Elizabeth Arena Sharma MAT 275 ONLINE B Summer 2019
Assignment Section 7.6 Complex Eigenvalues due 08/13/2019 at 11:59pm MST
1. (1 point) Suppose Ais a 2 ×2 real matrix with an eigen-
value λ=1+2iand 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:
•-cos(2t)-sin(2t)
(correct)
Correct Answers:
•<table border=’0’ cellspacing=’5’><tr><td> eˆ(1*t)*(-[cos(2*t)]-sin(2*t)) </td></tr><tr><td> eˆ(1*t)*(-[sin(2*t)]) </td></tr></table>
<br />
<table border=’0’ cellspacing=’5’><tr><td> eˆ(1*t)*[cos(2*t)-sin(2*t)] </td></tr><tr><td> eˆ(1*t)*cos(2*t) </td></tr></table>
2. (1 point)
Consider the Initial Value Problem:
x0
1=6x1+3x2
x0
2=−15x1−6x2
,x1(0) = 8
x2(0) = 6
(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
6−λ3
−15 −6−λ
=λ2+32. Thus the eigenvalues are λ=±3i.
A possible eigenvector associated to λ=3iis −2−i
5=
−2
5+i−1
0
One solution is then given by
−2
5cos3t−−1
0sin3t=−2 cos 3t+sin3t
5cos3t
while the other is
−2
5sin3t+−1
0cos3t=−2 sin 3t−cos3t
5sin3t
The general solution is
x(t) = c1−2 cos 3t+sin3t
5cos3t+c2−2 sin 3t−cos3t
5sin3t
Using the initial condition gives x(0) = −2c1−c2
5c1=8
6
Solving the system yields c1=6
5and c2=−52
5.
Substituting c1and c2into the general solution yields the solu-
tion
x1=8cos3t+22sin3t
x2=6cos3t−52sin3t
Since the real part of the eigenvalues is zero, the trajectory
is an ellipse.
The ellipse has clockwise orientation.
Answer(s) submitted:
•3i
•8cos(3t)+22sin(3t)
•6cos(3t)-52sin(3t)
•Ellipse clockwise
(correct)
Correct Answers:
•<table border=’0’ cellspacing=’0’>
<tr>
<td> 3i, </td>
<td><table border=’0’ cellspacing=’5’><tr><td> -2-i </td></tr><tr><td> 5 </td></tr></table></td>
<td> -3i,</td>
<td><table border=’0’ cellspacing=’5’><tr><td> -2+i </td></tr><tr><td> 5 </td></tr></table></td>
</tr>
</table>
•8*[2*sin(3*t)+cos(3*t)]+6*sin(3*t)
•6*[cos(3*t)-2*sin(3*t)]-8*[4*sin(3*t)+sin(3*t)]
•ELLIPSE CLOCKWISE
1
3. (1 point) Consider the Initial Value Problem
dx
dt =−4−5
5−4x,x(0) = 3
9
(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
−4−λ−5
5−4−λ
= (−4−λ)2+52. Thus the eigenvalues are
λ=−4±5i.
A possible eigenvector associated to λ=−4+5iis i
1=
0
1+i1
0
One solution is then given by e−4t−sin5t
cos5twhile the other is
e−4tcos5t
sin5t
The general solution is
x(t) = c1e−4t−sin5t
cos5t+c2e−4tcos5t
sin5t
Using the initial condition gives x(0) = c2
c1=3
9
Solving the system yields c1=9 and c2=3.
Substituting c1and c2into the general solution yields the solu-
tion
x(t) = e−4t(−9sin5t+3 cos 5t)
e−4t(9cos5t+3sin5t)
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:
•-4+5i
•eˆ(-4t)(6ieˆ(5it)-3ieˆ(-5it))
•6eˆ((-4-5i)t)+3ˆe((-4-5i)t)
•Spiral inward counterclockwise
(score 0.5)
Correct Answers:
•<table border=’0’ cellspacing=’0’>
<tr>
<td> -4+5i, </td>
<td><table border=’0’ cellspacing=’5’><tr><td> i </td></tr><tr><td> 1 </td></tr></table></td>
<td> -4-5i,</td>
<td><table border=’0’ cellspacing=’5’><tr><td> -i </td></tr><tr><td> 1 </td></tr></table></td>
</tr>
</table>
•[3*cos(5*t)-9*sin(5*t)]*eˆ(-4*t)
•[3*sin(5*t)+9*cos(5*t)]*eˆ(-4*t)
•SPIRAL INWARD COUNTERCLOCKWISE
4. (1 point) Consider the initial value problem
dx
dt =5−4
4 5 x,x(0) = 4
3
(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
5−λ−4
4 5 −λ
= (5−λ)2+42. Thus the eigenvalues are λ=
5±4i.
A possible eigenvector associated to λ=5+4iis i
1=0
1+
i1
0
One solution is then given by e5t−sin4t
cos4twhile the other is
e5tcos4t
sin4t
The general solution is
x(t) = c1e5t−sin4t
cos4t+c2e5tcos4t
sin4t
Using the initial condition gives x(0) = c2
c1=4
3
Solving the system yields c1=3 and c2=4.
Substituting c1and c2into the general solution yields the solu-
tion
x(t) = e5t(−3sin4t+4 cos 4t)
e5t(3cos4t+4sin4t)
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:
•5+4i
•eˆ(5t)(3cos(4t)+4sin(t))
•eˆ(5t)(3cos(4t)+4sin(t))
•Spiral outward counterclockwise
(score 0.5)
Correct Answers:
•<table border=’0’ cellspacing=’0’>
<tr>
<td> 5+4i, </td>
<td><table border=’0’ cellspacing=’5’><tr><td> i </td></tr><tr><td> 1 </td></tr></table></td>
<td> 5-4i,</td>
<td><table border=’0’ cellspacing=’5’><tr><td> -i </td></tr><tr><td> 1 </td></tr></table></td>
</tr>
</table>
•eˆ(5*t)*[4*cos(4*t)-3*sin(4*t)]
•eˆ(5*t)*[4*sin(4*t)+3*cos(4*t)]
•SPIRAL OUTWARD COUNTERCLOCKWISE
5. (1 point)
Consider the initial value problem
dx
dt =−3−3
6 3 x,x(0) = 8
2
(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
−3−λ−3
6 3 −λ
=λ2+32. Thus the eigenvalues are λ=±3i.
A possible eigenvector associated to λ=3iis −1+i
2=
−1
2+i1
0
One solution is then given by
−1
2cos3t−1
0sin3t=−1 cos 3t−sin3t
2cos3t
while the other is
−1
2sin3t+1
0cos3t=−1 sin 3t+cos3t
2sin3t
The general solution is
x(t) = c1−1 cos 3t−sin3t
2cos3t+c2−1 sin 3t+cos3t
2sin3t
Using the initial condition gives x(0) = −1c1+c2
2c1=8
2
Solving the system yields c1=1 and c2=9.
Substituting c1and c2into the general solution yields the solu-
tion
x1=8cos3t−10sin3t
x2=2cos3t+18sin3t
Since the real part of the eigenvalues is zero, the trajectory is an
ellipse.
The ellipse is traced counterclockwise.
Answer(s) submitted:
•3i
•8cos(3t)-10sin(3t)
•2cos(3t)+18sin(3t)
•Ellipse counterclockwise
(correct)
Correct Answers:
•<table border=’0’ cellspacing=’0’>
<tr>
<td> 3i, </td>
<td><table border=’0’ cellspacing=’5’><tr><td> -1+i </td></tr><tr><td> 2 </td></tr></table></td>
<td> -3i,</td>
<td><table border=’0’ cellspacing=’5’><tr><td> -1-i </td></tr><tr><td> 2 </td></tr></table></td>
</tr>
</table>
•8*[cos(3*t)-sin(3*t)]-2*sin(3*t)
•8*[sin(3*t)+sin(3*t)]+2*[cos(3*t)+sin(3*t)]
•ELLIPSE COUNTERCLOCKWISE
3
