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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.
Download (direct link): elementarydifferentialequations2001.pdf
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Hence one solution of the given D.E. is
By analogy with the scalar case
1
t
-3
considered in Section 5.5 and Example 2 of this section, we seek a second solution of the form x = nt-3lnt + Zt-3. Substituting this expression into the D.E. we find that n and Z satisfy the equations (A + 3I)n = 0 and
(A + 3I)Z = , where A =
and I is the identity
matrix. Thus n =
1
, from above, and Z is found to be
0
-1/4 .(2)
Thus a second solution is
' 1 ' 1
(t) = t-3lnt +
1 1
-
15. All solutions of the given system approach zero as
t ^ if and only if the eigenvalues of the coefficient matrix either are real and negative or else are complex with negative real part. Write down the determinantal equation satisfied by the eigenvalues and determine when the eigenvalues are as stated.
17a. The eigenvalues and eigenvectors of the coefficient
' 1- 1 '?1 ' ' 0
1
r
matrix satisfy 2 1-r -1 ?2 = 0 . The determinant
V-3 - v?3, v 0,
4
2
of coefficients is 8 - 12r + - r3 = (2-r) 3, so the
2
r
\D
eigenvalues are r1 = r2 = r3 = 2 . The eigenvectors
corresponding to this triple eigenvalue satisfy
Using row reduction we can reduce
this to the equivalent system ^ - ?2 - ?3 = 0, and ?2 + ?3 = 0. If we let ?2 = 1, then ?3 = -1 and = 0,
\ '?1 ' ' 0 '
-1 1 1
2 -1 -1 ?2 = 0
CN v?3, V 0 ,
2
3
-
0
so the only eigenvectors are multiples of ? =
a J
a j
1
3
t
1
154
Section 7.8
17b. From part a, one solution of the given D.E. is
0
x(1)(t) =
e2t, but there are no other linearly
independent solutions of this form.
17c. We now seek a second solution of the form
x
= ?te2t + ne
2t
Thus Ax = A?te2t + Ane
2t
and
' = 2?te2t + ?e2t + 2ne
2t
Equating like terms, we then
have (A-2I)? = 0 and (A-2I)n = ?. Thus ? is as in part a and the second equation yields
By row reduction this is
r-1 1 1 ^ / \ ' 0
n1
2 -1 -1 n2 = 1
V-3 2 2, , v-1 ,
equivalent to the system
' 1 - 1 -1 / \ ' 0
n1
0 1 1 n2 = 1
V 0 0 0, , V 0 ,
/
and n1 = 1, so n
If we
Hence
a second solution of the D.E. is
,2t.
' 0 ' 1 N
x(2)(t) = 1 te2t + 1
1 V 0,

17d. Assuming x = ?(t /2)e + nte + Ze , we have
Ax = A?(t2/2)e + Ante
2t
A?e and
x' = ?te2t + 2?(t2/2)e2t + ne2t + 2nte2t + 2?e2t and thus (A-2I)? = 0, (A-2I)n = ? and(A-2I)Z = n. Again, ? and n are as found previously and the last equation is equivalent to
/
1 1 1
2 -1 -1 V -3 2 2
\ 1 ' 1 ^
Z2 = 1
) vZ3 , v 0 ,
\ 1 '-1 ^
1 -1 -1
equivalent system 0 1 1 Z2 = 3
v 0 0 0, Z3 , V 0 ,
By row reduction we find the
If we let
1
^0 j
+
Section 7.8
155
17e.
17f.
19a.
19b.
Z2 = 0, then Z3 = 3 and Zi =2, so Z =
x(3) (t) =
' 0 ' 1 ' ' 2 N
1 (t2/2)e2t + 1 te2t + 0
1 V 0, V 3,
and
,2t
? is the matrix with x(1) as the first column, x the second column and x(3) as the third column.
(2)
as
1 1 0 -1 0 3
and using row operations on T and I, or a
computer algebra system, T
-3 3 2
3 -2 -2
and thus
' 2 1 0
T-1AT = 0 2 1 = J
v 0 0 2
J2 = JJ = 'x 1 'x 1 2x ^
=
V 0 x, V 0 x, V 0 x2 ,
J3 = JJ2 = c< 1 'x2 2x ^ "x3 3x 2 N
=
V 0 x, V 0 x2 , V 0 x3 ,
Based upon the results of part a, assume
/
=
xn nxn-1
J
v 0 x n+1 = jJ =
, then
rx 1 Wxn nxn"14
V
0 x
0
xn
, which is the same as Jn with n
0 xn
replaced by n+1. Thus, by mathematical induction, Jn has the desired form.
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