# Elementary differential equations 7th edition - Boyce W.E

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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