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

ISBN 0-471-31999-6

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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

T—1 1 T—1 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, t—1 1 t—1

9. We have

and

1

2-r 3/2

-3/2 —1—r

et/2 is one solution. For the second solution we

= (r—1/2) 2 = 0. For r = 1/2, the

is given by ( 3/2 3/2 ^ V^1 ' ( 1 1

Uy = 0, so ? = 1

v—3/2 —3/2 , v—1 )

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 v—2 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

*v—1 /

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

t—1 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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