only if a11 a22 − a12 a21 = 0. In other words,
λ 1= 0 is an eigenvalue if and only if det(A) = 0.
Cv1� + v2 + i3 = 0.
R
3.2It follows
Systems
of Two First Order Linear Differential Equathat
1
Cv1� = − v1 − i3 and Li�3 = v1 .
tions
R
� �
x
28.
Let
i
,
i
,
i
and
i
be
the
current
through
the
1
ohm
2 ohm resistor, inductor
1 2 3is autonomous,
4
1. The system
nonhomogeneous. Let u = resistor,
. Then
y
and capacitor, respectively. Assign v1 , v2 , v3 and v4 as the respective voltage drops. Based
� drops
�
� �each loop satisfy
on Kirchoff’s second law, the net voltage
0 1 around
0
�
u +OF TWO
.
152
CHAPTERu3.= SYSTEMS
FIRST ORDER EQUATIONS
1 0
4
v1 + v3 + v4 = 0, v1 + v3 + v2 = 0 and v4 − v2 = 0.
2.
The system
is nonautonomous,
Applying
Kirchoff’s
first law to thenonhomogeneous.
upper right node, we have
� �
x
3. The system is nonautonomous, homogeneous.
i1 − i3 = Let
0. u = y . Then
�
�
Likewise, in the remaining nodes, we � have −2t 1
u =
u.
3 −1
i2 + i4 − i1 = 0 and i3 − i4 − i2 = 0.
4. The system is autonomous, nonhomogeneous.
Combining the above equations, we have
� �
x
5. The system is autonomous, homogeneous. Let u =
. Then
v4 − v2 = 0, v1 + v3 + v4 = 0 and y i2 + i4 − i3 = 0.
�
�
Using the current-voltage relations, we
� have3 −1
u =
u.
1 2
v1 = R1 i1 , v2 = R2 i2 , Li�3 = v3 , Cv4� = i4 .
6. The system is nonautonomous, homogeneous.
� �
Combining these equations, we have
x
7. The system is nonautonomous, nonhomogeneous. Let u =
. Then
y1
�
�
R1 i3 + Li3 + v�4 = 0 and
4 =
� Cv�
�i3 − R2 v4 .
1
1
4
u� =
u+
.
3.2.
SYSTEMS
OFv4TWO
FIRST
ORDER
LINEAR
DIFFERENTIAL
EQUATIONS 153
−2system
sin
t of equations
0
Now set
i3 = i and
= v, to
obtain
the
1
� homogeneous.
8.
The system
is autonomous,
Liwe
= have
−R1 i − v and Cv � = i −
v.
Combining
these
equations,
R2
9.
1
Finally, using the fact that
=�31,+R
= 1 and
R 1 i3 R
+1 Li
v42 = 02, Land
Cv4� C
==
i3 1/2,
− we
v4 .have
R2
i� = −i − v and v � = 2i − v,
Now set i3 = i and v4 = v, to obtain the system of equations
as claimed.
1
Li� = −R1 i − v and Cv � = i −
v.
29. Let i1 , i2 , i3 and i4 be the current through the resistors, inductor
and capacitor, respecR2
tively. Assign v1 , v2 , v3 and v4 as the respective voltage drops. Based on Kirchoff’s second
30.
law, the net voltage drops around each loop satisfy
(a) Let Q1 (t) and Q
the amount of salt in the respective tanks at time t. Based on
v12 (t)
+ vbe
3 + v4 = 0, v1 + v3 + v2 = 0 and v4 − v2 = 0.
conservation of mass, the rate of increase of salt is given by
Applying Kirchoff’s first law to the upper right node, we have
rate of increase = rate in − rate out.
i3 − (i2 + i4 ) = 0.
For Tank 1, the rate of salt flowing in from Tank 2 is Q202 · 1.5 = .075Q2 ounces/minute.
Likewise,
in the remaining
nodes,
have
In addition,
salt is flowing
in we
from
a separate source at the rate of 1.5 ounces/minute.
Therefore, the rate of salt flowing in to Tank 1 is rin = .075Q2 + 1.5. The rate of flow
3 = 0 and i2 + i4 − i1 = 0.
out of Tank 1 is rout = Q30i11 ·−3 i=
0.1Q1 ounces/minute. Therefore,
Combining the above equations, dQ
we have
1
= −0.1Q1 + .075Q2 + 1.5.
v4 − v2 = 0, v1dt+ v3 + v4 = 0 and i2 + i4 − i3 = 0.
Similarly,
for Tank 2, relations,
salt is flowing
in from Tank 1 at the rate of
Using
the current-voltage
we have
Q1
30
· 3 = 0.1 oz/min.
= −0.1Q
= q2 +
− 2 +. 1.5.
1 + .075Q
dt dt
30
100
(f) The eigenvalue close to 0 corresponds0 to the slow temporal
change in the system, while
The initial conditions are Q1 (0) = Q1 and Q2 (0) = Q02 .
Q1
Similarly,
for Tank
flowing
in
from
Tank
1
at
the
rate
ofchange
·3 =
the eigenvalue
close 2,
to salt
−(kis
+
k
)
corresponds
to
the
fast
temporal
in 0.1
the oz/min.
system.
1
2
30
addition,
salt values
is flowing
from a by
separate
at +
the
3 oz/min.
Also, salt
(b) In
The
equilibrium
are in
obtained
solvingsource
−Q1 /15
Q2rate
/100of
+ 3q
1 = 0 and Q1 /30 −
26. is
E
E
flowing
out
of
Tank
2
at
the
rate
of
4Q
/20
=
.2Q
oz/min.
Therefore,
3Q2 /100 + q2 = 0. Its solution is given by2 Q1 = 54q12 + 6q2 and Q2 = 60q1 + 40q2 .
(a) For α = 0.5, the characteristic equation
is 2λ2 +4λ+1 = 0. Therefore, the eigenvalues are
dQ2 the system
√ to be able to solve
(c) We would need
= 0.1Q1 − 0.2Q
3.−1 + 1/√2 is v1 = �−√2 1�t .
2 +=
λ = −1 ± 1/ 2. The eigenvector
corresponding
to
λ
1
dt
√
�√
�t
+ 6q−2 1/
= 602 is v2 =
The eigenvector corresponding to λ254q
=1 −1
2 1 . Therefore, the
The initial conditions are Q1 (0) = 55 and Q2 (0) = 26.
general solution is
60q1 + 40q2 = 50.
�
�√ �
√as�
(b) This system can be written in matrix
form
√
√
−
2
2 a physically realistic
(−1+1/
2)t
(−1−1/
2)t
The solution of this
system
is
q
=
7/6
and
q
=
−1/2
which
is
not
2 + c2 e
x(t) = c1 e 1�
.
�
�
�
�
�
1 those
1equilibrium state.
solution. Therefore, it is not
possible
to
have
values
as
an
−0.1
0.075
Q
1.5
1
Q� (t) =
+
.
0.1the −0.2
Q2 point is
3 a stable node.
Since
both
eigenvalues
are
negative,
equilibrium
(d) We can write
(b)
For α =dQ
2, the
characteristic equation is λ2 + 2λ −E1 = 0. Therefore,
the eigenvalues
are
(c) Setting
satisfy
√1 /dt = 0 = dQ2 /dt, we see that anQequilibrium
√ solution must
�
√ �t the
1
q2 = −9q1 +
λ = −1 ± 2. The eigenvector corresponding
to λ = −1 + 2 is v1 = 1 − 2 . The
system
�
� � �61 �� �
√
√ �t
Q1vQ2 E= 1.5
eigenvector corresponding to 0.1
λ2 = −0.075
−1 − 32 is
1
2 . Therefore, the general
= − q1 Q
+2 2= 3 .
−0.1 q20.2
solution is
� 2 � 40
� �
√
√
1
1
(−1+ 2)t
(−1− 2)t √
√
Finding
thethe
inverse
of
the=
above
and qmultiplying,
we find
that
x(t)
ctwo
+−plane.
c2 e
.
1 e lines
which are
equations
ofmatrix
in −
the
q
The
intercepts
of the first line
21 2
2
E
E�
�
�
�
�
�
are (Q1 /54, 0) and (0, Q1 /6).
The intercepts
of the second
line are (QE
2 /60, 0) and
1 the 0.525
QE
42
SinceEthe
eigenvalues have opposite
sign,
equilibrium
point
is a saddle point.
1
=
=
(0, Q2 /40). Therefore, the system
will
have
a
unique
solution
in
the
first
quadrant as
QEE
0.45
36
0.0125
2
E
E
E
E
E
2
as Q1 /54
≤Q
≤ Q1 /6.
is, 10/9=≤0.QThe
(c) long
For general
α,√the
characteristic
equation
is λ That
+2λ+1−α
2 /60 or Q2 /40
2 /Qeigenvalues
1 ≤ 20/3. are given
√
by λ = −1 ± α. For 0.5 ≤ α ≤ 2, both eigenvalues √
are real and clearly −1 − α < 0.
31. Therefore, we just need to determine whether −1 + α = 0. We see this occurs when
3.3α = Homogeneous
Linearpoint
Systems
with
Constant
1. At that value the equilibrium
switches from
a saddle
point (for α >Co1) to
a stable
node (for α < 1).
efficients
27.
1. We look for eigenvalues and eigenvectors of
(a) For the given data, the system can be written
as
�
�
� �A =� 3 −2 . � � �
d i
−1/2 −1/2
i
= 2 −2
.
v
3/2
−5/2
v
dt
We see that det(A − λI) = λ2 − λ − 2 = (λ − 2)(λ +
1). Therefore, the eigenvalues are given
The characteristic equation for this system is λ2 + 3λ + 2 = 0. Therefore, the eigenvalues
by λ = 2, −1.
are
by v1 =
� given
�t by λ = −1, −2. For λ = −1, a corresponding eigenvector� is given
�t
1 1 . For λ = −2, a corresponding eigenvector is given by v2 = 1 3 . Therefore,
the general solution is
� �
� �
� �
i
−t 1
−2t 1
c1 e
+c e
.
186
CHAPTERv 3.= SYSTEMS
1 OF2 TWO3FIRST ORDER EQUATIONS
(b) Since both eigenvalues are real and negative, the equilibrium point (0, 0) is a stable node.
Therefore, i(t) → 0 and v(t) → 0 as t → ∞.
28.
(a) For the general equation (i), the characteristic equation is given by
�
�
L + CR1 R2
R1 + R2
2
λ +
λ+
= 0.
LCR2
LCR2
The eigenvalues are real and distinct provided that the discriminant is positive. That is,
�
L + CR1 R2
LCR2
�2
−4
�
R1 + R2
LCR2
�
> 0,
it follows that both roots are =
negative.
Therefore, i4= 0, v = 0 .is a stable node.
e
v
8 sin(t/2) + 3 cos(t/2)
(d) SinceComplex
the eigenvaluesEigenvalues
have negative real part, all solutions converge to the origin.
3.4
22.
1.
1 �3 −2
1�
2
(a)
The
characteristic
equation
is
λ
+
λ
+
= TWO
0. Therefore,
eigenvalues
are
208
CHAPTER 3. A
SYSTEMS
OF
FIRST the
ORDER
EQUATIONS
=
RC
4 CL
−1
�
1
1
4R2 C
implies
Plugging in for A − 1/2I, this equation
λ = − becomes
± 2
1−
.
2RC
det(A 2RC
− λI) =�
λ − 2λ
�
� + 5.
�L
3/4
3/4
4R2 C
The eigenvalues
are real
different
that1 1λ−
> 0. Otherwise, they are
Therefore,
the eigenvalues
areand
given
by λ =provided
1±w
2i.=Now,
1 .= 1
L + 2i implies
−3/4 −3/4
−1
complex conjugates.
�
�
2 − 2i
−2
We With
see that
− eigenvalues
λ1 I =
�are λ�= −1 ± i.. An eigenvector corresponding
(b)
the specified values, A
the
4 4/3 −2 − 2i
206 to λ1 = −1 + i is CHAPTER 3. SYSTEMS
TWO FIRST ORDER EQUATIONS
w = � OF
01 �
Therefore,
v1 =
�
� .
is a Equation
solution of(iii)
thiscan
equation.
Therefore,
(b)
be rewritten
as
1−4i
v1 =
� 1�
� �
−i
Therefore, one solution is
y)dx1+
(a11 xt/2
+ a4/3
21 x + a
22
12 y)dy = 0.
t/2
� � −(a
�
�
x2 (t) = te
+e
i
1 −1
0
= e(−1+i)t
To show that this equation
is exact,−1we+need
to show that for M = −(a21 x + a22 y) and
i
v
�y the
�
N a= solution
(a11 x + of
a12our
y), M
.� We see
that M
= −a
Nx = �a11
Then,
usingis the
is also
original
system.
Therefore,
general
of. the
system
y = Nx
22 andsolution
cos
t
sin
t
−t
−t
fact that a11 + a22 = 0, =wee see
that
N
=
a
=
−a
=
M
.
Therefore,
the
equation
is
y
� � x � 11 �+ ie�22
�
�� .
−
cos
t
−
sin
t
cos
t
−
sin
t
1
1
4/3
exact.
x(t) = c1 et/2
+ c2 tet/2
+ et/2
.
−1
−1
0
Therefore, the general solution is
(c) Since the equation
M�with respect�to x. We conclude
� is
� exact, we
� begin by integrating
�
t Then, taking
sin t
that ψ(x, y) = −ai21 x2=/2c −e−ta22 xy +cos
h(y).
+ c2 e−t the derivative
. with respect to
1
�
�
t y. Therefore,
cos th−(y)
sin=t a12 y which implies
y, we have −a22 x v+ h (y) = N −=cos
a11t x−+
a12
2sin
2
that h(y) = a12 y /2. Therefore, the solution of this exact equation is given implicitly by
2
2
2 1 that c1 = 2 and
(c) With
see
−ck.
c2 =
1. Therefore,
we
1 +
−a21 xthe
/ −given
a22 xyinitial
+ a12 yconditions,
/2 = k or we
a21x2x
= 2a22 xy − a12 y 2 =
The
discriminant
of the
conclude
quadraticthat
formc1is= 2 and c2 = 3. Therefore, the solution of the IVP is
� �
�
�
i –1 −t2 0 2 cos t +1 3 sin t2
–2
2
(2a22 ) − 4a21 (−a12 ) =
= e4a22 + 4a12 ax1
= 4(−a
. 11 a22 + a12 a21 ).
v
cos t − 21
5 sin t
–1
Since a11 a22 − a12 a21 > 0, by equation (ii), we can conclude that the discriminant of the
(d) Since
the eigenvalues
havenegative,
negativeand,
real therefore,
part, all solutions
converge
to an
theellipse.
origin.
quadratic
form is always
the graph
is always
–2
23.
3.5The Repeated
Eigenvalues
(a)
characteristic equation
is λ2 − (a11 + a22 )λ + a11 a22 − a12 a21 = 0. Therefore, the
The solution grows as t → ∞.
eigenvalues are
��
3.
�
1.
� �
(a11 + a22 ) ± −3/2
(a11
+
− 4(a
−λaλ22 )2 −4
1 11 a22 − a12 a21 )
3
−
λ=
.
A −AλI
−=
λI = −1/4 2 −1/2 − λ
1
−1 − λ
Therefore,
equation,
see
order λfor
eigenvalues
to eigenvalue.
be purely
2
implies
det(A −
−from
λI) =this
=λλ22−
+2λ
2λ++1 1=we
=(λ(λ
1). 2Therefore,
. in
Therefore,
isonly
the eigenvalue.
only
λI)
−+
1)that
=λ the
1=is−1
the
Now
imaginary,
we
need
a
+
a
=
0
and
a
a
−
a
a
>
0.
11
22
11
22
12
21
Now
λ
=
−1
implies
λ = 1 implies
��
��
−1/2
1
2
−4
A−
λIλI
==
A−
. .
−1/4
1 −21/2
Therefore,
is an eigenvector for λ = −1
and
1 and
� �
2
v1 =
1
�� ��
22
xx11(t)
(t)==ee
11
−t
t
is one solution.
Next, we need to look for a solution w of
� �
3.5. REPEATED EIGENVALUES
209
Next, we need to look for a solution w of
� �
2
(A + I) w =
.
1
Plugging in for A + I, this equation becomes
�
�
� �
−1/2 1
2
w=
.
−1/4 1/2
1
We see that
is a solution of this equation. Therefore,
� �
0
w=
2
x2 (t) = te
� �
� �
2
−t 0
+e
1
2
−t
is also a solution of our original system. Therefore, the general solution of the system is
� �
�
� �
� ��
−t 2
−t 2
−t 0
x(t) = c1 e
+ c2 te
+e
.
1
1
2
2
x2
–2
1
0
–1
1
x1
2
–1
–2
212
CHAPTER 3. SYSTEMS OF TWO FIRST ORDER EQUATIONS
The solutions approach the origin as t → ∞.
The solutions tend to the origin as t → ∞.
4.
�
�
−3 − λ 5/2�
6.
�
A − λI = 2 − λ 1/2
A − λI = −5/2 2 − λ
−1/2 1 − λ
implies det(A − λI) = λ22 + λ + 1/4 = (λ + 1/2)2 .2 Therefore, λ = −1/2 is the only eigenvalue.
implies
− λI)
= λ − 3λ + 9/4 = (λ − 3/2) . Therefore, λ = 3/2 is the only eigenvalue.
Now λ det(A
= −1/2
implies
�
�
Now λ = 3/2 implies
� −5/2 5/2 �
A − λI = 1/2
.
−5/2 1/2
5/2 .
A − λI =
−1/2 −1/2
Therefore,
� �
Therefore,
� 1�
v = 1
v1 1=
1
−1
is an eigenvector for λ = 3/2 and
x1 (t) = e
is one solution.
Next, we need to look for a solution w of
3t/2
�
1
−1
�
�
1
�
1
−1
is one solution.
Next, we need to look for a solution w of
(A − 3/2I) w =
�
�
1
.
−1
Plugging in for A − 3/2I, this equation becomes
218
CHAPTER
TWO� FIRST ORDER EQUATIONS
� 3. SYSTEMS
� OF �
1/2
1/2
1
w=
.
−1/2 −1/2
−1
We see that
3
2
1
–3
–2
–1
x1
2
1
3
4
0
is a solution of this equation. Therefore,
� �
2
w= 2
0
1
–1
x2 (t) = te
x2 –2
3t/2
–3
�
�
� �
2
10
24
3t/2
+e
–1
−1
0
6
8
10
t
is also a solution of our original system. Therefore,
the general solution of the system is
–2
� �
�
� �
� ��
–4
1
1
3t/2
3t/2
3t/2 2
x(t) = c1 e
+ c2 te
+e
.
−1
−1
0
1
2
4
t2
6
8
10
1
x1
2
0
x2
1
–1
–2
–2
–1
0
–1
–3
–2
3.5. REPEATED EIGENVALUES
The solutions tend to infinity as t → ∞.
213
7. The characteristic equation is (λ + 3)2 = 0. Therefore, the only eigenvalue is λ = −3. A
12.
From our answer
to problem
we see that the general solution of this system is given
corresponding
eigenvector
is given2,by
by
� �
� � v1 =
� 1 � �
� ��
1
1
1
3t/2
3t/2
3t/2 2
x(t) = c1 e
+ c2 te
+e
.
−1
−1
0
and
� �
−3t 1
� �x1 (t) = e
1
1
Now
initial of
condition
x(0) Next,
=
implies
is oneour
solution
the system.
3 we need to look for a solution w of
� �
1
(A�+ 3I)
� w =� 1� . � �
1
2
1
x(0) = c1
+ c2
=
.
−1
0
3
Plugging in for A + 3I, this equation becomes
�
�
� �
4 −4
1
w=
.
4
−4
1
Therefore, c1 = −3 and c2 = 2. Therefore,
We see that
� �
�
� 1/4 � �
� ��
1 w = 3t/2
1
3t/2
3t/2 2
0
x(t) = −3e
+ 2 te
+e
.
−1
−1
0
is a solution of this equation. Therefore,
� �
� �
�
3.5. REPEATED EIGENVALUES
0
1
2
3
4
219
5
t
1
t
2
3
4
5
0
20000
5
–2000
4
–4000
3
–6000
x2
10000
2
–8000
1
–1
0
15000
–10000
1
5000
–12000
2
3
x1
4
5
–1
0
1
2
3
4
5
t
13. The characteristic equation0 is
–2000
–4000
1
2
t
3
4
5
λ2 + (a + d)λ + ad − bc = 0.
Therefore, the eigenvalues –6000
are
a + d 1�
±
(a + d)2 − 4(ad − bc).
–10000
2
2
–8000
λ=
–12000
In order to guarantee that
the solution approaches zero, we need both eigenvalues to be
negative. As we can see from the equation above for λ, we need a + d < 0 and ad − bc > 0.
14.
13. The characteristic equation is
(a) The eigenvalues of this system are given by
λ2 + (a + d)λ +√ad2− bc =2 0.
1
L − 4R CL
λ=−
±
.
2RC
2RCL
Therefore, the eigenvalues are
The eigenvalues will be realaand
vanishes. The discriminant
+ d equal
1 �if the discriminant
2
λ 2=
will vanish when L = 4R
C. 2 ± 2 (a + d) − 4(ad − bc).
(b)
conditions,
the solution
system isapproaches zero, we need both eigenvalues to be
In For
orderthetogiven
guarantee
that the
negative. As we can see from the equation
above for λ,
� � �
� �we�need a + d < 0 and ad − bc > 0.
d i
0 1/4
i
= OF TWO FIRST
. ORDER EQUATIONS
220 14.
CHAPTER 3. SYSTEMS
−1 −1
v
dt v
(a) characteristic
The eigenvalues
of this of
system
are given
The
equation
this system
is by
λ2 + λ + 1/4 = 0. Therefore, the only
√ is given by
eigenvalue is λ = −1/2. An associated eigenvector
1� � L2 − 4R2 CL
λ=−
±
.
2RC 1
2RCL
v1 =
.
−2
The eigenvalues will be real and equal if the discriminant vanishes. The discriminant
Therefore,
one solution
is
will vanish
when Lof=this
4R2system
C.
� �
1
−t/2
(b) For the given conditions, thex1system
(t) = eis
.
−2
� � �
�� �
d
i
0
1/4
i
Next, we need to look for a solution w of =
.
−1
v
� dt v �
� −1
�
1
1
A+ I w =
.
−2
2
Plugging in for A + 12 I, this equation becomes
�
�
� �
1/2 1/4
1
w=
.
−1 −1/2
−2
We see that
� �
0
1/2 1/4
1
. 2
y) =wx2=− 2xy
−1 H(x,
−1/2
−2 + 2y .
We
(c) see that
� �
0
w=
4
is a solution of this equation. Therefore,
x2 (t) = te−t/2
�
�4
� �
1y
−t/2 0
+e
−2 2
4
is also a solution of our original system.
Therefore,
the2 general
solution of the system is
–4
–2
4
� �
�
� �x
� ��
1
1
–2
−t/2
−t/2
−t/2 0
x(t) = c1 e
+ c2 te
+e
.
4
−2
−2
–4
The initial condition implies that c1 = 1 and c2 = 1. Therefore, the solution of the IVP
is
�
�
1+t
x(t) =
e−t/2 .
2
−
2t
in the counter-clockwise
direction SYSTEMS
along the trajectories shown above.
3.6. Solutions
A BRIEFtravel
INTRODUCTION
TO NONLINEAR
221
(d) Solutions
orbitequation
the critical
doThe
not eigenvalues
tend towards
15. The
characteristic
is λ2point.
− pλ +They
q = 0.
areor away from the critical
3.6point.A Brief Introduction
to Nonlinear
Systems
�
√
p ± p2 − 4q
p± ∆
λ=
=
.
2
2
1.
10.
The results can be verified using Table 3.5.1.
(a) To find all critical points, we need to solve the following system of equations.
16.
dx
dyx� = −x + y + dy
x2 = 02y
(a) If q > 0 and p < 0, then the
roots are
either=⇒
complex
= −x,
=
−2y
= conjugates with negative real
dt
dty � = y − 2xy =�dx
x
0.
�
parts, or both real and negative.
1
1
=⇒
dy
=
2
dx or y = 0. Plugging
Solving
we see
thatimaginary.
we musty have x = x1/2
(b) If q >
0 and the
p = second
0, then equation,
the roots are
purely
these values into the first equation, we see=⇒
thatlnthe
points
|y|critical
= 2 ln |x|
+ C.are: (0, 0), (1, 0) and
(c) if q <(1/2,
0, then
the
roots
are
real
with
λ
·
λ
>
0.
If
p
>
0,
then
either
the roots are real
1
2
1/4).
withApplying
λ1 · λ2 ≥ the
0 orexponential
the roots are
complex
conjugates
with
positive
real
parts.
function to this equation, we conclude that
y = Cx2 for any
(b) From our equations for dx/dt and dy/dt, we see
that
solution of the system. That is, H(x, y) = y/x2 .
dy
y − 2xy
=
.
(b)
dx
−x + y + x2
Rewriting this equation as (2xy − y)dx + (x2 − x + y)dy = 0, we see that this is an exact
equation. For M = 2xy − y, we integrate M with respect to x. We conclude that the
4
level curves are given by H(x, y) = C where H(x, y) = x2 y − xy + h(y). Differentiating
�
2
�
with respect to y, we have Hy = x2 − x y+ h
2 (y) = N = x − x + y. Therefore, h (y) = y
which implies that h(y) = y 2 /2. Therefore,
–0.6
–0.4
–0.2
0
0.2
0.4
H(x, y) = x2 y − xy +x y 2 /2.
–2
–4
(c)
3
2
y
1
–4
–2
0
2
4
x
–1
0.6
17.
behave like a saddle, and, therefore, (−1/2, 1) is also unstable.
(a)14.
The critical points are given by the solutions of the equations
(a) To find the critical points, we need to solve
y(2 − x − y) = 0
�
−y =
− x x− y=−2 2xy
= 00.
�
2
y = y − x = 0.
The solutions of the first equation are y = 0 and x + y = 2. Plugging these
values into√
By the first equation, we must have y √
= 2. By the second
equation, we √
have x =ñ 2.
√
the second
we conclude that the critical points are (0, 0), (1 − 2, 1 + 2),
√ equation,
Therefore,
the √
two critical points are ( 2, 2) and (− 2, 2).
and (1 + 2, 1 − 2).
(b)
(b)
3
4
2
y
y
2
1
–4
0
–2
2
4
x
–2
–1
–2
1
x
2
–1
234
–4
CHAPTER 3. SYSTEMS
OF TWO FIRST ORDER EQUATIONS
√
(c) The trajectories near the critical point
√ ( 2, 2) tend away from the critical point. The
trajectories near the critical point (− 2, 2) spiral into the critical point.
(c) Based on the phase
(0, 0) is an asymptotically stable spiral
√ portrait,
√ we conclude
√ that √
point,
while
(1
−
2,
1
+
2)
and
(1
+
2,
1
−
2) are saddle points.
15.
236
CHAPTER 3. SYSTEMS OF TWO FIRST ORDER EQUATIONS
18.(a) The critical points are given by the solutions of the equations
(a) The critical points are given by the
solutions
of 0the equations
x(1 −
x − y) =
�
�
1 + x)(y
1 −3x) = 0
(2
− y − x = 0.
y
2 + 4x − x42 ) = 0.
y(2
(b)
(b)
The solutions
theequations,
second equation
y=
x = 2 and
x = are
−1. given
Plugging
these
Solving
these of
two
we see are
that
the0, critical
points
by (0,
0), values
(0, 2),
into
the
first
equation,
we
conclude
that
the
critical
points
are
(0,
0),
(2,
2),
(−1,
−1)
(1/2, 1/2) and (1, 0).
and (−2, 0).
4
4
y
y
–4
–4
–2
–2
2
2
0
0
2
2 x
x
4
4
–2
–2
–4
–4
and (−2, 0)
are saddle
and
(c) Based on the phase portrait,
portrait, we
we conclude
conclude that
that (0,
all0)
trajectories
starting
near points
the origin
(2, 2) and Therefore,
(−1, −1) are
points. point (0, 2), the trajectories
diverge.
(0,asymptotically
0) is unstable. stable
Near spiral
the critical
approach the critical point. Similarly, near (1, 0), trajectories approach the critical
19.