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174 CHAPTER3LINEARSYSTEMS
EXERCISES FOR SECTION 3.1
1. Since a > 0, Paul’s making a proÀt(x > 0) has a beneÀcial effect on Paul’s proÀts in the future
because the ax term makes a positive contribution to dx/dt. However, since b < 0, Bob’s making
aproÀt(y > 0) hinders Paul’s ability to make proÀt because the by term contributes negatively to
dx/dt. Roughlyspeaking, business is good for Paul if his store is proÀtable and Bob’s is not. In fact,
since dx/dt = x y,Paul’sproÀts will increase whenever his store is more proÀtable than Bob’s.
Even though dx/dt = dy/dt = x y for this choice of parameters, the interpretation of the
equation is exactly the opposite from Bob’s point of view. Since d < 0, Bob’s future proÀts are hurt
whenever he is proÀtable because dy < 0. But Bob’s proÀts are helped whenever Paul is proÀtable
since cx > 0. Once again, since dy/dt = x y, Bob’s proÀts will increase whenever Paul’s store is
moreproÀtable than his.
Finally, note that both x and y change by identical amounts since dx/dt and dy/dt are always
equal.
2. Since a = 2, Paul’s making a proÀt(x > 0) has a beneÀcial effect on Paul’s future proÀts because
the ax term makes a positive contribution to dx/dt. However, since b =1, Bob’s making a proÀt
(y > 0) hinders Paul’s ability to make proÀt because the by term contributes negatively to dx/dt.
In some sense, Paul’s proÀtability has twice the impact on his proÀts as does Bob’s proÀtability. For
example, Paul’s proÀts will increase whenever his proÀts are at least one-half of Bob’s proÀts since
dx/dt = 2x y.
Since c = d = 0, dy/dt = 0. Consequently, Bob’s proÀts are not affected by the proÀtability of
either store, and hence his proÀts are constant in this model.
3. Since a = 1andb = 0, we have dx/
dt = x. Hence, if Paul is making a proÀt(x > 0), then those
proÀts will increase since dx/dt is positive. However, Bob’s proÀts have no effect on Paul’s proÀts.
(Note that dx/dt = x is the standard exponential growth model.)
Since c = 2andd = 1, proÀts from both stores have a positive effect on Bob’s proÀts. In some
sense, Paul’s proÀts have twice the impact of Bob’s proÀts on dy/dt.
4. Since a =1andb = 2, Paul’s making a proÀt has a negative effect on his future proÀts. However,
if Bob makes a proÀt, then Paul’s proÀts beneÀt. Moreover, Bob’s proÀtability has twice the impact
as does Paul’s. In fact, since dx/dt =x + 2y,Paul’sproÀts will increase if x + 2y > 0or,in
other words, if Bob’s proÀts are at least one-half of Paul’s proÀts.
Since c = 2andd =1, Bob is in the same situation as Paul. His proÀts contribute negatively
to dy/dt since d =1. However, Paul’s proÀtability has twice the positive effect.
NotethatthismodelissymmetricinthesensethatbothPaulandBobperceiveeachothersproÀts
in the same way. This symmetry comes from the fact that a = d and b = c.
x
dY 21 x dY 03
5. Y = , dt = Y6.Y= , dt= Y
y 11 y 0.33π
⎛ p ⎞ dY ⎛ 3 2 7 ⎞
7. Y = ⎜ ⎟, =⎜ ⎟Y
q 206
⎝ ⎠ dt ⎝ ⎠
r 07.32
3.1 Properties of Linear Systems and The Linearity Principle 175
8. dx =3x +2πy 9. dx =βy
dt dt
dy =4x y dy =γx y
dt dt
10. (a) y (b) y (c) x, y
2 2 40 x(t)
❅
20 ❅❘
❅■
x ❅y(t)
x 22 t
22 12
2 2
11. (a) y (b) y (c) x, y x(t) y(t)
2 2 3
✠ ✠
10 20 t
x x 3
22
22
2 2
12. (a) y (b) y (c) y(t)x, y
2 20
2 ✠
10
x t
x 11
10
22
❅■
22 ❅x(t)
2 2
176 CHAPTER3LINEARSYSTEMS
13. (a) y (b) y (c) x, y y(t)
2 2 x(t)
1
✠ ✠
t
x 1 123
x 22
22
2 2
14. (a) If a = 0, then detA = ad bc = bc. Thus both b and c are nonzero if detA = 0.
(b) Equilibrium points (x , y ) are solutions of the simultaneous system of linear equations
0 0
⎧
⎨ ax +by =0
0 0
⎩ cx +dy =0.
0 0
If a = 0, the Àrst equation reduces to by = 0, and since b = 0, y = 0. In this case, the
0 0
second equation reduces to cx = 0, so x = 0 as well. Therefore, (x , y ) = (0,0) is the only
0 0 0 0
equilibrium point for the system.
15. ThevectorÀeldatapoint(x ,y )is(ax +by ,cx +dy ),soinorderforapointtobeanequilibrium
0 0 0 0 0 0
point, it must be a solution to the system of simultaneous linear equations
⎧
⎨ ax +by =0
0 0
⎩ cx +dy =0.
0 0
If a = 0, we know that the Àrst equation is satisÀed if and only if
b
x = y .
0 a 0
Nowweseethat any point that lies on this line x = (b/a)y also satisÀes the second linear
0 0
equation cx + dy = 0. In fact, if we substitute a point of this form into the second component of
0 0
the vector Àeld, we have
b
cx +dy =c y +dy
0 0 a 0 0
bc
= +d y
a 0
=
adbcy
a 0
= detAy
a 0
=0,
3.1 Properties of Linear Systems and The Linearity Principle 177
since we are assuming that detA = 0. Hence, the line x = (b/a)y consists entirely of equilib-
rium points. 0 0
If a = 0andb = 0, then the determinant condition detA = ad bc = 0 implies that c = 0.
Consequently, the vector Àeld at the point (x , y ) is (by ,dy ).Sinceb = 0, we see that we get
0 0 0 0
equilibrium points if and only if y = 0. In other words, the set of equilibrium points is exactly the
x-axis. 0
Finally, if a = b = 0, then the vector Àeld at the point (x , y ) is (0,cx + dy ). In this case,
0 0 0 0
weseethatapoint(x ,y )isanequilibriumpoint if and only if cx +dy = 0. Since at least one of
0 0 0 0
c or d is nonzero, the set of points (x , y ) that satisfy cx + dy = 0 is precisely a line through the
origin. 0 0 0 0
16. (a) Let v = dy/dt.Thendv/dt = d2y/dt2 =qy p(dy/dt) =qy pv. Thusweobtainthe
system
dy =v
dt
dv =qy pv.
dt
In matrix form, this system is written as
⎛ dy ⎞
⎜ dt ⎟= 01y .
⎝ dv ⎠ q p v
dt
(b) The determinant of this matrix is q. Hence, if q = 0, we know that the only equilibrium point
is the origin.
(c) If y is constant, then v = dy/dt is identically zero. Hence, dv/dt = 0.
Also, the system reduces to
⎛ dy ⎞
⎜ dt ⎟= 01y ,
⎝ dv ⎠ q p 0
dt
which implies that dv/dt =qy.
Combining these two observations, we obtain dv/dt =qy = 0, and if q = 0, then
y = 0.
17. The Àrst-order system corresponding to this equation is
dy =v
dt
dv =qy pv.
dt
(a) If q = 0, then the system becomes dy
dt =v
dv =pv,
dt
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