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Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
Download (direct link): elementarydifferentialequat2001.pdf
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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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