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Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
Download (direct link): elementarydifferentialequat2001.pdf
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Let t = t0 be a fixed value of t in the interval 0 < t < 1. To determine whether x(1) (t0) and x(2) (t0) are
Section 7.3
131
15.
linearly dependent we must solve c1x<1)(t0)+c2x<2)(t0}=0. We have the augmented matrix
t0e
t0
0
Multiply the first row by (-t0) and add to the second row
to obtain
0
0
0
0
Thus, for example, we can choose c1 = 1 and c2 = -e 0, and hence the given vectors are linearly dependent at t0. Since t0 is arbitrary the vectors are linearly dependent at each point in the interval. However, there is no linear relation between x(1) and x(2) that is valid throughout the interval 0 < t < 1. For example, if t1 ^ t0, and if c1 and c2 are chosen as above, then
c,x(1) (t,) + c2x(2) (t )
( tl ? e 1 + t0 -e 0 ( 1 ' ( t. t0 ^ e 1 - e 0 0 ]
t1et1 1 v t1 , t1et1 - t1et0 11 v 0 y
Hence the given vectors must be linearly independent on 0 < t < 1. In fact, the same argument applies to any interval.
To find the eigenvalues and eigenvectors of the given
matrix we must solve
5-X -1 3 1-X
x.
V 2 J
0
The
determinant of coefficients is (5-X) (1-X) - (-1)(3) = 0, or X2 - 6X + 8 = 0. Hence X 1 = 2 and X2 = 4 are the eigenvalues. The eigenvector corresponding to X1 must
satisfy
x2
2
0
, or 3x1 - x2 = 0. If we let
x, = 1, then x2 = 3 and the eigenvector is x
(1)
1
or
any constant multiple of this vector. Similarly, the eigenvector corresponding to X2 must satisfy
t
1
0
e
t
1
e
x
1
132
Section 7.3
18.
21.
' 1 -1 ' / \ x1 ' 0 ' (2) Hence x = ' 1N
= , or x1 - x2 = 0.
V 3 -3, x2 2 V 0 , t1
, or
a multiple thereof.
Since a12 = a21, the given matrix is Hermitian and we
know in advance that its eigenvalues are real. To find the eigenvalues and eigenvectors we must solve
' 1-X i N - i 1-X
^1
x2
V 2 /
The determinant of coefficients
is (1-X)2 - i(-i) = X2 - 2X, so the eigenvalues are X2 = 0 and X2 = 2; observe that they are indeed real even though the given matrix has imaginary entries. The eigenvector corresponding to X1 must satisfy
1 i
-i 1
x
2
o
o
, or x2 + ix2 = 0. Note that the second
equation -ix1 + x2 = 0 is a multiple of the first. If x1 = 1, then x2 = i, and the eigenvector is
.(1)
1
(2)
1
. In a similar way we find that the eigenvector associated with X2 is x(
The eigenvalues and eigenvectors satisfy
^ 1-X 0 0 " / \ x1 ' 0N
2 1-X -2 x2 = 0
v 3 2 1-X , x3 3 V 0 ,
The determinant of coefficients is
(1-X)[(1-X) + 4] = 0, which has roots X = 1, 1 2i. For X = 1, we then have 2x2 - 2x3 = 0 and 3x2 + 2x2 = 0. Choosing
x2 = 2 then yields
2
-3
as the eigenvector corresponding to
X = 1. For X = 1 + 2i we have
2x3 = 0 and 3x2 + 2x2 - 2ix3 = 0,
-2ix2 = 0, 2x2 - 2ix2
yielding x2 = 0 and x3 = -ix2. Thus
is the eigenvector
corresponding to X = 1 + 2i. A similar calculation shows that
0
x
1
Section 7.3
133
"0"
is the eigenvector corresponding to X = 1 - 2i.
24. Since the given matrix is real and symmetric, we know that the eigenvalues are real. Further, even if there are repeated eigenvalues, there will be a full set of three linearly independent eigenvectors. To find the eigenvalues and eigenvectors we must solve
The determinant of
coefficients is (3-X)[-X(3-X)-4] - 2[2(3-X) -8] + 4[4+4X] = -X3 + 6X2 + 15X + 8. Setting this equal to zero and solving we find X1 = X2 = -1, X3 = 8. The eigenvectors
corresponding to X1 and X2 must satisfy
' 3-X 2 4 " / \ x1 ' 0 N
2 -X 2 x2 = 0
V 4 2 3-Xy x3 3 V 0 ,
' 4 2 4 " / \ x1 ' 0 '
2 1 2 x2 = 0 ; hence there is only the single
v 4 2 4 y x3 3 V 0 ,
relation 2x1 + x2 + 2x3 = 0 to be satisfied.
Consequently, two of the variables can be selected arbitrarily and the third is then determined by this equation. For example, if x1 = 1 and x3 = 1, then x2 =
1
(1)
4, and we obtain the eigenvector x if x1 = 1 and x2 = 0, then x3 = -1, and we have the (2)
Similarly,
1
eigenvector x
(1)
0
, which is linearly independent of
x . There are many other choices that could have been made; however, by Eq.(38) there can be no more than two linearly independent eigenvectors corresponding to the eigenvalue -1. To find the eigenvector corresponding to X3 we must solve
134
Section 7.4
1 U1 2 4 " / \ x1 ' 0 N
2 -8 2 x2 = 0
V 4 2 -5, x3 3 V 0 ,
Interchange the first and
second rows and use row reduction to obtain the equivalent system x1 - 4x2 + x3 = 0, 2x2 - x3 = 0. Since there are two equations to satisfy only one variable can be assigned an arbitrary value. If we let x2 = 1, then
k2k
x3 = 2 and x1 = 2, so we find that x
(3)
27. We are given that Ax = b has solutions and thus we have (Ax,y) = (b,y). From Problem 26, though,
(Ax,y)= (x, A*y) = 0. Thus (b,y) = 0. For Example 2,
A* = A =
-2 1 -1 V 3 -2 3y
and, using row reduction, the augmented
and
' 1 -1 2 0 ' ' 1N
matrix for A*y = 0 becomes 0 1- 3 0 . Thus y = c 3
V 0 0 0 0, t1
hence (b,y) = b3 + 3b2 + b3 = 0.
1
Section 7.4, Page 371
Use Mathematical Induction. It has already been proven that if x(1) and x(2) are solutions, then so is
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