Austin Cholley Jones MAT 275 ONLINE B Spring 2021
Assignment Section 7.8 Repeated Eigenvalues due 04/21/2021 at 11:59pm MST
1. (1 point)
Solve the initial value problem
dx
dt =1−4
4−7x,x(0) = 3
2
Give your solution in real form.
x(t) = .
Use the phase plotter pplane9.m in MATLAB to determine
how the solution curves (trajectories) of the system x0=Ax be-
have.
•A. All of the solution curves converge towards 0. (Sta-
ble node)
•B. The solution curves race towards zero and then veer
away towards infinity. (Saddle)
•C. The solution curves converge to different points.
•D. All of the solution curves run away from 0. (Unsta-
ble node)
Solution: The characteristic polynomial is det(A−λI) =
1−λ−4
4−7−λ
= (1−λ)(−7−λ) + 16 =λ2+6λ+9=
(λ+3)2.
Thus there is one repeated eigenvalue, λ=−3.
A−λI=A+3I=4−4
4−4. A possible eigenvector associated
to λ=−3 is v=1
1.
Note that there is only one linearly independent eigenvector as-
sociated with the double eigenvalue.
One solution of the system is given by x(1)(t) = veλt=1
1e−3t.
The other solution has the form vteλt+weλt, where wis a solu-
tion to (A−λI)w=v. Substituting the values of λand v, yields
the system
4−4
4−4w=1
1and a possible solution is w=5/4
1.
Thus x(2)(t) = 1
1te−3t+5/4
1e−3tand the general solution is
x(t) = c11
1e−3t+c2t+5/4
t+1e−3t
The initial conditions yields the system 3
2=c1+5c2
4
c1+c2, with
solution c1=−2,c2=4.
Substituting the values of c1and c2in the general solution,
yields the solution to the initial value problem:
x(t) = (3+4t)e−3t
(2+4t)e−3t
Since the eigenvalue is negative, all solution curves converge
toward zero.
Answer(s) submitted:
•2eˆ(-3t)(1)+4eˆ(-3t)(t(1)+((1/4)))
•2eˆ(-3t)(1)+4eˆ(-3t)(t(1)+0)
•A
(correct)
Correct Answers:
•eˆ(-3*t) * (3+4*t)
•eˆ(-3*t) * (2+4*t)
•A
2. (1 point)
Solve the initial value problem
dx
dt =2 1.5
−1.5−1x,x(0) = 3
−2
Give your solution in real form.
x(t) = .
Use the phase plotter pplane9.m in MATLAB to determine
how the solution curves (trajectories)of the system x0=Axbe-
have.
•A. The solution curves race towards zero and then veer
away towards infinity. (Saddle)
•B. All of the solution curves converge towards 0. (Sta-
ble node)
•C. All of the solution curves run away from 0. (Unsta-
ble node)
•D. The solution curves converge to different points.
Solution: The characteristic polynomial is det(A−λI) =
2−λ1.5
−1.5−1−λ
= (2−λ)(−1−λ) + 1.52=λ2−λ+0.25 =
(λ−0.5)2.
Thus there is one repeated eigenvalue, λ=0.5.
A−λI=A−0.5I=1.5 1.5
−1.5−1.5. A possible eigenvector
associated to λ=0.5 is v=1
−1.
Note that there is only one linearly independent eigenvector as-
sociated with the double eigenvalue.
1
One solution of the system is given by x(1)(t) = veλt=
1
−1e0.5t.
The other solution has the form vteλt+weλt, where wis a solu-
tion to (A−λI)w=v. Substituting the values of λand v, yields
the system
1.5 1.5
−1.5−1.5w=1
−1and a possible solution is w=
−1/3
1.
Thus x(2)(t) = 1
−1te0.5t+−1/3
1e0.5tand the general solu-
tion is
x(t) = c11
−1e0.5t+c2t−1/3
−t+1e0.5t
The initial conditions yields the system 3
−2=c1−c2
3
−c1+c2,
with solution c1=7/2,c2=3/2.
Substituting the values of c1and c2in the general solution,
yields the solution to the initial value problem:
x(t) = (3+1.5t)e0.5t
(−2−1.5t)e0.5t
Since the eigenvalue is positive, all solution curves run away
from zero.
Answer(s) submitted:
•-2(-1)eˆ((t/2))-(3/2)eˆ((t/2))((-1)t+(-(2/3)))
•-2(1)eˆ((t/2))-(3/2)eˆ((t/2))(1t)
•C
(correct)
Correct Answers:
•exp(t/2) * (3+1.5*t)
•-exp(t/2) * (2+1.5*t)
•C
3. (1 point)
Solve the initial value problem
dx
dt =−2.5 1.5
−1.5 0.5x,x(0) = 3
−1
Give your solution in real form.
x(t) = .
Use the phase plotter pplane9.m in MATLAB to determine
how the solution curves of the system x0=Axbehave.
•A. The solution curves race towards zero and then veer
away towards infinity. (Saddle)
•B. The solution curves converge to different points.
•C. All of the solution curves run away from 0. (Unsta-
ble node)
•D. All of the solution curves converge towards 0. (Sta-
ble node)
Solution: The characteristic polynomial is det(A−λI) =
−2.5−λ1.5
−1.5 0.5−λ
= (−2.5−λ)(0.5−λ) + 1.52=λ2+2λ+
1= (λ+1)2.
Thus there is one repeated eigenvalue, λ=−1.
A−λI=A+I=−1.5 1.5
−1.5 1.5. A possible eigenvector associ-
ated to λ=−1 is v=1
1.
Note that there is only one linearly independent eigenvector as-
sociated with the double eigenvalue.
One solution of the system is given by x(1)(t) = veλt=1
1e−t.
The other solution has the form vteλt+weλt, where wis a solu-
tion to (A−λI)w=v. Substituting the values of λand v, yields
the system
−1.5 1.5
−1.5 1.5w=1
1and a possible solution is w=1/3
1.
Thus x(2)(t) = 1
−1te−t+−1/3
1e−tand the general solution
is
x(t) = c11
1e−t+c2t+1/3
t+1e−t
The initial conditions yield the system 3
−1=c1+c2
3
c1+c2, with
solution c1=5,c2=−6.
Substituting the values of c1and c2in the general solution,
yields the solution to the initial value problem:
x(t) = (3−6t)e−t
(−1−6t)e−t
Since the eigenvalue is negative, all solution curves converge
towardis zero.
Answer(s) submitted:
•-1eˆ(-t)-(4/(.66666))eˆ(-t)(1t+(-.66666))
•-1eˆ(-t)-(4/(.66666))eˆ(-t)(1t)
•D
(score 0.666667)
Correct Answers:
•exp(-t) * (3-6*t)
•-exp(-t) * (1+6*t)
•D
4. (1 point)
Solve the initial value problem
dx
dt =3 9
−1−3x,x(0) = 2
4
2
Give your solution in real form.
x(t) = .
Solution: The characteristic polynomial is det(A−λI) =
3−λ9
−1−3−λ
= (3−λ)(−3−λ) + 9=λ2.
Thus there is one double eigenvalue, λ=0.
A−λI=A−0I=A=3 9
−1−3. A possible eigenvector as-
sociated to λ=0 is v=−3
1.
Note that there is only one linearly independent eigenvector as-
sociated with the double eigenvalue.
One solution of the system is given by x(1)(t) = veλt=−3
1.
The other solution has the form vteλt+weλt, where wis a solu-
tion to (A−λI)w=v. Substituting the values of λand v, yields
the system
3 9
−1−3w=−3
1and a possible solution is w=−4
1.
Thus x(2)(t) = −3
1t+−4
1and the general solution is
x(t) = c1−3
1+c2−3t−4
t+1
The initial conditions yield the system 2
4=−3c1−4c2
c1+c2,
with solution c1=18,c2=−14.
Substituting the values of c1and c2in the general solution,
yields the solution to the initial value problem:
x(t) = 2+42t
4−14t
Answer(s) submitted:
•4(-3)-14(-3t+(-1))
•4-14(1t)
(correct)
Correct Answers:
•2+42*t
•4-14*t
3
