Elementary Differential Equations and Boundary Value Problems  Boyce W.E.
Download (direct link):
written as
V
2r 1
3 2r
eigenvectors are r1 = 1, E
For t Ô 0
\ ×1' ' 0 ^
/ v^2 , v 0 ,
= 1, E (1) =
. The eigenvalues and and r2 = 1,
1
:(2)
1
Substituting these in the assumed form we
' 1 ^ ' 1 ^
obtain the general solution x = c1 t + c2 t
V 1 , 3
25.
27.
r
v3 ^
Section 7.6
141
31c. The eigevalues are given by
1r 1
a 1r
r2 + 2r + 1  a = 0. Thus r1,2 = 1±\[~a
Note that in Part (a) the eigenvalues are both negative while in Part (b) they differ in sign. Thus, in this part, if we choose a = 1, then one eigenvalue is zero, which is the transition of the one root from negative to positive. This is the desired bifurcation point.
Section 7.6, Page 390
1. We assume a solution of the form x = ? ert thus r and ?
. The determinant of
2 '?1 ^ ' 0 '

r

3
are solutions of =
V4 1r, ? 00
ISJ
coefficients is (r22r3) + 8 = r2
eigenvalues are r = 1 ± 2i. The eigenvector
corresponding to 1 + 2i satisfies
22i 2 4 22i
\ '?1' ' 0 '
/ V?2 j v 0 ,
or (22i) ?1  2?2 = 0. If ?1 = 1, then ?2 = 1i and
:(1)
1 1
1i
and thus one
complexvalued solution of the D.E. is
x(1)(t) =
1
To find realvalued solutions (see Eqs.8 and 9) we
take the real and imaginary parts, respectively of
x(1) (t). Thus x(1) (t)
1i
t(cos2t + isin2t)
= e
cos2t + isin2t
cos2t + sin2t + i(sin2t  cos2t)
cos2t cos2t + sin2t j
Hence the general solution of the D.E. is
x = c1e
cos2t cos2t + sin2t j
+ c2e
sin2t sin2t  cos2t j
The
solutions spiral to ^ as t ^ ^ due to the et terms.
1
t
t
= e
+ ie
142
Section 7.6
7. The eigenvalues and eigenvectors of the coefficient
matrix satisfy
1r 0 0
2 1r 2
3 2 1r
\ ×1" Í0
^2 = 0
) v^3 , V 0 ,
The
determinant of coefficients reduces to (1r)(r2  2r + 5) so the eigenvalues are r1 = 1, r2 = 1 + 2i, and r3 = 1  2i. The eigenvector corresponding to rj_ satisfies
fV 0 0 0 Í ^ í
VV 2 0 2 ^2
3 2 0 , ÷^3 J
1 + 2^2 = 0.
3
0
; hence  ^3 = 0 and
2
so one solution of the D.E. is
3
et. The eigenvector
corresponding to r2 satisfies
^2i 0 2 2i 2 v3 2 2i
\ Í0
^2 = 0
) v^3 , V 0 ,
Hence = 0 and i^2 + ^3 = 0. If we let ^2 = 1, then ^3 = i. Thus a complexvalued solution is
0
1
i
ec(cos2t + i sin2t). Taking the real and imaginary
fV 0 Í fV 0 Í
cos2t et and sin2t e"
vsin2t, vcos2t,
parts, see prob. 1, we obtain
respectively. Thus the general solution is
, which spirals
to ro about the x^ axis in the xxx2x3 space as t ^ ro.
The eigenvalues and eigenvectors of the coefficient
fV 2 Í fV 0 Í fV 0 Í
x = cx 3 eb + c2et cos2t + c3et sin2t
V 2 Ó vsin2t, vcos2t ,
matrix satisfy
' 1r 5 Y^. Í ' 0ë
V 1 3r JC^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
Ê 2 J
Section 7.6
143
is given by
2i 5 Y ^i
1 2i j^2 ) thus one complexvalued solution is
(1+i)t
= 0 so that t)1 = (2+i)^2 and
x(1) (t)
1
Finding the real and complex
parts of x(1) leads to the general solution
c1e
^2cost  sint4
cost
^2sint + cost4
+ c2e
Setting
' 1' ' 2 ' ' 1 '
t = 0 we find x(0) = = c1 + c2
1 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 ë
sint
, which spirals to zero as
t ^ due to the e t term.
11a. The eigenvalues are given by 3/4r 2
5/4r
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 ^ ^.