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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
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Likewise
'1/4 1/2 " ^1 ' ' 0 ' , and thus ?(1) = '-2 N
=
v1/8 1/4, v^2 , V 0 , V 1 ,
.(2)
' 2 ' and thus x(1) (t) = '-2 ' e t and x(2) (t) = ' 2 N

V 1 , V 1 , t—1
>-t/2
To
find the first column of ® we choose ci and c2 so that
cix(1) (0) + c2x(2) (0) =
(2)
1
, which yields -2c1 + 2c2 = 1 and
ci + c2 = 0. Thus c2 = -1/4 and c2 = 1/4 and the first column 1/2e-t/2 + 1/2et
of O is
1/4e-t/2 - 1/4e
-t/2
The second colunm of O is
determined by d1x(1)(0) + d2x(2)(0) =
0
1/2 and thus the second column of O is
which yields d1 = d2 =
e-t/2 - e-t 1/2e-t/2 + 1/2e-
4. From Problem 4 of Section 7.5 we have the two linearly
independent solutions x(1) (t) =
e2t. Hence a fundamental matrix ? is given
1
e-3t and
x(2)(t) =
1
by ?(t) =
' e-3t e2t N -4e-3t e2t
To find the fundamental matrix
O(t) satisfying the I.C. O(0) = I we can proceed in either of two ways. One way is to find ?(0), invert it to obtain ?-1(0), and then to form the product ?(t)?-1(0), which is O(t). Alternatively, we can find the first column of O by determining the linear combination
c1x(1)(t) + c2x(2)(t) that satisfies the I.C.
1
This
requires that c1 + c2 = 1, -4c1 + c2 = 0, so we obtain c1 = 1/5 and c2 = 4/5. Thus the first column of ®(t) is
10 /
1
a 7
148
Section 7.7
(1/5)e-3t + (4/5)e2t -(4/5)e-3t + (4/5)e2t,
Similarly, the second column of
® is that linear combination of x(1) (t) and x(2) (t) that
satisfies the I.C.
Thus we must have
c1 + c2 = 0, -4c1 + c2 = 1; therefore c1 = -1/5 and
c2 = 1/5. Hence the second column of O(t) is
" ? (1/5)e-3t + (1/5)e2t'
(4/5)e-3t + (1/5)e2t,
Two linearly independent real-valued solutions of the given D.E. were found in Problem 2 of Section 7.6. Using
the result of that problem, we have
?(t) =
-2e sin2t 2e COs2t v e-tCOS2t e-tsin2t ;
To find O(t)
' 1 ' ' 0 N
that satisfy the I.C. and
V 0 , V 1 ,
we determine the linear combinations of the columns of
\
, respectively.
/
In the first case ci and c2 satisfy 0ci + 2c2 = 1 and
ci + 0c2 = 0. Thus c1 = 0 and c2 = 1/2. In the second
case we have 0c1 + 2c2 = 0 and c1 + 0c2 = 1, so c1 = 1
and c2 = 0. Using these values of c1 and c2 to form the
first and second columns of O(t) respectively, we obtain
O(t) =
etcos2t -2e tsin2t
(1/2)e sin2t e cos2t
10. From Problem 14 Section 7.5 we have x
(1)
1
-4
' 1 ' ' 1N
x(2) = -1 e 2t and x(3) = 2
V 1 , V 1 ,
,3t
For the first column
of O we want to choose ci, c2, c3 such that cix(1) (0) +
Thus c1 + c2 + c3 = 1,
c2x(2)(0) + c3x(3)(0) =
1
0
-4c1 - c2 + 2c3 = 0 and -c1 - c2 + c3 = 0, which yield c1 = 1/6, c2 = 1/3 and c3 = 1/2. The first column of ®
is then (1/6e + 1/3e + 1/2e
-1/6et
3t
2/3et - 1/3e
2t
+ e
3t
1/3e 2t + 1/2e3t) T. Likewise, for the second
,3t\ T
11 /
t
e
10/
Section 7.8
149
column we have d1x(1) (0) + d2x(2) (0) + d3x(3) (0) =
which yields d1 = -1/3, d2 = 1/3 and d3 = 0 and thus
-2t\ T
is
(-1/3et + 1/3e , 4/3et - 1/3e t, 1/3et - 1/3e c)
the second column of O(t). Finally, for the third column
we have e1x(1) (0) + e2x(2) (0) + e3x(3) (0) =
0
0
which
gives e3 = 1/2, e2 = -1 and e3 = 1/2 and hence (1/2et - e-2t + 1/2e3t, -2et + e-2t + e3t,
-1/2et + e-2t + 1/2e3t) T is the third column of O(t).
11. From Eq. (14) the solution is given by ®(t)x°. Thus
3/2et-1/2e-t 3/2et-3/2e-t 7/2et-3/2et ^ 7/2et-9/2e-t
1/2et+1/2e-t
1/2et+3/2e-t
3
2
1
7
e - — 2
-t
1
^0 J
<1J
x =
Section 7.8, Page 407
1. The eigenvalues and eigenvectors of the given coefficient
matrix satisfy
3-r
S1
V^2 7
0
The determinant of
coefficients is (3-r)(-1-r) + 4 = r2 - 2r + 1 = (r-1)2 so r3 = 1 and r2 = 1. The eigenvectors corresponding to
this double eigenvalue satisfy
'2 -4 " ^1 ' ' 0 N
CM 1 T 1 vk2 7 V 0 7
, or
k1 - 2^2 = 0. Thus the only eigenvectors are multiples
of k
(1)
x(1) (t) =
2
2
One solution of the given D.E. is
e, but there is no second solution of this
form. To find a second solution we assume, as in Eq. (13), that x = ktet + net and substitute this expression into the D.E. As in Example 2 we find that k
is an eigenvector, so we choose k =
2
Then n must
i1 -1-r/
10 )
<1J
a 7
1
150
Section 7.8
satisfy
ISO 1 K ] ' 2 '
1 -2 ; CM P v 1,
, which verifies Eq.(16).
Solving these equations yields n1 - 2n2 = 1. If n2 = k, where k is an arbitrary constant, then n1 = 1 + 2k. Hence the second solution that we obtain is
x(2) (t) =
tet +
1 + 2k k
2
tet +
1
+ k
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 ' 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 ^ ^. 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
1-r 1 1
2 1-r -1
0 -1 1-r
„3
\ ^1 ' ' 0 '
^2 = 0
/ ^3 , v 0 ,
The determinant of coefficients
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