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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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where
/2Xn+tnYnA
fi =
XnYn
/
, x0 = 1 and y0 = l.Thus f0 =
2-0 (1)(1) y
1+.1(2) 1+.1(1)
1.2
1.1
^ 2.4+ .1(1.1) ^ V 2. 51 ^
v (1.2)(1.1) 7 v 1. 32 y
?1.2+.1(2.51) ^ ^1.451^ (.2) N
x2 = =
?1.1+.1(1.32) y v1.232 7 vV(.2) ,
2b. Eqs. (7) give:
k01 =
1,1) ^ ' 2 + 0 ^ r 2 ?
1,1) 7 , (1)(1) 7 V 1y
k
k
02
03
^2.51^
V1.32 y \
2.4+.1(1.1)
(1.2)(1.1)
2.502+.1(1.132)
(1.251)(1.132)
3.04608+.2(1.28323)
(1.52304)(1.28323)
Using Eq. (6) in scalar form, we then have
X! = 1+(.2/6)[2+2(2.51)+2(2.6152)+3.30273] = 1.51844
Y! = 1+(.2/6)[1+2(1.32)+2(1.41613)+1.95440] = 1.28089.
k04 =
2.6152
1.41613
3.30273 y v1.95441 y
fn =
x
1
v1y
7. Write a computer program to do this problem as there are twenty steps or more for h < .05.
8. If we let y = x', then y' = x" and thus we obtain the system x' = y and y' = t-3x-t2y, with x(0) = 1 and y(0) = x'(0) = 2. Thus f(t,x,y) = y,
g(t,x,y) = t - 3x - t2y, t0 = 0, x0 = 1 and y0 = 2. If a program has been written for an earlier problem, then its best to use that. Otherwise, the first two steps are as follows:
k01 =
2
-3
Section 8.6
173
k02 -
03
"04
2+(-.15)
.05-3(1.1) (.05) 2(1.85)
2+(.16273)
.053(1.0925)(.05)2(1.83727) 2+(.32321)
.13(1.18373) (.1) 2(1.67679)
1.85
3.25463
A /
1.83727
3.23209
1.67679 j v3.46796 j
and thus
x1 = 1+(.1/6)[2 + 2(1.85)+2(1.83727)+(1.67679)]=1.18419, Y1 = 2+(.1/6)[-3-2(3.25463)-2(3.23209)-3.46796]=1.67598,
which are approximations to x(.1) and y(.1) - x'(.1).
In a similar fashion we find
kn -
1.67598
3.46933
k12 -
1.50251
3.68777
k13 -
1.49159
3.66151 and thus
k14 -
1.30983
3.85244
x2-x1+(.1/6)[1.67598+2(1.50251)+2(1.49159)+1.30983]=1.33376 Y2-y1(.1/6)[3.46933+2(3.68777)+2(3.66151)+3.85244]=1.30897.
Three more steps must be taken in order to approximate x(.5) and y(.5) = x'(.5). The intermediate steps Yield x(.3) @ 1.44489, y(.3) @ .9093062 and x(.4) @ 1.51499, Y(.4) @ .4908795.
174
CHAPTER 9
Section 9.1, Page 468
For Problems 1 through 16, once the eigenvalues have been found, Table 9.1.1 will, for the most part, quickly yield the type of critical point and the stability. In all cases it can be easily verified that A is nonsingular.
1a. The eigenvalues are found from the equation det(A-rI)=0.
3-r -2
Substituting the values for A we have
2 -2-r
r2 - r - 2 = 0 and thus the eigenvalues are r1 = -1 and
( 4 -2 YX11 ( 0 '
r2 = 2. For r1 = -1, we have | = and thus
4 1 ISJ p1 ] v 01
v 2 -1 1 VX2 ) v 0 ,
X (1) X (2)
1 ] V1 -21 ] V 01
and for r2 we have =
V 2 y v 2 -4 A^2 ) v 0 y
and thus
1b. Since the eigenvalues differ in sign, the critical point is a saddle point and is unstable.
1d.
4a. Again the eigenvalues are given by
1-r -4
4 -7-r
are solutions of
just one eigenvector X
0 and thus r1 = r2 =
? 4 -4 ^ '?1 ] f 01
of =
v 4 )v -4 \ v X2 y V 0 y
and hence there is
1
1
4b. Since the eigenvalues are negative, (0,0) is an improper node which is asymptotically stable. If we had found that there were two independent eigenvectors then (0,0) would have been a proper node, as indicated in Case 3a.
a Y
2
Section 9.1
175
4d.
7a. In this case det(A - rl) = r2 - 2r + 5 and thus the
eigenvalues are r12 = 1 2i. For rx = 1 + 2i we have
ISO 1 ISO H- -2 1 rX
1 ISO ISO H- vX
X (1) = 1 >
1-iy
f 2+2i -2 1 rX
^4 -2+2i y vX
2-2i -2
Xi
X2 7
and thus
Similarly for r2 = 1-2i we have and hence X(2)
0
1
1+1
7b. Since the eigenvalues are complex with positive real part, we conclude that the critical point is a spiral point and is unstable.
7d.
10a. Again, det(A-rl) = r2 + 9 and thus we have r12 = 3i.
\ A
For r1 = 3i we have
1-3i 2
v -5 -1-3iy
X1
X2 7
0
and thus
:(1)
-1 + 3i
V /
Likewise for r2 = -3i,
1 + 3i 2 V-5 -1+3i
1 'X1' ( 0 ] so that X(2) = ( 2 1
/ vX2 j v 0 , v-1-3i,
1
0
0
2
10b. Since the eigenvalues are pure imaginary the critical point is a center, which is stable.
176
Section 9.1
10d.
13.
17.
0 ' /
If we let x = x + u then x = u and thus the system
which will be in
/ f 1 1 ] + O X f 1 1 ? f 2 )
becomes u = u -
V 1 -1 y V 1 -1 y v 0 y
the form of Eq.(2) if
1 1 ^ X O II f 2 )
V 1 -1 y v 0 y
. Using row
operations, this last set of equations is equivalent to
f i i
v 0 -2 /
u =
x0 =
2
-2
and thus xl = 1 and x2 = 1. Since
1 1
u has (0,0) as the critical point, we
conclude that (1,1) is the critical point of the original system. As in the earlier problems, the eigenvalues are
1-r 1 2 /--------------------------------------
given by = r2 - 2 = 0 and thus r12 = y 2 .
1 -1-r
Hence the critical point (1,1) is an unstable saddle point.
The equivalent system is dx/dt = y,dy/dt = -(k/m)x-(c/m)y which is written in the form of Eq.(2) as
d
dt
/ \ x
V yJ
0 1 -k/m -c/m
\ / \ x
The point (0,0) is clearly a
V YJ
critical point, and since A is nonsingular, it is the only one. The characteristic equation is r2+(c/m)r+k/m=0
so r1,r2 = [-c (c2
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