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Mathematics LibreTexts

4.4E: Exercises

This page is a draft and is under active development. 

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In Exercises (4.4E.1) to (4.4E.15), find the general solution.

Exercise 4.4E.1

y=[1221]y

Exercise 4.4E.2

y=14[5335]y

Answer

14|54λ3354λ|=(λ+1/2)(λ+2).
Eigenvectors  associated with λ1=1/2  satisfy
[3344][x1x2]=[00],
so x1=x2.  Taking x2=1 yields
y1=[11]et/2.
Eigenvectors  associated with λ2=2
satisfy
[343411][x1x2]=[00],
so x1=x2.  Taking x2=1 yields
y2=[11]e2t. Hence
 y=c1[11]et/2+c2[11]e2t.

Exercise 4.4E.3

y=15[43211]y

Exercise 4.4E.4

y=[1411]y

Answer

|1λ411λ|=(λ1)(λ+3).
Eigenvectors  associated with λ1=3 satisfy
[2412][x1x2]=[00],
so x1=2x2.  Taking x2=1 yields
y1=[21]e3t.
Eigenvectors  associated with λ2=1  satisfy
[2412][x1x2]=[00],
so x1=2x2.  Taking x2=1 yields
y2=[21]et.
Hence  y=c1[21]e3t+c2[21]et.

Exercise 4.4E.5

y=[2411]y

Exercise 4.4E.6

y=[4321]y

Answer

|4λ321λ|=(λ2)(λ1).
Eigenvectors  associated with λ1=2  satisfy
[2323][x1x2]=[00],
so x1=32x2.  Taking x2=2 yields
y1=[32]e2t.
Eigenvectors  associated with λ2=1 satisfy
[3322][x1x2]=[00],
so x1=x2.  Taking x2=1 yields
y2=[11]et.
Hence
 y=c1[32]e2t+c2[11]et.

Exercise 4.4E.7

y=[6312]y

Exercise 4.4E.8

y=[112123411]y

Answer

|1λ1212λ3411λ|=(λ+3)(λ+1)(λ2).
The eigenvectors associated with
 with λ1=3 satisfy the system with  augmented matrix
[412011304120],
which is row equivalent to
[101001200000].
Hence x1=x3 and x2=2x3.  Taking x3=1 yields
y1=[121]e3t.
The eigenvectors associated with
 with λ2=1 satisfy the system with  augmented matrix
[212011304100],
which is row equivalent to
[101001400000].
Hence x1=x3 and x2=4x3.  Taking x3=1 yields
y2=[141]et.
The eigenvectors associated with
 with λ3=2 satisfy the system with  augmented matrix
[112014304130],
which is row equivalent to
[101001100000].
Hence x1=x3 and x2=x3.  Taking x3=1 yields
y3=[111]e2t. Hence
 y=c1[121]e3t+c2[141]et+c3[111]e2t.

Exercise 4.4E.9

y=[648404846]y

Exercise 4.4E.10

y=[358112111]y

Answer

|3λ5811λ2111λ|=(λ1)(λ+2)(λ2).
The eigenvectors associated with
 with λ1=1 satisfy the system with  augmented matrix
[258012201120],
which is row equivalent to [10230014300000]

Hence x1=23x3 and x2=43x3.  Taking x3=3
yields
y1=[243]et.
The eigenvectors associated with
 with λ2=2 satisfy the system with  augmented matrix
[558011201110],
which is row equivalent to
[110000100000].
Hence x1=x2 and x3=0.  Taking x2=1 yields
y2=[110]e2t.
The eigenvectors associated with
 with λ3=2 satisfy the system with  augmented matrix
[158013201130],
which is row equivalent to
[10740015400000].
Hence x1=74x3 and x2=54x3.  Taking x3=4
yields
y3=[754]e2t. Hence
y=c1[243]et+c2[110]e2t+c3[754]e2t.

 

Exercise 4.4E.11

y=[11212410617]y

Exercise 4.4E.12

y=[414432111]y

Answer

|4λ1443λ2111λ|=(λ3)(λ+2)(λ+1).
The eigenvectors associated
 with λ1=3 satisfy the system with  augmented matrix
[114046201140],
which is row equivalent to
[1011001700000].
Hence  x1=11x3 and x2=7x3.  Taking x3=1 yields
y1=[1171]e3t.
The eigenvectors associated
 with λ2=2 satisfy the system with  augmented matrix
[614041201110],
which is row equivalent to
[101001200000].
Hence x1=x3 and x2=2x3.  Taking x3=1 yields
y2=[121]e2t.
The eigenvectors associated
 with λ3=1 satisfy the system with  augmented matrix
[514042201100],
which is row equivalent to
[101001100000].
Hence  x1=x3 and x2=x3.  Taking x3=1 yields
y3=[111]et. Hence
 y=c1[111711]e3t+c2[121]e2t+c3[111]et.

Exercise 4.4E.13

y=[226262222]y

Exercise 4.4E.14

y=[32227210105]y

Answer

|3λ2227λ210105λ|=(λ+5)(λ5)2.
The eigenvectors associated
 with λ1=5 satisfy the system with  augmented matrix
[822021220101000],
which is row equivalent to
[10150011500000].
Hence  x1=15x3 and x2=15x3.  Taking
x3=5 yields
y1=[115]e5t.
The eigenvectors associated
 with λ2=5 satisfy the system with  augmented matrix
[222022201010100],
which is row equivalent to
[111000000000].
Hence x1=x2x3.  Taking x2=0 and x3=1 yields
y2=[101]e5t.
 Taking x2=1 and x3=0 yields
y3=[110]e5t. Hence
 y=c1[115]e5t+c2[101]e5t+c3[110]e5t.

Exercise 4.4E.15

y=[311351624]y

In Exercises (4.4E.16) to (4.4E.27), solve the initial value problem.

Exercise 4.4E.16

y=[7467]y,y(0)=[24]

Answer

|7λ467λ|=(λ5)(λ+5).
Eigenvectors  associated with λ1=5  satisfy
[12462][x1x2]=[00],
so x1=x23.  Taking x2=3 yields
y1=[13]e5t.
Eigenvectors  associated with λ2=5
satisfy
[24612][x1x2]=[00],
so x1=2x2.  Taking x2=1 yields
y2=[21]e5t.
The general solution is
y=c1[13]e5t+c2[21]e5t.
Now y(0)=[24]c1[13]+c2[21]=[24],
so c1=2 and c2=2. Therefore,
y=[26]e5t+[42]e5t.

Exercise 4.4E.17

y=16[7222]y,y(0)=[03]

Exercise 4.4E.18

y=[21122415]y,y(0)=[53]

Answer

|21λ122415λ|=(λ9)(λ+3).
Eigenvectors  associated with λ1=9  satisfy
[12122424][x1x2]=[00],
so x1=x2.  Taking x2=1 yields
y1=[11]e9t.
Eigenvectors  associated with λ2=3
[24122412][x1x2]=[00],
so x1=12x2.  Taking x2=2 yields
y2=[12]e3t.
The general solution is
y=c1[11]e9t+c2[12]e3t.
Now  y(0)=[53]c1[11]+c2[12]=[53], so c1=7
and c2=2. Therefore,
y=[77]e9t[24]e3t.

Exercise 4.4E.19

y=[7467]y,y(0)=[17]

Exercise 4.4E.20

y=16[120410003]y,y(0)=[471]

Answer

|16λ1302316λ00012λ|=(λ+1/2)(λ1/2)2.
The eigenvectors associated with
 with λ1=1/2 satisfy the system with  augmented matrix
[2313002313000010],
which is row equivalent to
[1120000100000].
Hence x1=x22 and x3=0.  Taking x2=2 yields
y1=[120]et/2.
The eigenvectors associated with
 with λ2=λ3=1/2 satisfy the system with  augmented
matrix
[1313002323000000],
which is row equivalent to
[110000000000].
Hence x1=x2 and x3 is arbitrary.  Taking x2=1 and x3=0
yields
y2=[110]et/2.
Taking x2=0 and x3=1 yields
y3=[001]et/2.
 The general solution is
y=c1[120]et/2+c2[110]et/2+c3[001]et/2.
Now y(0)=[471]c1[120]+c2[110]+c3[001]et/2=[471],
so c1=1, c2=5, and c3=1. Hence
y=[120]et/2+[550]et/2+[001]et/2.

Exercise 4.4E.21

y=13[223443210]y,y(0)=[115]

Exercise 4.4E.22

y=[638212335]y,y(0)=[011]

Answer

|6λ3821λ2335λ|=(λ1)(λ+2)(λ3).
The eigenvectors associated
 with λ1=1 satisfy the system with  augmented matrix
[538020203360],
which is row equivalent to
[101001100000].
Hence  x1=x3 and x2=x3.  Taking x3= yields
y1=[111]et.
The eigenvectors associated
 with λ2=2 satisfy the system with  augmented matrix
[838023203330],
which is row equivalent to
[101001000000].
Hence x1=x3 and x2=0.  Taking x3=1 yields
y2=[101]e2t.
The eigenvectors associated
 with λ3=3 satisfy the system with  augmented matrix
[338022203380],
which is row equivalent to
[110000100000].
Hence  x1=x2 and x3=0.  Taking x2=1 yields
y3=[110]e3t.
The general solution is
y=c1[111]et+c2[101]e2t+c3[110]e3t.
Now y(0)=[011]c1[111]+c2[101]+c3[110]=[011], so
c1=2, c2=3, and c3=1. Therefore,
y=[222]et[303]e2t+[110]e3t.

Exercise 4.4E.23

y=13[247155441]y,y(0)=[413]

Exercise 4.4E.24

y=[3011127103]y,y(0)=[276]

Answer

|3λ01112λ7103λ|=(λ2)(λ+2)(λ4).
The eigenvectors associated with
 with λ1=2 satisfy the system with  augmented matrix
[1010114701010],
which is row equivalent to
[101001100000].
Hence x1=x3 and x2=x3.  Taking x3=1 yields
y1=[111]e2t.
The eigenvectors associated with
 with λ2=2 satisfy the system with  augmented matrix
[5010110701050],
which is row equivalent to
[100000100000].
Hence x1=x3=0 and x2 is arbitrary.  Taking x3=1 yields
y2=[010]e2t.
The eigenvectors associated with
 with λ3=4 satisfy the system with  augmented matrix
[1010116701010],
which is row equivalent to
[101001300000].
Hence x1=x3 and x2=3x3.  Taking x3=1 yields
y3=[131]e4t.
 The general solution is
y=c1[111]e2t+c2[010]e2t+c3[131]e4t.
Now y(0)=[276]c1[111]+c2[010]+c3[131]=[276], so c1=2, c2=3, and
c3=4. Hence
y=[222]e2t[030]e2t+[4124]e4t

Exercise 4.4E.25

y=[251411453]y,y(0)=[8104]

Exercise 4.4E.26

y=[310420442]y,y(0)=[7102]

Answer

|3λ1042λ0442λ|=(λ+1)(λ2)2.
The eigenvectors associated
 with λ1=1 satisfy the system with  augmented matrix
[410041004430],
which is row equivalent to
[1014001100000]
Hence x1=x2/4 and x2=x3.  Taking x3=4 yields
y1=[144]et.
The eigenvectors associated with
 with λ2=λ3=2 satisfy the system with  augmented
matrix
[110044004400],
which is row equivalent to
[110000000000].
Hence x1=x2 and x3 is arbitrary.  Taking x2=1 and x3=0
yields
y2=[110]e2t.
Taking x2=0 and x3=1 yields
y3=[001]e2t.
 The general solution is
y=c1[144]et+c2[110]e2t+c3[001]e2t.
Now y(0)=[7102]c1[144]+c2[110]+c3[001]=[7102],
 so c1=1, c2=6, and
c3=2. Hence
 y=[144]et+[662]e2t.

Exercise 4.4E.27

y=[226262222]y,y(0)=[6107]

Exercise 4.4E.28

Let A be an n×n constant matrix. Then Theorem (4.2.1) implies that the solutions of

y=Ay

are all defined on (,).

(a) Use Theorem (4.2.1) to show that the only solution of (???) that can ever equal the zero vector is y0.

(b) Suppose y1 is a solution of (???) and y2 is defined by y2(t)=y1(tτ), where τ is an arbitrary real number. Show that y2 is also a solution of (???).

(c) Suppose y1 and y2 are solutions of (???) and there are real numbers t1 and t2 such that y1(t1)=y2(t2). Show that y2(t)=y1(tτ) for all t, where τ=t2t1.
Hint: Show that y1(tτ) and y2(t) are solutions of the same initial value problem for (???), and apply the uniqueness assertion of Theorem (4.2.1).

Answer

{\bf(a)}  If y(t0)=0, then y is the solution
of the initial value problem y=Ay, y(t0)=0. Since y0 is a solution of this problem,
Theorem~4.2.1 implies the conclusion.

{\bf(b)} It is given that y1(t)=Ay1(t) for all t.
Replacing t by \)t-\tau\) shows that y1(tτ)=Ay1(tτ)=Ay2(t) for all t. Since y2(t)=y1(tτ) by the chain rule, this implies that y2(t)=Ay2(t) for all t.

{\bf(c)} If z(t)=y1(tτ), then z(t2)=y1(t1)=y2(t2); therefore z and y2 are
both solutions of the initial value problem y=Ay, y(t2)=k, where k=y2(t2).

In Exercises (4.4E.29) to (4.4E.34), describe and graph trajectories of the given system.

Exercise 4.4E.29

y=[1111]y

Exercise 4.4E.30

y=[43211]y

Exercise 4.4E.31

y=[93111]y

Exercise 4.4E.32

y=[11054]y

Exercise 4.4E.33

y=[54110]y

Exercise 4.4E.34

y=[7135]y

Exercise 4.4E.35

Suppose the eigenvalues of the 2×2 matrix A are λ=0 and μ0, with corresponding eigenvectors x1 and x2. Let L1 be the line through the origin parallel to x1.

(a) Show that every point on L1 is the trajectory of a constant solution of y=Ay.

(b) Show that the trajectories of nonconstant solutions of y=Ay are half-lines parallel to x2 and on either side of L1, and that the direction of motion along these trajectories is away from L1 if μ>0, or toward L1 if μ<0.

The matrices of the systems in Exercises (4.4E.36) to (4.4E.41) are singular. Describe and graph the trajectories of nonconstant solutions of the given systems.

Exercise 4.4E.36

y=[1111]y

Exercise 4.4E.37

y=[1326]y

Exercise 4.4E.38

y=[1313]y

Exercise 4.4E.39

y=[1212]y

Exercise 4.4E.40

y=[4411]y

Exercise 4.4E.41

y=[3131]y

Exercise 4.4E.42

Let P=P(t) and Q=Q(t) be the populations of two species at time t, and assume that each population would grow exponentially if the other didn't exist; that is, in the absence of competition,

P aPandQ=bQ,

where a and b are positive constants. One way to model the effect of competition is to assume that the growth rate per individual of each population is reduced by an amount proportional to the other population, so (???) is replaced by

P=aPαQQ=βP+bQ,

where α and β are positive constants. (Since negative population doesn't make sense, this system holds only while P and Q are both positive.) Now suppose P(0)=P0>0 and Q(0)=Q0>0.

(a) For several choices of a, b, α, and β, verify experimentally (by graphing trajectories of (???) in the P-Q plane) that there's a constant ρ>0 (depending upon a, b, α, and β) with the following properties:
(i) If Q0>ρP0, then P decreases monotonically to zero in finite time, during which Q remains positive.
(ii) If Q0<ρP0, then Q decreases monotonically to zero in finite time, during which P remains positive.

(b) Conclude from part (a) that exactly one of the species becomes extinct in finite time if Q0ρP0. Determine experimentally what happens if Q0=ρP0.

(c) Confirm your experimental results and determine γ by expressing the eigenvalues and associated eigenvectors of

A=[aαβb]

in terms of a, b, α, and β, and applying the geometric arguments developed at the end of this section.

Answer

The characteristic polynomial of \)A\) is
\)p(\lambda)=\lambda^2-(a+b)+ab-\alpha\beta\), so the eigenvalues of \)A\)
are λ1=a+bγ2 and
λ1=a+b+γ2, where
γ=(ab)2+4αβ; x1=\ctwocolba+γ2β and x2=\ctwocolbaγ2β are associated eigenvectors. Since
γ>|ba|, if \)L_1\) and \)L_2\) are lines through the origin
parallel to x1 and x2, then \)L_1\) is in the first and
third quadrants and \)L_2\) is in the second and fourth quadrants. The
slope of \)L_1\) is ρ=2βba+γ>0. If \)Q_0=\rho
P_0\) there are three possibilities: (i) if αβ=ab, then
λ1=0 and \)P(t)=P_0\), \)Q(t)=Q_0\) for all \)t>0\); (ii) if
αβ<ab, then λ1>0 and
limtP(t)=limtQ(t)= (monotonically);
(iii) if αβ>ab, then λ1<0 and
limtP(t)=limtQ(t)=0 (monotonically). Now
suppose \)Q_0\ne\rho P_0\), so that the trajectory cannot intersect
\)L_1\), and assume for the moment that (A) makes sense for all \)t>0\);
that is, even if one or the other of \)P\) and \)Q\) is negative. Since
λ2>0 it follows that either limtP(t)= or
limtQ(t)= (or both), and the trajectory is
asymptotically parallel to \)L_2\). Therefore,the trajectory must cross
into the third quadrant (so \)P(T)=0\) and \)Q(T)>0\) for some finite t)
if \)Q_0>\rho P_0\), or into the fourth quadrant (so \)Q(T)=0\) and
\)P(T)>0\) for some finite t) if \)Q_0<\rho P_0\).


This page titled 4.4E: Exercises is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by William F. Trench.

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