Baer_MAT_275_C_Fall_2021.iscoggin.Section_7.5_Homogeneous_systems_Constant_Coefficients
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Course
275
Subject
Mathematics
Date
Feb 20, 2024
Type
Pages
4
Uploaded by HighnessFlamingoPerson1018
Isabella Scoggins
Baer
MAT
275
C
Fall
2021
Assignment Section
7.5
Homogeneous
systems
Constant
Coefficients due 11/28/2021 at 11:59pm MST
Problem 1.
(1 point)
Solve the IVP
d
x
dt
=
-
16
-
16
-
12
-
12
x
,
x
(
0
) =
-
17
-
4
x
(
t
) =
.
Solution:
SOLUTION
The eigenvalues of the coefficient matrix are
λ
=
-
28 and
λ
=
0
with associated eigenvectors
4
3
and
-
1
1
respectively.
The general solution is then
x
(
t
) =
c
1
e
-
28
t
4
3
+
c
2
-
1
1
The initial conditions yield the system
-
17
=
4
c
1
-
c
2
-
4
=
3
c
1
+
c
2
Solving the system yields
c
1
=
-
3 and
c
2
=
5.
Substituting into the general solution gives
x
(
t
) =
-
3
e
-
28
t
4
3
+
5
-
1
1
Thus
x
(
t
) =
-
(
12
e
-
28
t
+
5
)
5
-
9
e
-
28
t
Correct Answers:
•
-3*eˆ(-28*t)*4 + 5*-1
•
-3*eˆ(-28*t)*3 + 5*1
Problem 2.
(1 point)
Consider the system of differential equations
dx
dt
=
-
1
.
6
x
+
y
,
dy
dt
=
1
.
25
x
-
3
.
6
y
.
For this system, the smaller eigenvalue is
and the
larger eigenvalue is
.
Use the phase plotter
pplane9.m
in MATLAB to determine how
the solution curves behave.
•
A. The solution curves converge to different points.
•
B. All of the solution curves run away from 0. (Unstable
node)
•
C. The solution curves race towards zero and then veer
away towards infinity. (Saddle)
•
D. All of the solution curves converge towards 0. (Stable
node)
The solution to the above differential equation with initial values
x
(
0
) =
5
,
y
(
0
) =
3 is
x
(
t
) =
,
y
(
t
) =
.
Solution:
SOLUTION
The characteristic polynomial of the coefficient matrix is
-
1
.
6
-
λ
1
1
.
25
-
3
.
6
-
λ
= (
-
1
.
6
-
λ
)(
-
3
.
6
-
λ
)
-
1
.
25
=
λ
2
+
5
.
2
λ
+
4
.
51
Using the quadratic formula, we find the eigenvalues
λ
1
=
-
4
.
1
and
λ
2
=
-
1
.
1.
A possible eigenvector associated to
λ
1
is
1
-
5
2
,
while a possible
eigenvector associated to
λ
2
is
1
1
2
.
The general solution is then given by
x
(
t
)
=
c
1
e
-
4
.
1
t
+
c
2
e
-
1
.
1
t
y
(
t
)
=
(
-
5
/
2
)
c
1
e
-
4
.
1
t
+(
1
/
2
)
c
2
e
-
1
.
1
t
The initial conditions yield the system
5
=
c
1
+
c
2
3
=
(
-
5
/
2
)
c
1
+(
1
/
2
)
c
2
1
Solving the system gives
c
1
=
-
1
6
and
c
2
=
31
6
.
Subsituting into the general solution yields the solution
x
(
t
)
=
-
1
6
e
-
4
.
1
t
+
31
6
e
-
1
.
1
t
y
(
t
)
=
5
12
e
-
4
.
1
t
+
31
12
e
-
1
.
1
t
Since the eigenvalues are real and both negative , All of the solu-
tion curves converge towards 0. (Stable node)
Correct Answers:
•
-4.1
•
-1.1
•
D
•
-0.166666666666667*1*exp(-4.1*t)+5.16666666666667*1*exp(-1.1*t)
•
-0.166666666666667*(-2.5)*exp(-4.1*t)+5.16666666666667*(0.5)*exp(-1.1*t)
Problem 3.
(1 point)
Consider the systems of differential equations
dx
dt
=
0
.
4
x
+
0
.
5
y
,
dy
dt
=
1
.
5
x
-
0
.
6
y
.
For this system, the smaller eigenvalue is
and the
larger eigenvalue is
.
Use the phase plotter
pplane9.m
in MATLAB to determine how
the solution curves behave.
•
A. The solution curves race towards zero and then veer
away towards infinity. (Saddle)
•
B. All of the solution curves run away from 0. (Unstable
node)
•
C. All of the solution curves converge towards 0. (Stable
node)
•
D. The solution curves converge to different points.
The solution to the above differential equation with initial values
x
(
0
) =
7
,
y
(
0
) =
4 is
x
(
t
) =
,
y
(
t
) =
.
Solution:
SOLUTION
The characteristic polynomial of the coefficient matrix is
0
.
4
-
λ
1
2
1
.
5
-
0
.
6
-
λ
= (
0
.
4
-
λ
)(
-
0
.
6
-
λ
)
-
0
.
75
=
λ
2
+
0
.
2
λ
-
0
.
99
Using the quadratic formula, we find the eigenvalues
λ
1
=
-
1
.
1
and
λ
2
=
0
.
9.
A possible eigenvector associated to
λ
1
is
1
-
3
,
while a possible
eigenvector associated to
λ
2
is
1
1
.
The general solution is then given by
x
(
t
)
=
c
1
e
-
1
.
1
t
+
c
2
e
0
.
9
t
y
(
t
)
=
-
3
c
1
e
-
1
.
1
t
+
c
2
e
0
.
9
t
The initial conditions yield the system
7
=
c
1
+
c
2
4
=
-
3
c
1
+
c
2
Solving the system gives
c
1
=
3
4
and
c
2
=
25
4
.
Subsituting into the general solution yields the solution
x
(
t
)
=
3
4
e
-
1
.
1
t
+
25
4
e
0
.
9
t
y
(
t
)
=
-
9
4
e
-
1
.
1
t
+
25
4
e
0
.
9
t
2
Since the eigenvalues are real and of opposite signs, The solu-
tion curves race towards zero and then veer away towards infinity.
(Saddle)
Correct Answers:
•
-1.1
•
0.9
•
A
•
1.5*0.5*exp(-1.1*t)+12.5*0.5*exp(0.9*t)
•
1.5*(-1.5)*exp(-1.1*t)+12.5*(0.5)*exp(0.9*t)
Problem 4.
(1 point)
Consider the systems of differential equations
dx
dt
=
0
.
3
x
-
0
.
4
y
,
dy
dt
=
-
0
.
4
x
+
0
.
9
y
.
For this system, the smaller eigenvalue is
and the
larger eigenvalue is
.
Use the phase plotter
pplane9.m
in MATLAB to determine how the solution curves behave.
•
A. All of the solution curves converge towards 0. (Stable
node)
•
B. The solution curves converge to different points.
•
C. The solution curves race towards zero and then veer
away towards infinity. (Saddle)
•
D. All of the solution curves run away from 0. (Unstable
node)
The solution to the above differential equation with initial values
x
(
0
) =
8
,
y
(
0
) =
7 is
x
(
t
) =
,
y
(
t
) =
.
Solution:
SOLUTION
The characteristic polynomial of the coefficient matrix is
0
.
3
-
λ
-
2
5
-
0
.
4
0
.
9
-
λ
= (
0
.
3
-
λ
)(
0
.
9
-
λ
)
-
0
.
16
=
λ
2
-
1
.
2
λ
+
0
.
11
Using the quadratic formula, we find the eigenvalues
λ
1
=
0
.
1 and
λ
2
=
1
.
1.
A possible eigenvector associated to
λ
1
is
2
1
,
while a possible
eigenvector associated to
λ
2
is
-
1
2
.
The general solution is then given by
x
(
t
)
=
2
c
1
e
0
.
1
t
-
c
2
e
1
.
1
t
y
(
t
)
=
c
1
e
0
.
1
t
+
2
c
2
e
1
.
1
t
The initial conditions yield the system
8
=
2
c
1
-
c
2
7
=
c
1
+
2
c
2
Solving the system gives
c
1
=
23
5
and
c
2
=
6
5
.
Subsituting into the general solution yields the solution
x
(
t
)
=
46
5
e
0
.
1
t
-
6
5
e
1
.
1
t
y
(
t
)
=
23
5
e
0
.
1
t
+
12
5
e
1
.
1
t
3
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Since the eigenvalues are real and both positive, All of the so-
lution curves run away from 0. (Unstable node)
Correct Answers:
•
0.1
•
1.1
•
D
•
-23*-0.4*exp(0.1*t)+3*-0.4*exp(1.1*t)
•
-23*(-0.2)*exp(0.1*t)+3*(0.8)*exp(1.1*t)
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