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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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of is
1/4e-t/2 - 1/4e
-t/2
The second colunm of is
determined by d1x(1)(0) + d2x(2)(0) =
0
1/2 and thus the second column of 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 4 -4e-3t e2t
To find the fundamental matrix
(t) satisfying the I.C. (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 ^). Alternatively, we can find the first column of 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 ^) is
1
a J
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.
0
Thus we must have
c1 + c2 = 0, -4c1 + c2 = 1; therefore c1 = -1/5 and
c2 = 1/5. Hence the second column of ^) 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 )
' 1 ' 0
that satisfy the I.C. and
V 0 , V 1 ,
we determine the linear combinations of the columns of
4
, respectively.
/
In the first case c1 and c2 satisfy 0c1 + 2c2 = 1 and
c1 + 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
( =
etcos2t -2e tsin2t
(1/2)e sin2t e cos2t
10. From Problem 14 Section 7.5 we have x
(1)
1
-4
' 1 ' 1
x(2) = -1 e-2t and x(3) = 2
V 1 , V 1 ,
3t
For the first column
of we want to choose c1, c2, c3 such that c1x(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
,3tx T
t
e
\0)
Section 7.8
149
0
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-2t, 4/3et - 1/3e t , 1/3et - 1/3e-2t) the second column of ^). Finally, for the third column
we have eix(1) (0) + e2x(2) (0) + e3x(3) (0) =
0
0
which
gives e1 = 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 ^).
11. From Eq. (14) the solution is given by ^^0. Thus
3/2et-1/2e-t
3/2et-3/2e-t
7/2et-3/2et 4 7/2et-9/2e-t
1/2et+1/2e-t
1/2et+3/2e-t
3
2
1
e - 2
-t
1
00
10
x =
10
Section 7.8, Page 407
1. The eigenvalues and eigenvectors of the given coefficient
matrix satisfy
3-r
?1 V?2 J
0
The determinant of
coefficients is (3-r)(-1-r) + 4 = r2 - 2r + 1 = (r-1)2 so r1 = 1 and r2 = 1. The eigenvectors corresponding to
this double eigenvalue satisfy
2 '?1 ' ' 0 ^
1
4
11 -2 J v?2 j v 0 ,
, or
?1 - 2?2 = 0. Thus the only eigenvectors are multiples
of ?
(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 = ?tet + net and substitute this expression into the D.E. As in Example 2 we find that ?
is an eigenvector, so we choose ? =
2
Then n must
00
10
10
1
150
Section 7.8
satisfy
f 2 -4^ fm 1 ' 2 '
v 1 -2 > U2 1 v 1 ,
, which verifies Eq.(16).
Solving these equations yields n1 - 2n2 = 1. If 2 = k, where k is an arbitrary constant, then n1 = 1 + 2k. Hence the second solution that we obtain is
x(2) (t) =
2
tet +
1 + 2k k
2
tet +
1
+ k
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