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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
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3 - 2y = 0 which gives y = 3/2; y = 0 and 1 - x = 0 which
give x = 1; and 1 - x - y = 0, 3 - x - 2y = 0 which give
x = -1, y = 2. Thus the critical points are (0,0), (0,3/2), (1,0) and (-1,2).
6b, For the critical point (0,0) the D.E. is already in the
6c. form of an almost linear system; and the corresponding linear system is du/dt = u, dv/dt = 3v which has the eigenvalues r1 = 1 and r2 = 3. Thus the critical point (0,0) is an unstable node. Each of the other three critical points is dealt with in the same manner; we consider only the critical point (-1,2). In order to translate this critical point to the origin we set x(t) = -1 + u(t), y(t) = 2 + v(t) and substitute in the D.E. to obtain
du/dt = -1 + u - (-1+u)2 - (-1+u)(2+v) = u + v - u2 - uv and
dv/dt = 3(2+v) - (-1+u)(2+v) - 2(2+v)2 = -2u - 4v - uv - 2v2.
Writing this in the form of Eq.(4) we find that
A =
and g =
2
vuv + 2v y
which is an almost
linear system. The eigenvalues of the corresponding linear system are r = (-3 ± "\J~9 + 8 )/2 and hence the critical point (-1,2), of the original system, is an unstable saddle point.
22
10a. The critical points are solutions of x + x + y = 0 and y(1-x) = 0, which yield (0,0) and (-1,0).
10b. For (0,0) the D.E. is already in the form of an almost
linear system and thus du/dt = u and dv/dt = v. For
(-1,0) we let u = x+1, v = y so that substituting x = u-1
du 2 2
and y = v into the D.E. we obtain — = -u + u + v and
dt
v1y
2
u
+ uv
184
Section 9.3
dv
dt
u'
= 2v - uv. Thus the corresponding linear system is = -u and v 10c. For (0,0) A =
= -2v.
1 ox
r = 1, so that
2
which has r v 0 1) 1
(0,0), for the nonlinear system, will be either a node or spiral point, depending on how the roots bifurcate. In any case, since r^ and r^ are positive, the system will
be unstable. For (-1,0) A
/ _ _ \ -1 0
and thus
r = -1 and r = 2, and hence the nonlinear system, from Table 9.3.1, has an unstable saddle point at (-1,0).
d / \ x r 1 0 ] / \ x V 0 1
18a. The system is = + 3 V x 1
dt I yj V 0 -2 , I yj
and thus is
almost linear using the procedures outlined in the earlier problems. The corresponding linear system has the eigenvalues r1 = 1, r2 = -2 and thus (0,0) is an unstable saddle point for both the linear and almost linear systems.
18b. The trajectories of the linear system are the solutions of dx/dt = x and dy/dt = -2y and thus x(t) = c1et and y(t) = c2e-2t. To sketch these, solve the first equation for et and substitute into the second to obtain Y = cfc2/x2, Ci ^ 0.
Several trajectories are shown in the figure.
Since x(t) = c1et, we must pick c1 = 0 for x ^ 0 and t ^ •. Thus x = 0, y = c2e-2t (the vertical axis) is the only trajectory for
?
WF
18c
which x ? 0, y ? 0 as t ? •.
For x n 0 we have dy/dx = (dy/dt)/(dx/dt) = (-2y+x3)/x.
This is a linear equation, and the general solution is y = x3/5 + k/x2, where k is an arbitrary constant. In addition the system of equations has the solution x = 0, y = Be-2t. Any solution with its initial point on the y-axis (x=0) is given by the latter solution. The trajectories corresponding to these solutions approach
Section 9.3
185
the origin as t —> oo. The trajectory that passes through the origin and divides the family of curves is given by k = 0, namely y = x3/5. This trajectory corresponds to the trajectory y = 0 for the linear problem.
Several trajectories
are sketched in the figure
22a.
2n
22b. From the graphs in part a, we see that vc is between v = 2
and v = 5. Using several values for v, we estimate vc @ 4.00.
23a.
For v = 2, the motion is damped oscillatory about x = 0. For v = 5, the pendulum swings all the way around once and then is a damped oscillation about x = 2p (after one full rotation). For problem 22, this later case is not damped, so x continues to increase, as shown earlier.
27a. Setting c = 0 in Eq.(10) of Section 9.2 we obtain
mL2d20/dt2 + mgLsin0 = 0. Considering d0/dt as a function of 0 and using the chain rule we have
d ( d0 ) d ( d0 ) d0 1 d
( ) = ( ) = dt dt d0 dt dt 2
(1/2)mL2d[(d0/dt) 2]/d0 = -mgLsin0. Now integrate both sides from a to 0 where d0/dt = 0 at 0 = a:
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 = -V2g/L yjcos9 - 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 = -^/ 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
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