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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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which can be written in scalar form as xi = ciXi1) + c2Xi2)er2t and x2 = ciX21) + c2?22)er2t.
Assuming ?i2) 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 ?<2) .
Note that the family of lines is independent of c2. If ?i2) = 0, then the lines are vertical. If r2 > 0, the direction of motion will be in the same direction as indicated for ?<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 Tx 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 +
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 J
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 \[D )/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 = 2, so x = 4e-t and y = 2e-2t. Solving the first equation for e-t 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.
3. 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
\ x
dt
V y )
equation
0 -1 >
1 0
-r -1
1 -r
r \ x
which has the characteristic
V y)
+ 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 j and v(t) =
f , \ cost
sint
f \ sint
J
-cost
\ x
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
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).
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.
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