not the ODE
1
+
e
−
2
y
+
2
x
e
−
2
y
y
′
=
0
is exact.
-
Determine whether or not the ODE
2
y
(
y
−
1
)
+
x
(
2
y
−
1
)
y
′
=
0
is exact.
-
Determine a value of the integer
n
so that the ODE
t
n
(
t
2
+
x
2
+
t
)
+
t
n
+
1
x
x
′
=
0
is exact.
-
Determine a value of the integer
n
so that the ODE
t
n
x
n
+
1
+
t
n
+
1
x
n
(
1
−
3
t
2
x
2
)
x
′
=
0
is exact.
-
State (and prove) a necessary and sufficient condition for exactness of the
ODE
A
t
+
B
x
+
E
+
(
C
t
+
D
x
+
F
)
x
′
=
0
,
where
A
,
B
,
C
,
D
,
E
,
and
F
are constants.
-
Solve the exact ODE
t
2
−
x
−
t
x
′
=
0.
-
Solve the exact ODE
t
2
+
x
+
(
t
+
x
2
)
x
′
=
0.
-
Solve the exact ODE
3
(
x
+
1
)
2
−
2
y
y
′
=
0.
[Note:
x
y
-variables
are used instead of
t
x
-variables.
Also observe that this is a separable ODE.]
-
Solve the exact ODE
x
+
cos
(
x
+
y
)
+
cos
(
x
+
y
)
y
′
=
0.
[Note:
x
y
-variables
are used instead of
t
x
-variables.]
-
Solve the exact ODE
(
cos
t
sin
t
−
t
x
2
)
+
x
(
1
−
t
2
)
x
′
=
0
-
Calculate an integrating factor for the ODE
e
−
t
−
cos
x
+
(
sin
x
)
x
′
=
0.
-
Calculate an integrating factor for the ODE
2
t
x
+
(
x
2
−
t
2
)
x
′
=
0.
-
Calculate an integrating factor for the general linear ODE
x
′
+
p
(
t
)
x
=
f
(
t
)
by the methods of Lecture 02. Does this agree with the integrating factor
obtained by the method of Lecture 04?
-
Check for exactness, if necessary calculate an integrating factor, and solve
the ODE
x
+
(
2
t
−
e
x
)
x
′
=
0.
-
Check for exactness, if necessary calculate an integrating factor, and solve
the ODE
3
t
x
3
+
x
2
+
(
1
−
t
x
)
x
′
=
0.
-
Solve the ODE
t
x
′
+
2
x
=
8
t
2
.
-
Solve the ODE
x
(
ⅆ
y
/
ⅆ
x
)
=
1
−
x
y
−
y
.
-
Solve the ODE
t
x
′
+
(
t
+
1
)
x
=
e
−
t
.
-
Solve the ODE
t
2
x
′
=
t
x
−
x
.
-
Solve the ODE
x
′
+
x
=
2
+
2
t
.
-
Solve the ODE
x
′
=
2
t
x
+
t
.
-
Solve the ODE
x
′
=
x
+
2
sin
t
.
-
Solve the ODE
t
x
′
=
t
sin
t
−
x
.
-
Solve the ODE
(
cos
t
)
x
′
+
(
sin
t
)
x
=
1.
-
Solve the ODE
(
1
+
t
2
)
x
′
=
2
t
x
+
1
+
t
2
.
-
Solve the ODE
ⅆ
r
/
ⅆ
θ
+
r
tan
θ
=
cos
2
θ
.
-
Solve the ODE
t
x
′
=
1
−
t
x
−
x
.
-
Solve the ODE
(
1
−
t
3
)
y
′
=
3
t
2
y
-
Solve the ODE
x
′
+
x
=
cos
t
.
-
Solve the ODE
x
′
=
x
−
1
2
t
x
2
-
Solve the ODE
x
′
−
x
=
e
−
t
x
2
.
-
Solve the ODE
x
′
+
1
2
x
t
=
t
x
3
.
-
Solve the ODE
2
t
x
′
=
x
−
(
cos
t
)
x
3
.
-
Solve the IVP
x
′
=
1
/
(
t
+
x
+
1
)
,
x
(
0
)
=
0.
[Hint: Invert the ODE to get
ⅆ
t
/
ⅆ
x
=
t
+
x
+
1
,
a linear ODE with
x
as the independent variable and
t
as the dependent variable. The resulting solution will be an implicit one for
x
.
]
-
Use inversion to solve the ODE
x
′
=
1
/
(
t
+
sin
x
)
.
-
Use inversion to solve the ODE
(
x
2
+
2
t
x
+
1
)
x
′
=
1
=
x
2
.
-
Use inversion to solve the ODE
x
′
=
t
/
(
t
2
+
x
)
.
-
Use inversion to solve the IVP
x
′
=
x
/
(
x
sin
x
−
t
)
,
x
(
1
)
=
π
.
-
Solve the ODE
x
′
=
1
+
t
e
−
x
by making the substitution
v
=
e
x
to get the linear ODE
ⅆ
v
/
ⅆ
t
=
v
+
t
,
where
t
is the independent variable and
v
is the dependent variable. Solve this linear ODE for
v
and replace
v
with the substitution
e
x
to obtain a solution for
x
in terms of
t
.
-
Solve the ODE
t
x
′
+
x
=
e
t
x
by making the substitution
v
=
t
x
to get the separable ODE
ⅆ
v
/
ⅆ
t
=
e
v
,
with
t
as the independent variable and
v
as the dependent variable. Solve this separable ODE for
v
and replace
v
with the substitution
t
x
to obtain a solution for
x
in terms of
t
.
-
Solve the ODE
(
t
2
cos
x
)
x
′
=
2
t
sin
x
−
1
by making the substitution
v
=
sin
x
to get a linear ODE with
t
as the independent variable and
v
as the dependent variable. Solve this linear ODE for
v
and replace
v
with the substitution
sin
x
to obtain a solution for
x
in terms of
t
.
-
The temperature
U
of a cup of coffee as it sits cooling in its cup at time
t
is given by an ODE called Newton's Law of
Cooling
ⅆ
U
ⅆ
t
=
−
k
(
U
−
U
a
)
where
k
is a positive constant that depends upon the coffee and the shape of the cup,
and
U
a
is the temperature of the room (the ambient
temperature). If
U
(
0
)
=
U
0
,
Derive the
solution
U
(
t
)
=
U
a
+
(
U
0
−
U
a
)
e
−
k
t
[This
model is good only for relatively small differences in temperature between the
coffee and the surrounding room.)
-
(NEW) Suppose
p
(
t
)
is continuous and
T
-periodic
function; that is,
T
is a positive number for which
p
(
t
+
T
)
=
p
(
t
)
for all
t
∈
R
.
Prove that all solutions to
x
′
+
p
(
t
)
x
=
0
are
T
-periodic
if and only if
∫
0
T
p
(
t
)
ⅆ
t
=
0
.
-
Consider the
ODE
x
′
=
x
2
1
−
t
x
2
−
2
t
x
-
Verify that
(
t
x
2
−
1
)
e
x
=
c
is an implicit solution.
-
Use the
GCalc3
Implicit Function Plugin to confirm that there is a solution
through
(
1
,
1
)
.
-
What can you say about the maximal interval of definition of the solution you
identified in part b.
-
Consider the
ODE
ⅆ
y
ⅆ
x
=
x
2
y
4
−
2
-
Calculate a general solution to the ODE. [Warning: the general solution is
implicit.]
-
Determine the implicit solution that satisfies the initial condition
y
(
0
)
=
0.
-
Use the
GCalc3
Implicit Function Plugin to
-
Sketch a solution to part b. Be careful identify what part of the graph of the
implicit solution is actually a solution to the IVP through
(
0
,
0
)
.
-
Estimate the maximal (i.e., largest) interval of definition of the solution in
part b and identify it on your sketch for part (c) ii.
-
The plot produced by GCalc 3 suggests that there are two other solutions [with
y
(
0
)
≈
1.8
and with
y
(
0
)
≈
−
1.8
]
whose graphs are part of the solution to part (b). Explain what is going on.
-
Consider the
ODE
ⅆ
y
ⅆ
t
=
1
+
cos
t
y
(
1
+
e
y
)
-
Calculate a general solution to the ODE. [Warning: the general solution is
implicit.]
-
Determine the implicit solution that satisfies the initial condition
y
(
π
)
=
1.
-
Use the
GCalc3
Implicit Function Plugin to
-
Sketch a solution to part b. Be careful identify what part of the graph of the
implicit solution is actually a solution to the IVP through
(
π
,
1
)
.
-
Estimate the maximal (i.e., largest) interval of definition of the solution in
part b and identify it on your sketch for part c ii).
-
Consider the
ODE
ⅆ
x
ⅆ
t
=
t
cos
t
6
x
5
−
1
-
Calculate a general solution to the ODE. [Warning: the general solution is
implicit.]
-
Determine the implicit solution that satisfies the initial condition
x
(
0
)
=
1.
-
Use the GCalc3
Implicit Function Plugin to
-
Sketch a solution to part b. Be careful identify what part of the graph of the
implicit solution is actually a solution to the IVP through
(
0
,
1
)
.
-
Estimate the maximal (i.e., largest) interval of definition of the solution in
part b and identify it on your sketch for part c ii).
-
Consider the
ODE
ⅆ
x
ⅆ
t
=
sin
(
x
)
-
Calculate a general solution to the ODE. [Warning: the general solution is
implicit.]
-
Determine the implicit solution that satisfies the initial condition
x
(
0
)
=
1
2
π
.
-
Use the
GCalc3
Implicit Function Plugin to
-
Sketch a solution to part b. Be careful identify what part of the graph of the
implicit solution is actually a solution to the IVP through
(
0
,
π
/
2
)
.
-
Estimate the maximal (i.e., largest) interval of definition of the solution in
part b and identify it on your sketch for part c ii).
-
Consider the ODE
x
′
=
0.6
x
−
0.1
x
2
-
Compute the slope of the line tangent to the solution at each of the last six
points in the table below. The first six entries are provided as an aid.
(
t
,
x
)
Slope
(
1
,
−
2
)
−
1.6
(
1
,
−
1
)
−
0.7
(
1
,
0
)
0.0
(
1
,
1
)
0.5
(
1
,
2
)
0.8
(
1
,
3
)
0.9
(
1
,
4
)
(
1
,
5
)
(
1
,
6
)
(
1
,
7
)
(
1
,
8
)
(
1
,
9
)
-
Use the information from the table to construct a slope field by
hand on
t
x
-window
0
≤
t
≤
6
,
−
2
≤
x
≤
10
with a
1
×
1
grid.
-
Use Rychlik's
Slope
Field Calculator to graph a slope field for the ODE in the
t
x
-window
0
≤
t
≤
6
,
−
2
≤
x
≤
10
with a
1
×
1
grid.
-
Consider the ODE
x
′
=
−
x
2
.
-
Sketch a slope field by
hand
(no
software) in the
t
x
-window
−
2
≤
t
≤
2
,
−
2
≤
x
≤
2
with a 1/2
×
1/2 grid.
-
Sketch all equilibrium solutions.
-
Sketch solution curves corresponding to the following initial values (be sure
to identify each of the five curves):
-
x
(
0
)
=
1
-
x
(
0
)
=
0
-
x
(
2
)
=
−
2
-
Consider the ODE
x
′
=
1
−
x
2
.
-
Sketch a slope field by
hand
(no
software) in the
t
x
-window
−
2
≤
t
≤
2
,
−
2
≤
x
≤
2
with a 1/2
×
1/2 grid.
-
Sketch all equilibrium solutions.
-
Sketch solution curves corresponding to the following initial values (be sure
to identify each of the five curves):
-
x
(
−
1
)
=
2
-
x
(
0
)
=
1
-
x
(
−
2
)
=
0
-
x
(
0
)
=
0
-
x
(
0
)
=
−
2
-
Consider the ODE
x
′
=
sin
x
.
-
Sketch a slope field by
hand
(no
software) in the
t
x
-window
−
8
≤
t
≤
8
,
−
8
≤
x
≤
8
with a 1
×
1 grid.
-
Sketch all equilibrium solutions.
-
Sketch solution curves corresponding to the following initial values (be sure
to identify each of the four curves):
-
x
(
−
1
)
=
4
-
x
(
0
)
=
1
-
x
(
1
)
=
−
1
-
x
(
−
5
)
=
−
2
π
-
Consider the ODE
x
′
=
sin
x
−
1
.
-
Sketch a slope field by
hand
(no
software) in the
t
x
-window
−
8
≤
t
≤
8
,
−
8
≤
x
≤
8
with a 1
×
1 grid.
-
Sketch all equilibrium solutions.
-
Sketch solution curves corresponding to the following initial values (be sure
to identify each of the four curves):
-
x
(
−
1
)
=
4
-
x
(
0
)
=
1
-
x
(
1
)
=
−
1
-
x
(
−
5
)
=
−
2
π
-
Consider the ODE
x
′
+
x
=
f
(
t
)
,
where
f
(
t
)
=
{
0
,
if
−
∞
<
t
<
1
1
,
if
1
≤
t
<
∞
-
Sketch a slope field by
hand
(no
software) in the
t
x
-window
0
≤
t
≤
4
,
0
≤
x
≤
2
with a 1/4
×
1/4 grid.
-
Sketch solution curves corresponding to the following initial values (be sure
to identify each of the three curves):
-
x
(
0
)
=
0
-
x
(
0
)
=
1
-
x
(
0
)
=
2
-
Consider the ODE
x
′
+
x
=
f
(
t
)
,
where
f
(
t
)
=
{
0
,
if
−
∞
<
t
<
0
1
,
if
0
≤
t
<
1
0
,
if
1
≤
t
<
∞
-
Sketch a slope field by
hand
(no
software) in the
t
x
-window
0
≤
t
≤
4
,
−
2
≤
x
≤
2
with a 1/2
×
1/2 grid.
-
Sketch solution curves corresponding to the following initial values (be sure
to identify each of the three curves):
-
x
(
0
)
=
0
-
x
(
0
)
=
1
-
x
(
0
)
=
2
-
x
(
0
)
=
−
1
-
The slope field for the ODE
x
′
=
t
(
x
−
1
)
2
is shown below.
-
What is the slope for the direction line at the point
(
2
,
−
1
)
?
.
-
Sketch (on the slope field above) the two solution curves with initial
conditions
x
(
0
)
=
−
2
and
x
(
0
)
=
1.
-
Use Rychlik's
Slope
Field Calculator to graph a slope field for
x
′
=
1
−
t
x
in the
t
x
-window
0
≤
t
≤
6
,
−
2
≤
x
≤
2
.
Choose an appropriate grid to display solution behavior.
-
Use Rychlik's
Slope
Field Calculator to graph a slope field for
x
′
=
t
2
−
t
x
in the
t
x
-window
−
2
≤
t
≤
2
,
−
2
≤
x
≤
2
.
Choose an appropriate grid to display solution behavior.
-
Use Rychlik's
Slope
Field Calculator to graph a slope field for
x
′
=
x
−
cos
t
in the
t
x
-window
−
2
π
≤
t
≤
2
π
,
−
2
≤
x
≤
2.
Choose an appropriate grid to display solution behavior.
-
Use Rychlik's
Slope
Field Calculator to graph a slope field for
x
′
=
cos
(
t
x
)
in the
t
x
-window
0
≤
t
≤
10
,
−
2
≤
x
≤
2.
Choose an appropriate grid to display solution behavior of the solution curves
as
t
→
∞
.
-
Consider the ODE
x
′
=
x
2
−
t
2
.
Sketch enough isoclines by hand in an appropriately sized
t
x
-window
to allow you to sketch the integral curve through
(
0
,
0
)
.
-
Consider the ODE
x
′
=
x
−
cos
t
.
Sketch enough isoclines by hand in an appropriately sized
t
x
-window
to allow you to sketch the integral curve through
(
0
,
0
)
.
-
Consider the ODE
x
′
=
x
2
−
t
.
-
Use
d'Arbeloff's
java applet to view the isoclines for
m
=
−
3
,
−
2
,
−
1
,
0
,
1
,
2
,
&
3
in the
t
x
-window
−
4
≤
t
≤
4
,
−
4
≤
x
≤
4.
-
Conjecture what happens to the solution through
(
0
,
0
)
.
In particular, identify a curve
x
=
u
(
t
)
to which the solution through
(
0
,
0
)
is asymptotic as
t
→
∞
-
Provide a mathematical proof of your conjecture in part (b).
-
Consider the IVP
x
′
=
sin
t
x
,
x
(
2
)
=
2.
-
Use appropriate software to graph a slope field for the ODE in the
t
x
-window
−
10
≤
t
≤
10
,
−
5
≤
x
≤
5.
-
Use your slope field to conjecture what happens to the solution to the IVP as
t
→
∞
.
-
Use isoclines to prove your conjecture.
-
Consider the ODE
x
′
=
x
3
/
2
.
-
Sketch by hand a slope field in the
t
x
-window
−
3
≤
t
≤
3
,
−
1
≤
x
≤
5
and plot the integral curve through
(
2
,
1
)
.
[You may check your slope field by using Rychlik's
Slope
Field Calculator.]
-
Calculate the general solution to
x
′
=
x
3
/
2
.
[This equation is separable.]
-
Use your general solution from part (b) to calculate the particular solution
through
(
2
,
1
)
.
-
Plot your solution to part (c) on the slope field you created in part (a).
-
Explain any discrepencies between the integral curve you sketched in part (a)
and the plot you made in part (d).
-
(NEW) Consider the
ODE
x
′
=
2
t
(
1
+
1
x
2
)
-
Calculate a general solution to the ODE. [Warning: the general solution is
implicit.]
-
Determine the implicit solution that satisfies the initial condition
x
(
0
)
=
−
1.
-
Use the
GCalc3
Implicit Function Plugin to
-
Sketch a solution to part b. Be careful identify what part of the graph of the
implicit solution is actually a solution to the IVP through
(
0.
−
1
)
.
-
Estimate the maximal (i.e., largest) interval of definition of the solution in
part b and identify it on your sketch for part c ii.
-
Consider the IVP
x
′
+
x
=
f
(
t
)
,
x
(
0
)
=
0
,
where
f
(
t
)
=
{
0
,
if
−
∞
<
t
<
1
1
,
if
1
≤
t
<
∞
Solve
by each of the two following methods (The two methods must produce the same
piecewise defined solution.):
-
The integral formula method (e.g., Example 6.8).
-
The piecewise method (e.g., Example 6.9).
-
Use Rychlik's
Slope
Field Calculator to plot a graph of your solution. Make sure that your
window size is appropriate to show all important graphical features of the
solution.
-
Consider the IVP
x
′
+
x
=
f
(
t
)
,
x
(
0
)
=
1
,
where
f
(
t
)
=
{
0
,
if
−
∞
<
t
<
1
1
,
if
1
≤
t
<
∞
Solve
by each of the two following methods (The two methods must produce the same
piecewise defined solution.):
-
The integral formula method (e.g., Example 6.8).
-
The piecewise method (e.g., Example 6.9).
-
Use Rychlik's
Slope
Field Calculator to plot a graph of your solution. Make sure that your
window size is appropriate to show all important graphical features of the
solution.
-
Consider the IVP
x
′
+
x
=
f
(
t
)
,
x
(
0
)
=
0
,
where
f
(
t
)
=
{
0
,
if
−
∞
<
t
<
0
1
,
if
0
≤
t
<
1
0
,
if
1
≤
t
<
∞
Solve
by each of the two following methods (The two methods must produce the same
piecewise defined solution.):
-
The integral formula method (e.g., Example 6.8).
-
The piecewise method (e.g., Example 6.9).
-
Use Rychlik's
Slope
Field Calculator to plot a graph of your solution. Make sure that your
window size is appropriate to show all important graphical features of the
solution.
-
Consider the IVP
x
′
+
x
=
f
(
t
)
,
x
(
0
)
=
1
,
where
f
(
t
)
=
{
0
,
if
−
∞
<
t
<
0
1
,
if
0
≤
t
<
1
0
,
if
1
≤
t
<
∞
Solve
by each of the two following methods (The two methods must produce the same
piecewise defined solution.):
-
The integral formula method (e.g., Example 6.8).
-
The piecewise method (e.g., Example 6.9).
-
Use Rychlik's
Slope
Field Calculator to plot a graph of your solution. Make sure that your
window size is appropriate to show all important graphical features of the
solution.
-
Consider the IVP
x
′
+
x
=
f
(
t
)
,
x
(
0
)
=
1
,
where
f
(
t
)
=
{
0
,
if
−
∞
<
t
<
0
1
−
t
,
if
0
≤
t
<
1
0
,
if
1
≤
t
<
∞
Solve
by each of the two following methods (The two methods must produce the same
piecewise defined solution.):
-
The integral formula method (e.g., Example 6.8).
-
The piecewise method (e.g., Example 6.9).
-
Use Rychlik's
Slope
Field Calculator to plot a graph of your solution. Make sure that your
window size is appropriate to show all important graphical features of the
solution.
-
Consider the IVP
x
′
+
x
=
f
(
t
)
,
x
(
0
)
=
0
,
where
f
(
t
)
=
{
0
,
if
−
∞
<
t
<
0
t
,
if
0
≤
t
<
1
0
,
if
1
≤
t
<
∞
Solve
by each of the two following methods (The two methods must produce the same
piecewise defined solution.):
-
The integral formula method (e.g., Examples 6.7-6.9).
-
The piecewise method (e.g., Examples 6.10-6.11).
-
Use Rychlik's
Slope
Field Calculator to plot a graph of your solution. Make sure that your
window size is appropriate to show all important graphical features of the
solution.
-
Consider the IVP
x
′
+
x
=
f
(
t
)
,
x
(
0
)
=
0
,
where
f
(
t
)
=
{
0
,
if
−
∞
<
t
<
0
t
,
if
0
≤
t
<
1
1
,
if
1
≤
t
<
∞
Solve
by each of the two following methods (The two methods must produce the same
piecewise defined solution.):
-
The integral formula method (e.g., Examples 6.7-6.9).
-
The piecewise method (e.g., Examples 6.10-6.11).
-
Use Rychlik's
Slope
Field Calculator to plot a graph of your solution. Make sure that your
window size is appropriate to show all important graphical features of the
solution.
-
Consider the IVP
x
′
+
x
=
f
(
t
)
,
x
(
0
)
=
0
,
where
f
(
t
)
=
{
1
,
if
−
∞
<
t
<
0
1
−
t
,
if
0
≤
t
<
1
0
,
if
1
≤
t
<
∞
Solve
by each of the two following methods (The two methods must produce the same
piecewise defined solution.):
-
The integral formula method (e.g., Examples 6.7-6.9).
-
The piecewise method (e.g., Examples 6.10-6.11).
-
Use Rychlik's
Slope
Field Calculator to plot a graph of your solution. Make sure that your
window size is appropriate to show all important graphical features of the
solution.
-
Consider the IVP
x
′
+
x
=
f
(
t
)
,
x
(
0
)
=
0
,
where
f
(
t
)
=
{
0
,
if
−
∞
<
t
<
0
sin
π
t
,
if
0
≤
t
<
1
0
,
if
1
≤
t
<
∞
Solve
by each of the two following methods (The two methods must produce the same
piecewise defined solution.):
-
The integral formula method (e.g., Examples 6.7-6.9).
-
The piecewise method (e.g., Examples 6.10-6.11).
-
Use Rychlik's
Slope
Field Calculator to plot a graph of your solution. Make sure that your
window size is appropriate to show all important graphical features of the
solution.
-
Consider the IVP
x
′
+
x
=
f
(
t
)
,
x
(
0
)
=
1
,
where
f
(
t
)
=
{
1
,
if
−
∞
<
t
<
2
cos
π
t
,
if
2
≤
t
<
2.5
0
,
if
2.5
≤
t
<
∞
Solve
by each of the two following methods (The two methods must produce the same
piecewise defined solution.):
-
The integral formula method (e.g., Examples 6.7-6.9).
-
The piecewise method (e.g., Examples 6.10-6.11).
-
Use Rychlik's
Slope
Field Calculator to plot a graph of your solution. Make sure that your
window size is appropriate to show all important graphical features of the
solution.
-
Consider the IVP
x
′
+
p
(
t
)
x
=
1
,
x
(
0
)
=
0
,
where
p
(
t
)
=
{
0
,
if
−
∞
<
t
<
1
1
,
if
1
≤
t
<
∞
-
Solve by the piecewise method (e.g., Examples 6.10, 6.11).
-
Use Rychlik's
Slope
Field Calculator to plot a graph of your solution. Make sure that your
window size is appropriate to show all important graphical features of the
solution.
-
Consider the IVP
x
′
+
p
(
t
)
x
=
2
e
−
t
,
x
(
0
)
=
0
,
where
p
(
t
)
=
{
1
,
if
−
∞
<
t
<
1
2
,
if
1
≤
t
<
∞
-
Solve by the piecewise method (e.g., Examples 6.10, 6.11).
-
Use Rychlik's
Slope
Field Calculator to plot a graph of your solution. Make sure that your
window size is appropriate to show all important graphical features of the
solution.
-
Compute the values of the Euler approximation to the IVP
x
′
=
x
2
−
t
−
4
,
x
(
−
4
)
=
−
1
,
at
t
=
−
4
,
−
3
,
−
2
,
−
1
,
0
,
1
,
and
2.
Use stepsize
h
=
1.0
.
Fill in the table below with values
t
k
and
x
k
.
(Recall that
x
k
denotes the Euler approximation to
x
(
t
k
)
.
k
t
k
x
k
0
1
2
3
4
5
6
-
Consider the IVP
x
′
+
t
x
=
1
,
x
(
0
)
=
−
1.25
.
-
Compute an approximation for
x
(
3
)
using Euler's method with stepsize
h
=
0.5
.
Fill in the following table with values
t
k
and
x
k
.
(Recall that
x
k
denotes the Euler approximation to
x
(
t
k
)
.
k
t
k
x
k
0
1
2
3
4
5
6
-
Sketch a graph of your Euler approximate solution on an appropriately sized
t
x
-coordinate
system.
-
Consider the IVP
x
′
+
x
=
2
t
,
x
(
0
)
=
1
-
Compute an approximation for
x
(
1.0
)
,
using Euler's method with stepsize
h
=
0.1
.
Fill in the following table with values
t
k
and
x
k
.
Also calculate the values of
x
(
t
k
)
,
the actual values of the solution at times
t
k
,
k
=
0
,
1
,
…
,
10.
[Recall that
x
k
denotes the Euler approximation to the actual values
x
(
t
k
)
.]
k
t
k
x
k
x
(
t
k
)
0
1
2
3
4
5
6
7
8
9
10
-
What is the approximate value of
x
(
1.0
)
,
namely
x
1
0
?
-
A slope field for the ODE
x
′
=
x
2
−
t
−
1
is sketched below.
-
Without making any calculations, sketch the Euler approximation to the
solutuion with initial values
x
(
0
)
=
0
.
Use stepsize
h
=
1.0
.
Extend your Euler aproximation forward to
t
=
4
and backward to
t
=
−
4.
-
Just use your plot above to estimate the value of
x
(
4
)
.
-
Apply the algorithm for Euler's method to calculate the approximate value of
x
(
4
)
.
-
The following plot represents the slope field for the ODE
x
′
=
2
−
2
t
x
:
-
Sketch the Euler approximation to the solution with initial values
x
(
0
)
=
0
with step size
h
=
0.5
.
-
Estimate the value of
x
(
2.5
)
by Euler's method with step size
h
=
0.5
.
You must NOT calculate ANYTHING. All you need is the slope field, a pencil and
a straightedge.
-
Apply the algorithm for Euler's method to calculate the approximate value of
x
(
2.5
)
-
The following plot represents the slope field for the ODE
x
′
=
t
2
+
x
2
−
1
:
-
Sketch and label (on the slope field) the solution curve that satisfies the IC
x
(
−
1.5
)
=
−
2.
-
Estimate the value of
x
(
1.5
)
by Euler's method with step size
h
=
0.5
.
You must NOT calculate ANYTHING. All you need is the slope field, a pencil and
a straightedge.
-
Apply the algorithm for Euler's method to calculate the approximate value of
x
(
4
)
.
-
Fill in the following table using Euler's method to compute an approximate
solution to the IVP
x
′
=
t
−
x
2
+
2
,
x
(
0
)
=
−
1
on the interval
0
≤
t
≤
4
using
h
=
1.0
.
k
t
k
x
k
Use
DFIELD
2005.10 in each of Exercises 8, 9, & 10. In each case choose
the following from the DFIELD Direction Field
Window
(i)
"Solution/Keyboard Input of Initial Values" and set the initial
values
(ii)
"Options/ODE Solver/Euler" and select the Euler method and the step size
-
Consider the IVP
x
′
=
2
t
−
x
,
x
(
0
)
=
1.
Plot the Euler approximations of the IVP using
h
=
1.0
,
h
=
0.5
,
and
h
=
0.1
.
Plot all approximations on the same graph. Use a window size
0
≤
t
≤
4
and
0
≤
x
≤
6.
-
Consider the IVP
x
′
=
1
−
t
x
,
x
(
0
)
=
1.
Plot the Euler approximations of the IVP using
h
=
1.0
,
h
=
0.5
,
and
h
=
0.1
.
Plot all approximations on the same graph. Use a window size
0
≤
t
≤
4
and
0
≤
x
≤
2.
-
Consider the IVP
x
′
=
sin
(
t
x
)
,
x
(
0
)
=
1.
Plot the Euler approximations of the IVP using
h
=
1.0
,
h
=
0.5
,
and
h
=
0.1
.
Plot all approximations on the same graph. Use a window size
0
≤
t
≤
20
and
0
≤
x
≤
4.
Expain in words the apparent behavior of the approximations.
-
Determine the degree 3 Taylor polynomial approximation to the solution of the
IVP
x
′
=
e
−
t
x
,
x
(
0
)
=
1.
-
Determine the degree 3 Taylor polynomial approximation to the solution of the
IVP
x
′
=
1
/
(
t
+
x
+
1
)
,
x
(
0
)
=
0.
-
Consider the IVP
x
′
=
−
x
,
x
(
0
)
=
2
-
Determine the degree 2 Taylor polynomial solution of the IVP.
-
Determine the degree 5 Taylor polynomial solution of the IVP.
-
Compute the acutal solution of the IVP; i.e., solve the IVP.
-
Graph your result of parts (a), (b) and (c) on the same
t
x
-coordinate
system. Choose an appropriate
t
x
-window
for your plot. Be sure to label your graphs.
-
Consider the IVP
x
′
=
−
t
x
,
x
(
0
)
=
1
-
Determine the degree 2 Taylor polynomial solution of the IVP.
-
Determine the degree 5 Taylor polynomial solution of the IVP.
-
Compute the acutal solution of the IVP; i.e., solve the IVP.
-
Graph your result of parts (a), (b) and (c) on the same
t
x
-coordinate
system. Choose an appropriate
t
x
-window
for your plot. Be sure to label your graphs.
-
Consider the IVP
x
′
=
1
−
t
x
,
x
(
0
)
=
0
-
Determine the degree 3 Taylor polynomial approximate solution of the IVP.
-
Determine the degree 5 Taylor polynomial approximate solution of the IVP.
-
Determine the degree 7 Taylor polynomial approximate solution of the IVP.
-
Graph your result of parts (a), (b) and (c) on the same
t
x
-coordinate
system. Use a window size
−
4
≤
t
≤
4
and
−
2
≤
x
≤
2.
Be sure to label your graphs.
-
Compare your results of part (d) with that of either Use Rychlik's
Slope
Field Calculator or Polking's
DFIELD
2005.10 to generate an integral curve for the actual solution in
the
t
x
-window
−
4
≤
t
≤
4
and
−
2
≤
x
≤
2
.
-
Consider the IVP
x
′
=
cos
(
t
+
x
)
,
x
(
0
)
=
0
-
Determine the degree 3 Taylor polynomial approximate solution of the IVP.
-
Determine the degree 5 Taylor polynomial approximate solution of the IVP.
-
Compute the acutal solution of the IVP; i.e., solve the IVP. [Hint: Make the
substitution
u
=
x
+
t
.
]
-
Graph your result of parts (a), (b), (c) and (d) on the same
t
x
-coordinate
system. Use a window size
−
4
≤
t
≤
4
and
−
2
≤
x
≤
2.
Be sure to label your graphs.