Austin Cholley Jones MAT 275 ONLINE B Spring 2021
Assignment Section 7.6 Complex Eigenvalues due 04/21/2021 at 11:59pm MST
1. (1 point) Suppose Ais a 2 ×2 real matrix with an eigen-
value λ=4+5iand corresponding eigenvector
~v=−1+i
i.
Determine a fundamental set (i.e., linearly independent set) of
solutions for ~y0=A~y, where the fundamental set consists en-
tirely of real solutions.
Enter your solutions below. Use tas the independent variable in
your answers.
~y1(t) =
~y2(t) =
Answer(s) submitted:
•[eˆ((4t))(-cos(5t)-sin(5t))]
(correct)
Correct Answers:
•<table border=’0’ cellspacing=’5’><tr><td> eˆ(4*t)*(-[cos(5*t)]-sin(5*t)) </td></tr><tr><td> eˆ(4*t)*(-[sin(5*t)]) </td></tr></table>
<br />
<table border=’0’ cellspacing=’5’><tr><td> eˆ(4*t)*[cos(5*t)-sin(5*t)] </td></tr><tr><td> eˆ(4*t)*cos(5*t) </td></tr></table>
2. (1 point)
Consider the Initial Value Problem:
x0
1=−4x1+2x2
x0
2=−10x1+4x2
,x1(0) = 4
x2(0) = 4
(a) Find the eigenvalues and eigenvectors for the coefficient ma-
trix.
λ1=,~v1= , and λ2=,~v2=
(b) Solve the initial value problem. Give your solution in real
form.
x1=,
x2=.
? 1. Use the phase plotter pplane9.m in MATLAB to de-
scribe the trajectory.
Solution: The characteristic polynomial of the matrix is
−4−λ2
−10 4 −λ
=λ2+22. Thus the eigenvalues are λ=±2i.
A possible eigenvector associated to λ=2iis 2−i
5=2
5+
i−1
0
One solution is then given by
2
5cos2t−−1
0sin2t=2cos2t+sin2t
5cos2t
while the other is
2
5sin2t+−1
0cos2t=2sin2t−cos2t
5sin2t
The general solution is
x(t) = c12cos2t+sin2t
5cos2t+c22sin2t−cos2t
5sin2t
Using the initial condition gives x(0) = 2c1−c2
5c1=4
4
Solving the system yields c1=4
5and c2=−12
5.
Substituting c1and c2into the general solution yields the solu-
tion
x1=4cos2t−4sin2t
x2=4cos2t−12sin2t
Since the real part of the eigenvalues is zero, the trajectory
is an ellipse.
The ellipse has clockwise orientation.
Answer(s) submitted:
•2i
•
•
•Ellipse clockwise
(score 0.25)
Correct Answers:
•<table border=’0’ cellspacing=’0’>
<tr>
<td> 2i, </td>
<td><table border=’0’ cellspacing=’5’><tr><td> 2-i </td></tr><tr><td> 5 </td></tr></table></td>
<td> -2i,</td>
<td><table border=’0’ cellspacing=’5’><tr><td> 2+i </td></tr><tr><td> 5 </td></tr></table></td>
</tr>
</table>
•4*[cos(2*t)-2*sin(2*t)]+4*sin(2*t)
•4*[cos(2*t)+2*sin(2*t)]-4*[4*sin(2*t)+sin(2*t)]
•ELLIPSE CLOCKWISE
1
3. (1 point) Consider the Initial Value Problem
dx
dt =−3−3
3−3x,x(0) = 5
8
(a) Find the eigenvalues and eigenvectors for the coefficient ma-
trix.
λ1=,~v1= , and λ2=,~v2=
(b) Find the solution to the initial value problem. Give your
solution in real form.
x(t) = .
Use the phase plotter pplane9.m in MATLAB to help you
describe the trajectory:
? 1. Describe the trajectory.
Solution: The characteristic polynomial of the matrix is
−3−λ−3
3−3−λ
= (−3−λ)2+32. Thus the eigenvalues are
λ=−3±3i.
A possible eigenvector associated to λ=−3+3iis i
1=
0
1+i1
0
One solution is then given by e−3t−sin 3t
cos3twhile the other is
e−3tcos3t
sin3t
The general solution is
x(t) = c1e−3t−sin3t
cos3t+c2e−3tcos3t
sin3t
Using the initial condition gives x(0) = c2
c1=5
8
Solving the system yields c1=8 and c2=5.
Substituting c1and c2into the general solution yields the solu-
tion
x(t) = e−3t(−8sin3t+5 cos 3t)
e−3t(8cos3t+5 sin 3t)
Since the real part of the eigenvalues is negative, the trajec-
tory is a spiral, spiraling inward.
The spiral has counterclockwise orientation.
Answer(s) submitted:
•-3-3i
•8eˆ(-3t)(cos(3t)(0)-sin(3t)(1))+5eˆ(-3t)(cos(3t)1+sin(3t)(0))
•8eˆ(-3t)(cos(3t)(1)-sin(3t)(0))+5eˆ(-3t)(cos(3t)(0)+sin(3t)(1))
•Spiral inward counterclockwise
(correct)
Correct Answers:
•<table border=’0’ cellspacing=’0’>
<tr>
<td> -3+3i, </td>
<td><table border=’0’ cellspacing=’5’><tr><td> i </td></tr><tr><td> 1 </td></tr></table></td>
<td> -3-3i,</td>
<td><table border=’0’ cellspacing=’5’><tr><td> -i </td></tr><tr><td> 1 </td></tr></table></td>
</tr>
</table>
•[5*cos(3*t)-8*sin(3*t)]*eˆ(-3*t)
•[5*sin(3*t)+8*cos(3*t)]*eˆ(-3*t)
•SPIRAL INWARD COUNTERCLOCKWISE
4. (1 point) Consider the initial value problem
dx
dt =2−4
4 2 x,x(0) = 6
4
(a) Find the eigenvalues and eigenvectors for the coefficient ma-
trix.
λ1=,~v1= , and λ2=,~v2=
(b) Solve the initial value problem. Give your solution in real
form.
x(t) = .
Use the phase plotter pplane9.m in MATLAB to answer the
following question.
? 1. Describe the trajectory.
Solution: The characteristic polynomial of the matrix is
2−λ−4
4 2 −λ
= (2−λ)2+42. Thus the eigenvalues are λ=
2±4i.
A possible eigenvector associated to λ=2+4iis i
1=0
1+
i1
0
One solution is then given by e2t−sin4t
cos4twhile the other is
e2tcos4t
sin4t
The general solution is
x(t) = c1e2t−sin4t
cos4t+c2e2tcos 4t
sin4t
Using the initial condition gives x(0) = c2
c1=6
4
Solving the system yields c1=4 and c2=6.
Substituting c1and c2into the general solution yields the solu-
tion
x(t) = e2t(−4sin4t+6 cos 4t)
e2t(4cos4t+6 sin 4t)
Since the real part of the eigenvalues is positive, the trajectory
2
is a spiral, spiraling outward.
The spiral has counterclockwise orientation.
Answer(s) submitted:
•
•4eˆ(2t)(cos(4t)(0)-sin(4t)(1))+6eˆ(2t)(cos(4t)(1)+sin(4t)(0))
•4eˆ(2t)(cos(4t)(1)-sin(4t)(0))+6eˆ(2t)(cos(4t)(0)+sin(4t)(1))
•Spiral outward counterclockwise
(score 0.75)
Correct Answers:
•<table border=’0’ cellspacing=’0’>
<tr>
<td> 2+4i, </td>
<td><table border=’0’ cellspacing=’5’><tr><td> i </td></tr><tr><td> 1 </td></tr></table></td>
<td> 2-4i,</td>
<td><table border=’0’ cellspacing=’5’><tr><td> -i </td></tr><tr><td> 1 </td></tr></table></td>
</tr>
</table>
•eˆ(2*t)*[6*cos(4*t)-4*sin(4*t)]
•eˆ(2*t)*[6*sin(4*t)+4*cos(4*t)]
•SPIRAL OUTWARD COUNTERCLOCKWISE
5. (1 point)
Consider the initial value problem
dx
dt =−9−3
30 9 x,x(0) = 9
5
(a) Find the eigenvalues and eigenvectors for the coefficient ma-
trix.
λ1=,~v1= , and λ2=,~v2=
(b) Solve the initial value problem. Give your solution in real
form.
x(t) = .
Use the phase plotter pplane9.m in MATLAB to answer the
following question.
? 1. Describe the trajectory.
Solution: The characteristic polynomial of the matrix is
−9−λ−3
30 9 −λ
=λ2+32. Thus the eigenvalues are λ=±3i.
A possible eigenvector associated to λ=3iis −3+i
10 =
−3
10 +i1
0
One solution is then given by
−3
10 cos3t−1
0sin3t=−3cos3t−sin3t
10cos3t
while the other is
−3
10 sin3t+1
0cos3t=−3sin3t+cos3t
10sin3t
The general solution is
x(t) = c1−3cos3t−sin3t
10cos3t+c2−3sin3t+cos3t
10sin3t
Using the initial condition gives x(0) = −3c1+c2
10c1=9
5
Solving the system yields c1=1
2and c2=21
2.
Substituting c1and c2into the general solution yields the solu-
tion
x1=9cos3t−32sin3t
x2=5cos3t+105sin3t
Since the real part of the eigenvalues is zero, the trajectory is an
ellipse.
The ellipse is traced counterclockwise.
Answer(s) submitted:
•-3i
•(1/2)(cos(3t)(-3)-sin(3t)(1))+((21)/2)(cos(3t)(1)+sin(3t)(-3))
•(1/2)(cos(3t)(10)-sin(3t)(0))+((21)/2)(cos(3t)(0)+sin(3t)(10))
•Ellipse counterclockwise
(correct)
Correct Answers:
•<table border=’0’ cellspacing=’0’>
<tr>
<td> 3i, </td>
<td><table border=’0’ cellspacing=’5’><tr><td> -3+i </td></tr><tr><td> 10 </td></tr></table></td>
<td> -3i,</td>
<td><table border=’0’ cellspacing=’5’><tr><td> -3-i </td></tr><tr><td> 10 </td></tr></table></td>
</tr>
</table>
•9*[cos(3*t)-3*sin(3*t)]-5*sin(3*t)
•9*[9*sin(3*t)+sin(3*t)]+5*[cos(3*t)+3*sin(3*t)]
•ELLIPSE COUNTERCLOCKWISE
3
