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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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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 , m 0 , M 2 J
x
(1)
-t
The eigenvectors corresponding to the
double eigenvalue must satsify
-1 1 1 2 -1 -1
v 0 -1 -1 .
\ ^1 ' ' 0 ^
^2 = 0
/ V^3 y , 0,
t
t
t
a ;
0
1
4
Section 7.8
151
' 0 '
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.
1 -, -, -, 1 -1 1 1 / \ n1 ' 0 N
2 -1 -1 n2 = 1
T1 1 T1 1 o v^3 , v-1 ,
1 = 1 and ^2 + ^3 = 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 ' ( 1N
then n = 0 and thus x(3) = 1 te2t + 0
V 1, t1 1 t1
9. We have
and
1
2-r 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 ^ V^1 ' ( 1 1
Uy = 0, so ? = 1
v3/2 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 3^i/2 + 3n2/2 = 1 and 3^i/2 3n2/2 = 1. Choosing n2 = 0 yields n1 = 2/3 and hence
x = c1 ' 11 et/2 + c2 ' 2/3 ' et/2 + c2 ' 1 ' tet/2. x(0) = ' 3 N
V 1 y v 0 , V 1 y v2 y
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 y v 0 , M 1 y
solution is x
0 0 0 -4 0 0 3 6 1
0
(1)
0
1
e2t. For r = 1, we have
\ ^1 ' ' 0 '
^2 = 0
7 v^3 y v 0 ,
, which yields the second solution
.(2) =
-(3)
et. The third solution is of the form
0 ^ o o o ' 0 ^
1 tet + net, where -4 0 0 n = 1 and thus
-6y V 3 6 1 , V -6 y
n 1 = -1/4 and 6n2 + n3 = -21/4. Choosing n2 = 0 gives n 3 = -21/4 and hence
' 0 ^ r -1/4 " ' 0 ' r 0 '
X(t) = C1 1 et + C2 0 et + 1 tet ro U + 0
v-6 y v-21/4y v-6y V 1y
e2t. The
then yield C1 CM U CM II = 4 and c3 = 3 and hence
' -1 ^ ' 0 ' ' 0 ^
2 et + 4 1 tet + 3 0 2t e , which become unbounded
v-33 , v-6 y , 1 >
as t ^ ^.
Assuming x = ?tr and substituting into the given system,
"1-r -4'
,4 -7-ry
has the double eigenvalue r = -3 and single eigenvector
we find r and ^ must satisfy
\ ^1 ' ' 0 '
7 ^2 y V 0 y
, which
1
x =
Section 7.8
153
x(1)(t) =
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 = n, where A =
and I is the identity
matrix. Thus n =
1
, from above, and Z is found to be
-1/4 .(2)
Thus a second solution is
' 1 ' ' 1 N
(t) = t-3lnt +
V 1 , V-1/4 ,
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- r 1 1 ' 0 '
matrix satisfy 2 1-r -1 Z2 = 0 . The determinant
v-3 2 4-ry v^3, V 0 ,
of coefficients is 8 - 12r + 6r2 - r3 = (2-r) 3, so the
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 Z1 - Z2 - Z3 = 0, and Z2 + Z3 = 0. If we let Z2 = 1, then Z3 = -1 and Z1 = 0,
7 -, -, -1 1 1 ' 0 N
2 -1 -1 Z2 = 0
v-3 2 2, v^3 , V 0 ,
0
so the only eigenvectors are multiples of Z =
a 7
a 7
1
3
t
1
154
Section 7.8
17b. From part a, one solution of the given D.E. is
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 ' / \ n1 ' 0 N
2 -1 -1 n2 = 1
V-3 2 2, vn3, v-1 ,
equivalent to the system
' 1 - 1 -1 ' / \ n1 ' 0 N
0 1 1 n2 = 1
V 0 0 0, vn3, V 0 , /
and n1 = 1, so n
If we
Hence
a second solution of the D.E. is
,2t.
' 0 ' ' 1N
x(2)(t) = 1 te2t + 1
t1 1 V 0,
17d. Assuming x = ?(t /2)e + nte + ?e , 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
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