Austin Cholley Jones MAT 275 ONLINE B Spring 2021
Assignment Section 7.5 Homogeneous systems Constant Coefficients due 04/20/2021 at 11:59pm MST
1. (1 point) Suppose that the matrix A has the following
eigenvalues and eigenvectors:
λ1=4 with ~v1=−2
2.
and
λ2=−4 with ~v2=4
2.
Write the solution to the linear system~r0=A~rin the follow-
ing forms.
A. In eigenvalue/eigenvector form:
x(t)
y(t)=c1 et+c2 et
B. As two equations: (write ”c1” and ”c2” for c1and c2)
x(t) =
y(t) =
Note: if you are feeling adventurous you could use other
eigenvectors like 4~v1or −3~v2.
Solution: (a) In eigenvalue/eigenvector form, the solution is
given by
x(t)
y(t)=c1~v1eλ1t+c2~v2eλ2t=c1−2
2e4t+c24
2e−4t
(b) As two equations the solution is
x(t) = −2c1e4t+4c2e−4t
y(t) = 2c1e4t+2c2e−4t
Answer(s) submitted:
•-2
•
(score 0.5)
Correct Answers:
•<table border=’0’><tr>
<td> \(c_1\ \biggl\lbrack\) </td>
<td><table border=’0’ cellspacing=’10’><tr><td> -2 </td></tr><tr><td> 2 </td></tr></table></td>
<td> \(\biggr\rbrack\ eˆ{4 t} \ \ + \ \ c_2\ \biggl\lbrack\) </td>
<td><table border=’0’ cellspacing=’10’><tr><td> 4 </td></tr><tr><td> 2 </td></tr></table></td>
<td> \(\biggr\rbrack\ eˆ{-4 t} \)</td>
</tr></table>
•<table border=’0’>
<tr><td style=’text-align:left’>\(x(t) = \) 4*c2*eˆ(-4*t)-2*c1*eˆ(4*t) </td></tr>
<tr><td style=’text-align:left’>\(y(t) = \) 2*c1*eˆ(4*t)+2*c2*eˆ(-4*t) </td></tr> </table>
2. (1 point)
Solve the IVP
dx
dt =−16 0
−16 0x,x(0) = 6
12
x(t) = .
Solution: The eigenvalues of the coefficient matrix are λ=
−16 and λ=0 with associated eigenvectors 2
2and 0
2re-
spectively.
The general solution is then
x(t) = c1e−16t2
2+c20
2
The initial conditions yield the system
6=2c1
12 =2c1+2c2
Solving the system yields c1=3 and c2=3.
Substituting into the general solution gives
x(t) = 3e−16t2
2+30
2
Thus
x(t) = 6e−16t
6e−16t+6
Answer(s) submitted:
•18
•12
(incorrect)
Correct Answers:
•3*eˆ(-16*t)*2 + 3*0
•3*eˆ(-16*t)*2 + 3*2
3. (1 point) Consider the system of differential equations
dx
dt =−1.4x+0.5y,
dy
dt =2.5x−3.4y.
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.
1
•A. All of the solution curves converge towards 0. (Sta-
ble node)
•B. All of the solution curves run away from 0. (Unsta-
ble node)
•C. The solution curves converge to different points.
•D. The solution curves race towards zero and then veer
away towards infinity. (Saddle)
The solution to the above differential equation with initial
values x(0) = 6,y(0) = 3 is
x(t) = ,
y(t) = .
Solution: The characteristic polynomial of the coefficient
matrix is
−1.4−λ1
2
2.5−3.4−λ
= (−1.4−λ)(−3.4−λ)−1.25 =λ2+4.8λ+3.51
Using the quadratic formula, we find the eigenvalues λ1=−3.9
and λ2=−0.9.
A possible eigenvector associated to λ1is 1
−5,while a possi-
ble eigenvector associated to λ2is 1
1.
The general solution is then given by
x(t) = c1e−3.9t+c2e−0.9t
y(t) = −5c1e−3.9t+c2e−0.9t
The initial conditions yield the system
6=c1+c2
3=−5c1+c2
Solving the system gives c1=1
2and c2=11
2.
Subsituting into the general solution yields the solution
x(t) = 1
2e−3.9t+11
2e−0.9t
y(t) = −5
2e−3.9t+11
2e−0.9t
Since the eigenvalues are real and both negative , All of the
solution curves converge towards 0. (Stable node)
Answer(s) submitted:
•1.1
•1.1
•A
•
•
(score 0.2)
Correct Answers:
•-3.9
•-0.9
•A
•1*0.5*exp(-3.9*t)+11*0.5*exp(-0.9*t)
•1*(-2.5)*exp(-3.9*t)+11*(0.5)*exp(-0.9*t)
4. (1 point)
Consider the systems of differential equations
dx
dt =0.4x+1.5y,
dy
dt =0.5x−0.6y.
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. 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)
The solution to the above differential equation with initial
values x(0) = 9,y(0) = 8 is
x(t) = ,
y(t) = .
Solution: The characteristic polynomial of the coefficient
matrix is
0.4−λ3
2
0.5−0.6−λ
= (0.4−λ)(−0.6−λ)−0.75 =λ2+0.2λ−0.99
Using the quadratic formula, we find the eigenvalues λ1=−1.1
and λ2=0.9.
A possible eigenvector associated to λ1is 1
−1,while a possi-
ble eigenvector associated to λ2is 3
1.
The general solution is then given by
x(t) = c1e−1.1t+3c2e0.9t
y(t) = −c1e−1.1t+c2e0.9t
The initial conditions yield the system
9=c1+3c2
8=−c1+c2
Solving the system gives c1=−15
4and c2=17
4.
Subsituting into the general solution yields the solution
x(t) = −15
4e−1.1t+51
4e0.9t
y(t) = 15
4e−1.1t+17
4e0.9t
Since the eigenvalues are real and of opposite signs, The so-
lution curves race towards zero and then veer away towards
infinity. (Saddle)
2
Answer(s) submitted:
•-1.1
•.9
•A
•1.5eˆ(-1.1t)7(1.5)eˆ(.9t)
•
(score 0.6)
Correct Answers:
•-1.1
•0.9
•A
•-2.5*1.5*exp(-1.1*t)+8.5*1.5*exp(0.9*t)
•-2.5*(-1.5)*exp(-1.1*t)+8.5*(0.5)*exp(0.9*t)
5. (1 point)
Consider the systems of differential equations
dx
dt =0.5x−0.8y,
dy
dt =−0.2x+1.1y.
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 run away from 0. (Unsta-
ble node)
•B. All of the solution curves converge towards 0. (Sta-
ble node)
•C. The solution curves converge to different points.
•D. The solution curves race towards zero and then veer
away towards infinity. (Saddle)
The solution to the above differential equation with initial
values x(0) = 2,y(0) = 9 is
x(t) = ,
y(t) = .
Solution: The characteristic polynomial of the coefficient
matrix is
0.5−λ−4
5
−0.2 1.1−λ
= (0.5−λ)(1.1−λ)−0.16 =λ2−1.6λ+0.39
Using the quadratic formula, we find the eigenvalues λ1=0.3
and λ2=1.3.
A possible eigenvector associated to λ1is 4
1,while a possible
eigenvector associated to λ2is −1
1.
The general solution is then given by
x(t) = 4c1e0.3t−c2e1.3t
y(t) = c1e0.3t+c2e1.3t
The initial conditions yield the system
2=4c1−c2
9=c1+c2
Solving the system gives c1=11
5and c2=34
5.
Subsituting into the general solution yields the solution
x(t) = 44
5e0.3t−34
5e1.3t
y(t) = 11
5e0.3t+34
5e1.3t
Since the eigenvalues are real and both positive, All of the
solution curves run away from 0. (Unstable node)
Answer(s) submitted:
•.3
•1.3
•A
•((52)/5)eˆ(.3t)-((32)/5)eˆ(1.3t)
•((13)/5)eˆ(.3t)+((32)/5)eˆ(1.3t)
(score 0.6)
Correct Answers:
•0.3
•1.3
•A
•-11*-0.8*exp(0.3*t)+8.5*-0.8*exp(1.3*t)
•-11*(-0.2)*exp(0.3*t)+8.5*(0.8)*exp(1.3*t)
3
