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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.
Download (direct link): elementarydifferentialequations2001.pdf
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12a. The critical points are given by y = 0 and
x(1 - x2/6 - y/5) = 0, so (0,0), (/6,0) and (-^/6,0)
are the only critical points.
180
Section 9.2
12b.
12c.
15a.
15b.
19a.
y-2
Clearly (/6,0) and (-^/6,0) are spiral points, and are asymptotically stable since the trajectories tend to each point, respectively. (0,0) is a saddle point, which is unstable, since the trajectories behave like the ones for (1/2,1/2) in Problem 7.
dy dy/dt 8x 2 2
--- = ------- = ---, so 4xdx - ydy = 0 and thus 4x2 - y2 = c,
dx dx/dt 2y
which are hyperbolas for c 0 and straight lines y = 2x for c = 0.
dy y-2xy
so (y-2xy)dx + (x-y-x2)dy = 0, which is an
-x+y+x2
dx 2
exact D.E. Therefore f(x,y) = xy - x2y + g(y) and hence
df 2 / 2 /
= x - x + g (y) = x - y - x, so g (y) = -y and
dy
g(y) = -y2/2. Thus 2x2y - 2xy + y2 = c (after multiplying by
-2) is the desired solution.
dy -sinx 2
---- = -------, so ydy + sinxdx = 0 and thus y /2 - cosx = c.
dx y
Section 9.3 181
21b.
23. We know that f'(t) = F[f(t), y(t)] and
y/(t) = G[f(t), y(t)] for a < t < p. By direct substitution we have
F'(t) = f'(t-s) = F[f (t-s), y(t-s)] = F[F(t), Y(t)] and
Y'(t) = y'(t-s) = G[f(t-s), y(t-s)] = G[F(t), Y(t)] for
a < t-s < p or a+s < t < p+s.
24. Suppose that t1 > t0. Let s = t1 - t0. Since the system
is autonomous, the result of Problem 23, with s replaced by -s shows that x = f1(t+s) and y = y1(t+s) generates the same trajectory (C1) as x = f1(t) and y = y1(t). But at t = t0 we have x = f1(t0+s) = f1(t1) = x0 and
y = y1(t0 + s) = y1(t1) = y0. Thus the solution
x = f1(t+s), y = y1(t+s) satisfies exactly the same initial conditions as the solution x = f0(t), y = y0(t) which generates the trajectory C0. Hence C0 and C1 are the same.
25. From the existence and uniqueness theorem we know that if the two solutions x = f(t), y = y(t) and x = x0, y = y0 satisfy f(a) = x0, y(a) = y0 and x = x0, y = y0 at t = a, then these solutions are identical. Hence f(t) = x0 and y(t) = y0 for all t contradicting the fact that the trajectory generated by [f(t), y(t)] started at a noncritical point.
26. By direct substitution
F'(t) = f'(t+T) = F[f (t+T), y(t+T)] = F[F(t), Y(t)] and
Y'(t) = y'(t+T) = G[f (t+T), y (t+T)], G[F(t), Y(t)].
Furthermore F(t0) = x0 and Y(t0) = y0. Thus by the existence and uniqueness theorem F(t) = f(t) and
Y(t) = y(t) for all t.
Section 9.3, Page 487
In Problems 1 through 4, write the system in the form of
182
Section 9.3
Eq.(4). Then if g(0) = 0 we may conclude that (0,0) is a critical point. In addition, if g satisfies Eq.(5) or Eq.(6), then the system is almost linear. In this case the linear system, Eq.(1), will determine, in most cases, the type and stability of the critical point (0,0) of the almost linear system. These results are summarized in Table 9.3.1.
3.
d / \ 0 0 > / \ (1+x )siny
x x
- = + 1 -
dt IYJ v -1 0 I YJ cosy
J
In this case the system can be written as
However, the
coefficient matrix is singular and g (x,y) = (1+x)siny
does not satisfy Eq.(6). However, if we consider the Taylor series for siny, we see that (1+x)siny - y =
3 5 3
siny - y + xsiny = -y /3! + y /5! + + x(y - y /3! + ), which does satisfy Eq.(6), using x = rcos0, y = rsin0. Thus the first equation now becomes dx dt d dt
coefficient matrix is now nonsingular and
= y + [(1+x) siny-y] and hence
\ V 0 1 > / \ + ( ! 1 \ A , where the
x x (1+x)siny-y
V YJ v -1 0 I yy v 1-cosy )
g(x,y) =
f ! 1 \
(1+x)siny-y
satisfies Eq.(6).
4.
1-cosy
In this case the system can be written as
d dt g =
( \
x
vyy
( 2
y
v 0
1 0 v1 1j
V yy
and thus A =
1 0 v1 1j
and
. Since g(0) =
0
0
we conclude that (0,0) is a
critical point. Following the procedure of Example 1, we
let x = rcos0 and y = rsin0 and thus 22 r sin 0
g1(x,y)/r = ---------- ^ 0 as r ^ 0 and thus the system is
almost linear. Since det(A-rl) = (r-1) , we find that
the eigenvalues are r = r = 1. Since the roots are
12
equal, we must determine whether there are one or two eigenvectors to classify the type of critical point. The
eigenvectors are determined by
V 0 0 1 "x1 >
V 1 0, Vx2 v I 0J
and hence
+
0
r
Section 9.3
183
there is only one eigenvector X =
Thus the critical
point for the linear system is an unstable improper node. From Table 9.3.1 we then conclude that the given system, which is almost linear, has a critical point near (0,0)
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