# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

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of Ô is

1/4e-t/2 - 1/4e

-t/2

The second colunm of Ô is

determined by d1x(1)(0) + d2x(2)(0) =

0

1/2 and thus the second column of Ô 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

Ô(t) satisfying the I.C. Ô(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 Ô^). Alternatively, we can find the first column of Ô 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 Ô^) 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

Ô 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 Ô^) 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 ó

To find Ôè)

' 1ë ' 0 ë

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

Ô(è =

etcos2t -2e tsin2t

(1/2)e sin2t e cos2t

10. From Problem 14 Section 7.5 we have x

(1)

1

-4

' 1ë ' 1ë

x(2) = -1 e-2t and x(3) = 2

V 1 , V 1 ,

3t

For the first column

of Ô 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 Ô

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 Ô^). 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 Ô^).

11. From Eq. (14) the solution is given by Ô^^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 Ï2 = 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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