Books in black and white
 Books Biology Business Chemistry Computers Culture Economics Fiction Games Guide History Management Mathematical Medicine Mental Fitnes Physics Psychology Scince Sport Technics

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

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Previous << 1 .. 461 462 463 464 465 466 < 467 > 468 469 470 471 472 473 .. 609 >> Next

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
r1,2 = 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 = 4 9/24 = 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 -l-r Jf?2y v 0 ,
is (-1-r) 2 +2 r
r + 2r + 3, so the eigenvalues are -1 ± \j~2 i. The eigenvector corresponding to
ã f-^i -1 ¯?1 H î Í
-1 + V 2 i satisfies I = or
V 2 -\f~2 i j|?2j v 0)
If we let ?1 = 1, then ?2 = -/2i, and Thus a complex-valued solution of the
2i?1 + ?2
/
?(1) =
0.
1
-\/2 i
given D.E. is
t
1
-yf2 i J
Section 5.5 we have (since ^^2i
2 lnt) + isin
From Eq. (15) of
lnt^ _ e^T2 i lnt)
t-1^/2i = t-1 [cos
e^“w _ 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(ó 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, y3 = x'2 = y4, y2 _ -2y1 + y3, and
y4 _ y1 - 2y3. Thus
146
Section 7.7
Ó =
´ 0 1 0 0Ë
-2010
0 0 0 1
v 1 0 -2 0 ó
Ó.
4 2
29b. The eigenvalues are given by r4 + 4r2 + 3 = 0, which
yields r2 = -1, -V"3,
so r
= ±i, ±\J~3 i.
29c. For r = ± i the eigenvectors are given by
i 1 0 0 ë ×1ë
1
-2 -i 1 0 ?2
0 0 -i 1 ?3
V 1 0 -2 -i ó v?4 ,
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 = ± ypbi, we obtain ? = (1,V"3i, -1, -/3i) and
w3 = (cos^Tt, -ä/Isin ^t, -cos^Tt, ^sin^t) T and w4 = (sin^/Tt, ^cos t, -sin^/Tt, -^cos 1)T.
Thus y=c1w1 + c2w2 + c3w3 + c4w4, so yT(0) = (2, 1, 2, 1)
yields c1 + c3 = 2, c2 + y/~3 c4 = 1, c1 - c3 = 2, and c2 - y/~3 c4 = 1, which yields c1 = 2, c2 = 1, and
´ 2cost + sint 4
c3 = c4 = 0. Hence y =
-2sint + cost 2cost + sint
29e. The natural frequencies are Þ1 = 1 and Þ2 = \/~3 , 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 Ô^) satisfying Ô(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
Likewise
'1/4 1/2" ×1 ' ' 0ë , and thus ?(1) = '-2 N
=
V1/8 1/4, v^2 , v 0 , V 1 ,
.(2)
' 2 ' and thus x(1) (t) = '-2 ' e t and x(2) (t) = ' 2 N

V 1 , V 1 , 1I
>-t/2
To
find the first column of Ô we choose ci and c2 so that
cix(1) (0) + c2x(2) (0) =
(2)
1
, which yields -2c1 + 2c2 = 1 and
c1 + c2 = 0. Thus c1 = -1/4 and c2 = 1/4 and the first column 1/2e-t/2 + 1/2et
Previous << 1 .. 461 462 463 464 465 466 < 467 > 468 469 470 471 472 473 .. 609 >> Next