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# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
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Ê Ã J
132
Section 7.3
18.
21.
´ 1 -1ë / \ ' 0ë (2) ' 1 N
x1 Hence x =
= , or x1 - x2 = 0.
V 3 -3, x2 V 0 , I1
2
, 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-À, i 4 - i 1-À
x2
2
0
The determinant of coefficients
is (1-À)2 - i(-i) = X2 - 2X, so the eigenvalues are À1 = 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
0
0
, or x1 + 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-À 0 0 ë / \ ' 0 ^
x1
2 1-À -2 x2 = 0
v 3 2 1-À , x3 V 0 ,
3
The determinant of coefficients is
(1-À)[(1-À) + 4] = 0, which has roots X = 1, 1 ± 2i. For X = 1, we then have 2x1 - 2x3 = 0 and 3x1 + 2x2 = 0. Choosing
x1 = 2 then yields
2
-3
as the eigenvector corresponding to
X = 1. For X = 1 + 2i we have
2x3 = 0 and 3x! + 2x2 - 2ix3 = 0,
-2ix! = 0, 2x! - 2ix2
0
yielding x1 = 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 " / \ ' 0 N
x1
2 -X 2 x2 = 0
V 4 2 3-Xy x3 V 0 ,
3
2 4" / \ ' 0 ë
4 x1
2 1 2 x2 = 0 ; hence there is only the single
4 4 ó x3 V 0 ,
2 3
relation 2x1 + 2x3 = 0 to be satisfied.
2
x
+
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
m 2 4" / \ ' 0 N
- x1
2 -8 2 x2 = 0
V 4 2 -5, x3 V 0 ,
3
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
2
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 =
´ 1 -1 2 4
-2 1 -1 V 3 -2 3Ó
and, using row reduction, the augmented
and
' 1 -1 2 0ë ' 1ë
matrix for A*y = 0 becomes 0 1- 3 0 . Thus y = c 3
V 0 0 0 0, 1I
hence (b,y) = b1 + 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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