Elementary Differential Equations and Boundary Value Problems  Boyce W.E.
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2
The last term is a multiple of the first solution x(1) (t) and may be neglected, that is, we may set k = 0. Thus
x(2)(t) =
' 2 Ë tet + ' 1 "
v 1 > v 0 ,
et and the general solution is
x
c1x(1) (t) + c2x(2) (t). All solutions diverge to
infinity as t ^ ro. The graph is shown on the right.
3. The origin is attracting 1.
5.
Substituting x = ?ert into the given system, we find that the eigenvalues and eigenvectors satisfy
1r 1 1
2 1r 1
0 1 1r
„3
\ "?1' ' 0 '
?2 = 0
/ ,?3, V 0 ,
The determinant of coefficients
is r3 + 3r2  4 and thus r1 = 1, r2 = 2 and r3 = 2. The eigenvector corresponding to r1 satisfies
2 1 1 2 2 1
0 1 2
\ "?1' ' 0 ^ '3 ^
?2 = 0 which yields ?(1) = 4 and
/ ,?3, , 0, 4 2 j
3
x
(1)
t
The eigenvectors corresponding to the
double eigenvalue must satsify
1 1 1 2 1 1
v 0 1 1 î
\ "?1' ' 0 ^
?2 = 0
/ V?3 ó . 0,
t
t
t
0
1
4
ê 2 )
Section 7.8
151
' 0 ë
which yields the single eigenvector ?
(2)
and hence
x(2)(t) =
0
1
e2t. The second solution corresponding to the double eigenvalue will have the form specified by Eq.(13), which yields x
Substituting this into the given system, or using Eq.(16), we find that n satisfies
.(3)
0
1
te2t + ne2t.
¥ \ / \ ' 0 N
1 1 1 Ï1
2 1 1 Ï2 = 1
1 ÷Ï3, v1 ,

1

0
1 = 1 and Ï2 + Ï3 = 1,
where either n2 or n3 is arbitrary. If we choose n2 = 0,
e2t. The
general solution is then x = c1x(1) + c2x(2) + c3x(3)
( 1 ' ' 0 ë ( 11
then n = 0 and thus x(3) = 1 te2t + 0
V 1, v1, V 1,
9. We have
and
1
2r 3/2
3/2 1r
et/2 is one solution. For the second solution we
= (r1/2)2 = 0. For r = 1/2, the
is given by ( 3/2 3/2 ^ ¥$1ë ( 1 1
Uó = 0, so ? = 1
ó3/2 3/2 , V1)
have x = ?tet/2 + ne 2, where (A  — I)n = ?, A being
2
the coefficient matrix for this problem. This last equation reduces to 3n1/2 + 3n2/2 = 1 and 3Ï³/2  3n2/2 = 1. Choosing n2 = 0 yields Ï³ = 2/3 and hence
x = c1 ' 1ë et/2 + c2 '2/3' et/2 + c2 ' 1 ' e x(0) = ' 3 ë
ãò
2
V1 Ó v 0 , V1 Ó V2 Ó
gives c1 + 2c2/3 = 3 and c1 = 2, and hence c1 = 2,
c2 = 3/2. Substituting these into the above x yields the
solution.
1
'v1 )
152
Section 7.8
11.
12.
14.
The eigenvalues are r = 1,1,2. For r = 2, we have
, so one
'1 0 0' ×1' ' 0 ^ ' 0 ^
4 1 0 ?2 = 0 , which yields ? = 0
v 3 6 0, V?3 ó v 0 , , 1 Ó
0
solution is x
0 0 0 4 0 0 3 6 1
0
(1)
0
1
e2t. For r = 1, we have
\ ×1' r 0 '
?2 = 0
Ó v?3 ó v 0 ,
, which yields the second solution
.(2) =
(3)
et. The third solution is of the form
0 ^ ' 0 0 0 ^ ' 0 ^
1 tet + net, where 4 0 0 n = 1 and thus
6, 3 V 6 Ó
6
10
Ï1 = 1/4 and 6Ï2 + Ïç = 21/4. Choosing n2 = 0 gives Ï 3 = 21/4 and hence
' 0 ^ 4 ' 0 ' ' 0 ^
1

x(t) = c1 1 et + c2 0 et + 1 t + c3 0
e
t
v6 , 21/40 V 6Ó . 1>
e2t. The
then yield c1 = 2, c2 = 4 and c3 = 3 and hence
' 1" ' 0 ' ' 0 ^
2 et + 4 1 tet + 3 0 2t
e , which become unbounded
V33 , v6 , v 1,
as t ^ ro.
Assuming x = ?tr and substituting into the given system,
4r 4 '
4 7r 0
has the double eigenvalue r = 3 and single eigenvector
we find r and ? must satisfy
\ ×1' ' 0'
Ó v?2  v 0 ,
, which
1
x =
Section 7.8
153
x(1)(t) =