# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

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7b.

7c. For (0,0) since all trajectories leave this point, this is an unstable node. For (0,2) and (1,0) since the trajectories tend to these points, respectively, they are asymptotically stable nodes. For (1/2,1/2), one trajectory tends to (1/2,1/2) while all others tend to infinity, so this is an unstable saddle point.

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 n 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 ^ > t0. Let s = ^ - t0. Since the system

is autonomous, the result of Problem 23, with s replaced by -s shows that x = f ^t + s) and y = y^t + s) generates the same trajectory (C^ as x = f i(t) and y = yx(t). But at t = t0 we have x = f2(t0+s) = fi(ti) = 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 / \ x V 0 0 > / \ x ' (1+x )siny^

= + Vvv 1 -V.

dt IYJ v -1 0 7 1 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

/ \ x r 0 1 > / \ x + 7 , -i , 3 (1+x)siny-y , where the

V YJ Vv -1 0 7 V y 7 v 1-cosy 7

g(x,y) =

f 1i \ 3

(1+x)siny-y

satisfies Eq.(6).

4.

1-cosy

In this case the system can be written as

d

dt

g =

( \

x

iY7

( 2 3

ry

v 0 7

1 0 v1 1j

V y )

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

r 0 0 37 ^x1 ^ (r 0 37

V 1 0, Vx2 7 v 0,

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)

which is either a node or spiral point (depending on how

the roots bifurcate) which is unstable.

6a. The critical points are the solutions of x(1-x-y) = 0 and y(3-x-2y) = 0. Solutions are x = 0, y = 0; x = 0,

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