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

ISBN 0-471-31999-6

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0

1

-i

ec(cos2t + i sin2t). Taking the real and imaginary

' 0 ' ' 0 '

cos2t et and sin2t et

vsin2t, v-cos2t,

parts, see prob. 1, we obtain

respectively. Thus the general solution is

, which spirals

to ? about the x! axis in the x1x2x3 space as t ^ ?.

The eigenvalues and eigenvectors of the coefficient

' 2 ' ' 0 ' ' 0 N

x = cx -3 et + c2et cos2t + c3et sin2t

V 2 2 vsin2t, v-cos2t ,

matrix satisfy

' 1-r -5 Y^. ' ' 0N

v 1 -3-r J^^2 , V 0 ,

The determinant of

coefficients is r2 + 2r + 2 so that the eigenvalues are r = -1 ± i. The eigenvector corresponding to r = -1 + i

Section 7.6

143

is given by

2-i -5 Y

1 -2-i 1^2/ thus one complex-valued solution is

(-1+i)t

= 0 so that ^1 = (2+i)^2 and

x(1) (t)

2 +i

1

Finding the real and complex

parts of x(1) leads to the general solution

c1e

^2cost - sint A

cost

^2sint + costA

+ C2e

Setting

' 1 ' ' 2 ^ ' 1 '

t = 0 we find x(0) = = c1 + c2

V 1 , v 1, v 0 ,

equivalent to the system c2 = -1 and

1

2C1 + C2

Ci + 0 = 1

sint

, which is

Thus c1 = 1 and

x(t) = e

- ^-t

-t

2cost - sint cost cost - 3sint cost - sint

- e

-t

^2sint + cost A

sint

, which spirals to zero as

t ^ ^ due to the e t term.

11a. The eigenvalues are given by 3/4-r -2

-5/4-r

1

r2 + r/2 + 17/16

0.

11d. The trajectory starts at (5,5) in the x1x2 plane and spirals around and converges to the t axis as t ^ ^.

15a. The eigenvalues satisfy r1,r2 = ±V 4-5a .

2-r -5 a -2-r

= r2 - 4 + 5a = 0, so

t

t

x

15b. The critical value of a yields r1 = r2 = 0, or a = 4/5.

144

Section 7. 6

15c.

16a.

5/4-r 3/4

a 5/4-r

ri,2 = 5/4 ± \f3a /2.

r2 - 5r/2 + (25/16 - 3a/4) = 0, so

16b. There are two critical values of a. For a < 0 the

eigenvalues are complex, while for a > 0 they are real. There will be a second critical value of a when r2 = 0, or a = 25/12. In this case the second real eigenvalue goes from positive to negative.

16c.

18a. We have

3-r a -6 -4-r

r1(r2 = -1/2 ±V49-24a /2.

= r2 + r - 12 + 6a = 0, so

18b. The critical values occur when 49 - 24a = 1 (in which case

r2 = 0) and when 49 - 24a = 0, in which case r1 = r2 = -1/2. Thus a = 2 and a = 49/24 = 2.04.

18c.

Section 7.6

145

21. If we seek solutions of the form x = ?tr, then r must be an eigenvalue and ? a corresponding eigenvector of the coefficient matrix. Thus r and ? satisfy

The determinant of coefficients

"-1-r -1 Y?1^ " 0 ^

v 2 -1-r JV?2, v 0 ,

is (-1-r) 2 +2 r

r + 2r + 3, so the eigenvalues are -1 ± \j~2 i. The eigenvector corresponding to

r f-^i -i Y^i 1 01

-1 + V 2 i satisfies I = or

V 2 -\[2 i Jv^2) v0)

If we let ^1 = 1, then ?2 = -\j~2 i, and Thus a complex-valued solution of the

2i?i + ?2

/

?(1) =

0.

1

-\/2 i

given D.E. is

t

-1+/2 i

1

-yf2 i J

Section 5.5 we have (since ^^2i

2 lnt) + isin

From Eq. (15) of

lnt^ _ e^T? i lnt)

t-1^/2i = t-1[cos

e^^ _ e v ~ ^ ^“^)

2 lnt)] for t > 0. Separating the complex valued solution into real and imaginary parts, we obtain the two real-valued solutions

t

-1

cos(y 2 lnt)

2 sin(A/2 lnt)

and v

t

-1

sin(y2 lnt)

2 cos(^/2lnt)

u

23a. The eigenvalues are given by (r+1/4)[(r+1/4) 2 + 1] = 0.

5

23c. Graph starts in the first octant and spirals around the x3 axis, converging to zero.

2 9a. We have y1 = x1 = y2, y'3 = x'2 = y4, y 2 _ -2y1 + y3, and

y4 _ y1 - 2y3. Thus

146

Section 7.7

y =

r 0 1 0 0^

-2010 0 0 0 1

v 1 0 -2 0 y

y.

4 2

29b. The eigenvalues are given by r4 + 4r2 + 3 = 0, which

yields r2 = -1, -V"3,

so r

= ±i, ±a/T i .

29c. For r = ± i the eigenvectors are given by

?H 1 1 0 0 ^ T1 '

-2 -i 1 0 ^2

0 0 -i 1 ^3

V 1 0 -2 -i y v^4 y

and choosing ^3 = 1

= 0. Choosing ^ = 1 yields ?2 = i

(1, i, 1, i) (cost + isint) is a solution. Finding the real and imaginary parts yields w1 = (cost, -sint, cost, -sint)T and

w2 = (sint, cost, sint, cost)T as two real solutions.

In a similar fashion, for r = ± yp3i, we obtain ? = (1,V"3i, -1, -^Ti) and

w3 = (cosA^Tt, -^sin ?/Tt, -cos^/Tt, A/Tsi^^Tt) T and w4 = (sinA^Tt, A^Tcos t, -sinA^Tt, -^cos a/T t) T.

Thus y=c1w1 + c2w2 + cTwT + c4w4, so yT(0) = (2, 1, 2, 1) yields ci + cT = 2, c2 + a/Tc4 = 1, ci - cT = 2, and c2 - ypTc4 = 1, which yields c1 = 2, c2 = 1, and

r 2cost + sint ^

c3 = c4 = 0. Hence y =

-2sint + cost 2cost + sint

29e. The natural frequencies are ra1 = 1 and ra2 = AAT, which

are the absolute value of the eigenvalues. For any other choice of I.C., both frequencies will be present, and thus another mode of oscillation with a different frequency (depending on the I.C.) will be present.

Section 7.7, Page 400

^-2sint + cost )

Each of the Problems 1 through 10, except 2 and 8, has been solved in one of the previous sections. Thus a fundamental matrix for the given systems can be readily written down.

The fundamental matrix O(t) satisfying 0(0) = I can then be

Section 7.7

147

found, as shown in the following problems.

2. The characteristic equation is given by

-3/4-r 1/2

1/8 -3/4-r

r2 + 3r/2 + 1/2 = 0, so r = 1, 1/2. For r = 1 we have

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