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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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4km) 1/2]/2m. In the underdamped
case c2 - 4km < 0, the characteristic roots are complex with negative real parts and thus the critical point (0,0) is an asymptotically stable spiral point. In the overdamped case c2 - 4km > 0, the characteristic roots are real, unequal, and negative and hence the critical point (0,0) is asymptotically stable node. In the critically damped case c2 - 4km = 0, the characteristic roots are equal and negative. As indicated in the solution to Problem 4, to determine whether this is an improper or proper node we must determine whether there
a -1j
Section 9.1
177
r c/2m 1 ^ '%1' ( 0 ^
v -k/m -c/2my ?2 j V 0,
are one or two linearly independent eigenvectors. The eigenvectors satisfy the equation
which has just one solution if
-k/m -c/2mj (X2 ) V. 0)
c2 - 4km = 0. Thus the critical point (0,0) is an asymptotically stable improper node.
18a. If A has one zero eigenvalue then for r = 0 we have
det(A-rl) = detA = 0. Hence A is singular which means Ax = 0 has infinitely many solutions and consequently there are infinitely many critical points.
18b. From Chapter 7, the solution is x(t) = c1X<1) + c2?<2)er2t, which can be written in scalar form as xi = ciXi1) + c2Xi2)er2t and X2 = ciX21) + c2^22)er2t. Assuming Xi2) n 0, the first equation can be solved for
r t
c2e 2 , which is then substituted into the second equation to yield x2 = c1X21) + [X22)/Xi2)][x1-c1Xi1)].
These are straight lines parallel to the vector X(2) .
Note that the family of lines is independent of c2. If Xi2) = 0, then the lines are vertical. If r2 > 0, the direction of motion will be in the same direction as indicated for X(2) . If r2 < 0, then it will be in the opposite direction.
19a. Det(A-rI) = r2 - (a11+a22)r + a11a22 - a21a12 = 0. If an + a22 = 0, then r2 = -(aiia22 - a2iai2) < 0 if a11a22 - a21a12 > 0.
19b. Eq.(i) can be written in scalar form as dx/dt = ax +
11
a y and dy/dt = a x + a y, which then yields Eq.(iii). 12 21 22
Ignoring the middle quotient in Eq.(iii), we can rewrite
that equation as (a x + a y)dx - (a x + a y)dy = 0,
21 22 11 12
which is exact since a = -a from Eq.(ii)..
22 11
19c. Integrating fx = a x + a y we obtain f = a x2/2 + a xy x 21 22 21 22
/ /
+ g(y) and thus a x + g = -a x - a y or g = -a y 22 11 12 12
using Eq.(ii). Hence a x2/2 + a xy - a y2/2 = k/2 is
21 22 12
the solution to Eq.(iii). The quadratic equation Ax2 +
22 Bxy + Cy = D is an ellipse provided B - 4AC < 0. Hence
for our problem if
178
Section 9.2
2
a + a a < 0 then Eq.(iv) is an ellipse. Using
22 21 12
2
a + a = 0 we have a = -a a and hence
11 22 22 11 22
-a a + a a < 0 or a11a22 11 22 21 12
a a > 0, which is true by
21 12
Eqs.(ii). Thus Eq.(iv) is an ellipse under the conditions of Eqs.(ii).
20.
The given system can be written as ---------
dt
Thus the eigenvalues are given by
/ \ a
x 11
v yJ a O 1
a V \ 12 x
a v y J
22 )
r2-(a
+a)r + a a -a a = 0 and using the given
11 22 11 22 12 21
definitions we rewrite this as r2
-1,2
pr + q = 0 and thus = (p ± VP2 -4q)/2 = (p ± \!~K )/2. The results are
now obtained using Table 9.1.1.
Section 9.2, Page 477
1. Solutions of the D.E. for x are y are x = Ae-t and y = Be-2t respectively. x(0) = 4 and y(0) = 2 yield
A
4 and B
and y = 2e
-2t
2, so x = 4e-Solving the
first equation for e and then substituting into the
second yields y = 2[x/4] 2 = x2/8, which is a parabola. From the original D.E., or from the parametric solutions, we find that 0 < x ? 4 and 0 < y ? 2 for t > 0 and thus only the portion of the parabola shown is the trajectory, with the direction of motion indicated.
Utilizing the approach indicated in Eq.(14), we have dy/dx = -x/y, which separates into xdx + ydy = 0. Integration then yields the circle x2 + y2 = c2, where c2 = 16 for both sets of I.C. The direction of motion can be found from the original D.E. and is counterclockwise for both I.C. To obtain the parametric equations, we write the system in the form
d
dt
y
equation
0 -1 >
1 0 y
-r -1
1 -r
r \ x
which has the characteristic
V yJ
+ 1 = 0, or r = ± i.
Following
2
r
Section 9.2
179
the procedures of Section 7.6, we find that one solution
of the above system is
it
cost + isint
and thus
v sint - icost y and v(t) =
f , \
cost
sint
f ? 4- A
sint
y
-cost
7a.
two real solutions are u(t) =
The general solution of the system is then
= c1u(t) + c2v(t) and hence the first I.C. yields
V YJ
c1 = 4, c2 = 0, or x = 4cost, y = 4sint. The second I.C. yields c1 = 0, c2 = -4, or x = -4sint, y = 4cost. Note that both these parametric representations satsify the form of the trajectories found in the first part of this problem.
The critical points are given by the solutions of x(1-x-y) = 0 and y(1/2 - y/4 - 3x/4) = 0. The solutions
corresponding to either x = 0 or y = 0 are seen to be x = 0, y = 0; x = 0, y = 2; x = 1, y = 0. In addition, there is a solution corresponding to the intersection of the lines 1 - x - y = 0 and 1/2 - y/4 - 3x/4 = 0 which is the point x = 1/2, y = 1/2. Thus the critical points are (0,0), (0,2), (1,0), and (1/2,1/2).
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