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

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d0

d0 ( dt ) .

--- ) . Thus

(1/2)mL2(d0/dt)2 = mgL(cos0 - cosa). Thus

(d0/dt)2 = (2g/L)(cos0 - cosa). Since we are releasing

the pendulum with zero velocity from a positive angle a,

186

Section 9.4

the angle 9 will initially be decreasing so d9/dt < 0.

If we restrict attention to the range of 9 from 9 = a to

9 = 0, we can assert d9/dt = -ä/2g/L ä/cos9 - cosa .

Solving for dt gives dt = - VL/2g d9/\/cos9 - cosa .

27b.Since there is no damping, the pendulum will swing from

its initial angle a through 0 to -a, then back through 0 again to the angle a in one period. It follows that 9(T/4) = 0. Integrating the last equation and noting that as t goes from 0 to T/4, 9 goes from a to 0 yields

T/4 = -ä/ L/2g 10 (1/V cos9 - cosa )d9.

Ja

dx

28a. If — = y, then dt

d2x

dt2

dy

= — = -g(x) dt

c(x)y.

28b.Under the given assumptions we have g(x) = g(0) + g'(0)x 2

+ g"(X )x /2 and c(x) = c(0) + c'(X )x, where 1 2

0 < x , x < x and g(0) = 0.

1 2

dy

= (-g(0) - g' (0)x)

dt

c(0)y

and thus the system can be written as

Hence

[g"(Xi)x2/2 - c'(X2)xy]

d / \ 1 ¥ \ Ã 0 ¯

x 0 x

dt I YJ v -g' (0) -c(0) j I YJ v-g"(^1)x2/2 - c'(^2)xy ,

from which the results follow.

Section 9.4, Page 501

3b. x(1.5 - .5x - y) = 0 and y(2 - y - 1.125x) = 0 yield (0,0),

(0,2) and (3,0) very easily. The fourth critical point is the intersection of .5x + y = 1.5 and 1.125x + y = 2, which is (.8,1.1).

3c. From Eq (5)

d

we get — dt

/ \

u

V vV

For

1.5-x0-y0 -x0 Yu

-1.125y0 2-2y0-1.125x0ylv v

(0,0) we get u' = 1.5u and v' = 2v, so r = 3/2 and r = 2, and thus (0,0) is an unstable node. For (0,2) we have u' = -.5u and v' = -2.25u-2v, so r = -.5, -2 and thus (0,2) is an asymptotically stable node. For (3,0) we get u' = -1.5u-3v and v' = -1.375v, so r = -1.5, -1.375 and hence (3,0) is an symptotically stable node. For (.8,1.1) we have u' = -.4u -.8v and v' = -1.2375u - 1.1v which give r = -1.80475, .30475 and thus (.8,1.1) is an unstable saddle

point.

Section 9.4

187

3e.

5b. The critical points are found by setting dx/dt = 0 and

dy/dt = 0 and thus we need to solve x(1 - x - y) = 0 and

y(1.5 - y - x) = 0. The first yields x = 0 or y = 1 - x

and the second yields y = 0 or y = 1.5 - x. Thus (0,0), (0,3/2) and (1,0) are the only critical points since the two straight lines do not intersect in the first quadrant (or anywhere in this case). This is an example of one of

the cases shown in Figure 9.4.5 a or b.

6b. The critical points are found by setting dx/dt = 0 and

dy/dt = 0 and thus we need to solve x(1-x + y/2) = 0 and

y(5/2 - 3y/2 + x/4) = 0. The first yields x = 0 or

y = 2x - 2 and the second yields y = 0 or y = x/6 + 5/3. Thus we find the critical points (0,0), (1,0), (0,5/3)

and (2,2). The last point is the intersection of the two

straight lines, which will be used again in part d.

6c. For (0,0) the linearized system is x' = x and y' = 5y/2,

which has the eigenvalues r1 = 1 and r2 = 5/2. Thus the

origin is an unstable node. For (2,2) we let x = u + 2 and y = v + 2 in the given system to find

(since x' = u' and y' = v') that

du/dt = (u+2)[1 - (u+2) + (v+2)/2]

dv/dt = (v+2)[5/2 - 3(v+2)/2 + (u+2)/4]

Hence the linearized equations are

f Y u

= (u+2)(-u+v/2) and

= (v+2)(u/4 - 3v/2).

´ -2 1 '

1/2 -3,

which has the eigenvalues r1,2 = (-5 ± <\/~3)/2. Since these are both negative we conclude that (2,2) is an asymptotically stable node. In a similar fashion for

188

Section 9.4

system

Ã Y u

/ \

u

¨ vJ

This has

(1,0) we let x = u + 1 and y = v to obtain the linearized ^-1 1/2 0 11/4,

^ = -1 and r2 = 11/4 as eigenvalues and thus (1,0) is an unstable saddle point. Likewise, for (0,5/3) we let

11/6 0'

.5/12 -5/2,

x = u, y = v + 5/3 to find

Ã Y u

r \

u

¨ vJ

as the

corresponding linear system. Thus ^ = 11/6 and

r2 = -5/2 and thus (0,5/3) is an unstable saddle point.

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