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

ISBN 0-471-31999-6

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/

1 1 1

2 -1 -1 V -3 2 2

\ ' 1 N

Z2 = 1

) v^3 , v 0 ,

T 1 1 T—1 1 T 1 '-1 N

equivalent system 0 1 1 Z2 = 3

o o o v^3 , V 0 ,

By row reduction we find the

If we let

1

^0 j

+

Section 7.8

155

17e.

17f.

19a.

19b.

Z2 = 0, then Z3 = 3 and Z1 = 2, so Z =

x(3) (t) =

' 0 ' ' 1 ' ' 2 N

1 (t2/2)e2t + 1 te2t + 0

t—1 1 V 0, co

and

,2t

? is the matrix with x(1) as the first column, x the second column and x(3) as the third column.

(2)

as

1 1 0 v-1 0 3,

and using row operations on T and I, or a

computer algebra system, T

-3 3 2

3 -2 -2

and thus

' 2 1 0 '

T-1AT = 0 2 1 = J

v 0 0 2 y

J2 = JJ = 'l 1 ' 'l 1 ' 2l ^

=

V 0 l, V 0 V V 0 l2 ,

J3 = JJ2 = 'l 1 ' V 2l" 3l2 N

=

V 0 V V 0 l2 , V 0 l3 ,

Based upon the results of part a, assume

/ ZS

ln nln-1

J

V 0 l

n+1 = jJ =

, then

rl 1?fln nln-1^

v

0 l

0

ln

V+1 (n+1)ln

, which is the same as Jn with n

v 0 Xn

replaced by n+1. Thus, by mathematical induction, Jn has the desired form.

0

13 ^

T

1-1 1 17

n

19c. From Eq.(23), Section 7.7, we have

156

Section 7.9

exp(Jt) = I +

= I +

n=1

n=1

n!

Xntn nXn-1tn

1 +

n!

0

Xntn

n!

X ntn

n!

X — X

n!

n=1

n=1

Xn-1tn

(n-1)!

1 +

X

n=1

X ntn

n!

(eXt teXt" V 0 eXt y

Since

X

n=1

Xn-1tn

(n-1)!

= t( 1 + X

n=1

nn

— ) = teXt.

n!

19d. From Eq. (28), Section 7.7, we have

x = exp(Jt)x0 =

V

O 0 'i ( 0 A

= x1 eXt + x2

v x2 , v 0 ,

Section 7.9, Page 417

1. From Section 7.5

(eXt teXtYx1x 0 eXt x2.

(x0eXt+x2teXt^ x02eXt

te

Xt

x(c) = ci

g (t) =

1

et +

1

c2

-t

Note that

' 1 ' et + ' 0N

V 0, t—1

t and that r = 1 is an eigenvalue of

the coefficient matrix. Thus if the method of undetermined coefficients is used, the assumed form is given by Eq.(18).

2. Using methods of previous sections, we find that the

eigenvalues are rj_ = 2 and r2 = -2, with corresponding

V3

.-73.

eigenvectors

Thus

jntn

0

Section 7.9

157

x(c) = c1

2t

+ C2

1

-2t

Writing the

' 1 ' ' 0 N

term as et + V a/"3 j

V 0 ,

can assume x(p) = aet + be D.E., we obtain

e t we see that we Substituting this in the

ael - be p = Aae1 + Abe p + ' 1 ' et + ' 0 N

V a/"3 j

V 0,

' 1 ' ' 0N

V 0 , V 0 ,

e , where A

is the given coefficient matrix. All the terms involving

et must add to zero and thus we have Aa - a +

This is equivalent to the system

V3" a2 = -1 and V3a1 - 2a2 = 0, or a2 = -2/3 and

a2 = -1/V"3. Likewise the terms involving e-t must add

The solution

of this system is b1 = -1 and b2 = 2/^/3. Substituting these values for a and b into x(p) and adding x (p) to

' 0 ' ' 0 N

zero, which yields Ab + b + V V3 J =

V 0 ,

x

(c)

yields the desired solution.

3. The method of undetermined coefficients is not straight forward since the assumed form of x(p) = acost + bsint leads to singular equations for a and b. From Problem 3 of Section 7.6 we find that a fundamental matrix is

?(t) =

5cost 5sint

v2cost + sint -cost + 2sint matrix is

The inverse

?-1(t) =

cost - 2sint

2cost + sint 5

sint

cost

, which may be found as

in Section 7.2 or by using a computer algebra system. Thus we may use the method of variation of parameters where x = ?(t) u(t) and u(t) is given by u'(t) = ?-1(t)g(t) from Eq.(27). For this problem

g (t) =

f 4-1

-cost

and thus

e

5

158

Section 7.9

u'(t) =

cost - 2sint

2cost + sint 5

sint

-cost

1 4-1

-cost

V sint y

1 2 - 3cost2t + sin2t

5 ^ -1 - cos2t - 3sin2t J after multiplying and using appropriate trigonometric identities. Integration and multiplication by V yields the desired solution.

5

4. In this problem we use the method illustrated in Example 1. From Problem 4 of Section 7.5 we have the

transformation matrix T =

1 1

Inverting T we find

that T

-1

-1

1

If we let x = Ty and substitute

into the D.E., we obtain

y' =

1 ' 1 -1 ' ' 1 1 ' / -, - ^ 1 1 1 7 -, -, 1 1 -1 -2t e 2t

— y + — v -2e y

5 V 4 1y V 4 -2y t—1 1 5 v 4 1 ,

'-3 0 ' 1 / e -2t + 2et 1

y + — 2t - 2e t y This corresponds

V 0 2 y 5 V 4e -

the two scalar equations

yi + 3yi = (1/5)e-2t + (2/5)et, y2 - 2y2 = (4/5)e-2t - (2/5)et, which may be solved by the methods of Section 2.1. For

the first equation the integrating factor is e3t and we obtain (e3tyi)' = (1/5)et + (2/5)e

4t

so

e yi = (1/5)e + (1/10)e + ci. For the second equation

the integrating factor is e (e-2ty2)' = (4/5)e-4t - (2/5)e

•2t

so

Hence

-2t.

Y2 = -(1/5)e-4t + (2/5)e

-t

y =

1/5

-1/5

e-2t +

^1/10^

2/5

et +

+ c2. Thus

Finally,

/ -3t ^

cxe 3t

c2e2t /

multiplying by T, we obtain

x = Ty = ' 0 ' e-2t + ' 1/2 1 et + c1 ' 11 e-3t + c2 ' 11

V -1 y V 0 , V-4 y v 1 y

2t

The last two terms are the general solution of the corresponding homogeneous system, while the first two terms constitute a particular solution of the nonhomogeneous system.

4-4 1 y

e

Section 7.9

159

12. Since the coefficient matrix is the same as that of Problem 3, use the same procedure as done in that problem, including the ?-1 found there. In the interval n/2 < t < n sint > 0 and cost < 0; hence |sint| = sint, but |cost| = -cost.

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