The
salt isFIRST
flowingORDER
out at the
rate of Q/100 oz/gallon · 2 gallons/minute 51
=
2.2. gal/minute.
MODELING
WITH
EQUATIONS
Q/50 oz/minute. Therefore,
−t/60
The solution of this equation is Q(t)dQ
= 120γ
+ Ce
. Since
Q(0) = 0 grams, C = −120γ.
1
1
Q
2.2 Modeling
with
Order
+ Q(t)
sin t→−Equations
.
Therefore,
Q(t) = 120γ[1 −
e−t/60 ].First
As t =
→ ∞,
120γ.
dt
2 4
50
3. Let Q(t) be the quantity of salt in the tank. We know that
t/50
1. Let
Q(t)
the equation
quantitywith
of dye
in the tank.
Weµ(t)
know
This
is a be
linear
integrating
factor
= ethat
. The solution of this equation
−t/50
is Q(t) = (12.5 sin t − 625 cos dQ
tdQ
+ 63150e
)/2501 + c. The initial condition, Q(0) = 50
=
rate
in
−
rate
out.
=
rate
in
−
rate
oz implies C = 25. Therefore,dt
Q(t) = 25 + (12.5
sinout.
t − 625 cos t + 63150e−t/50 )/2501 oz.
dt
Here,
containing
1/2 lb/gallon
of saltnoisdye
flowing
in at in.
a rate
2 gallons/minute.
(b)
Here, water
fresh water
is flowing
in. Therefore,
is coming
Theofdye
is flowing out atThe
the
salt
outgrams/liters
at the rate of
lb/gallon·2
gallons/minute
= Q/50
gallons/minute.
rateisofflowing
(Q/200)
· 2(Q/100)
liters/minute
= Q/100
liters/minute.
Therefore,
Therefore,
50
dQ
Q
dQ
=1−− Q ..
=
100
dtdt
50
40
The
50 −t/100
+ Ce.−t/50
. Since
0 grams,
C 200.
= −50.
The solution
solution of
of this equation is Q(t) = Ce
Since
Q(0) Q(0)
= 200=grams,
C =
We
−t/50
30
Therefore,
Q(t)
=
50[1
−
e
]
for
0
≤
t
≤
10
minutes.
After
10
minutes,
the
amount
of
need to find the time T when the amount of dye present is 1% of what it is initially. That
Q
−1/5
−T
/100
salt
in need
the tank
is Q(10)
= 50[1
− e Q(T
] ≈) 9.06
lbs. Starting
at the
thatequation
time (and
resetting
is, we
to find
the time
T when
= 2 grams.
Solving
2=
200e the,
20
time
variable),
theTnew
equation
is given by
we conclude
that
= 100
ln(100)for
≈ dQ/dt
460.5 minutes.
2. Let Q(t) be the quantity of10salt in the
dQtank. We
Q know that
=− ,
dt
50
0 dQ
20 = rate
40
60
80 rate
100 out.
120 140
in −
t
dt
since fresh water is being added. The solution of this equation is Q(t) = Ce−t/50 . Since we
−t/50
are
now
starting
with
lbs of salt,
Q(0)
=
c. in
Therefore,
Q(t)
9.06e
. After
Here,
water
containing
γapproaches
grams/liter
of
salt=state,
is9.06
flowing
a rate
of 2 =
liters/minute.
The
(c)
The
amount
of salt9.06
a steady
which
isatan
oscillation
of
amplitude
1/4
−1/5 ∼
10
Q(10)
=
9.06e
7.42
lbs.
=
saltminutes,
is
flowing
out
at
the
rate
of
(Q/120)
grams/liter
·
2
liters/minute
=
Q/60
liters/minute.
about a level of 25 oz.
Therefore,
4.
Let Q(t) be the quantity of salt in the tank. We know that
dQ
Q
6.
= 2γ − .
dQ dt
60
= rate
in
− rateweout.
(a) Using the Principle of Conservation
of
Energy,
know that the kinetic energy of a
dt
particle after it has fallen from a height h is equal to its potential energy
at a height t.
√
Here,
water
containing
1
lb/gallon
of
salt
is
flowing
in
at
a
rate
of
3
gallons/minute.
The
2
Therefore, mv /2 = mgh. Solving this equation for v, we have v = 2gh.
salt is flowing out at the rate of (Q/(200 + t)) lb/gallon · 2 gallons/minute = 2Q/(200√
+ t)
(b)
The volumetric
outflow rate is (outflow cross-sectional area) × (outflow velocity): αa 2gh.
lb/minute.
Therefore,
2Q is:
The volume of water in the tankdQ
at any
= 3 instant
−
.
dt
+t
�200
h
This is a linear equation with integrating
factor A(u)
µ(t) du
= (200 + t)2 . The solution of this
V (h) =
equation is Q(t) = 200 + t + C(200 + t)−2 . 0Since Q(0) = 100 lbs, C = −4, 000, 000.
Therefore,
Q(t) =
t − [100(200)2area
/(200
t)2 tank
]. Since
the tank
netchain
gain rule,
of 1 gallon
where A(u)
is 200
the +
cross-sectional
of+the
at height
u. has
By athe
of water every minute, the tank will reach its capacity after 300 minutes. When t = 300, we
dV the
dV dh
dh
see that Q(300) = 484 lbs. Therefore,
= concentration
·
= A(h) of .salt when it is on the point of
dt concentration
dh dt
dt is given by Q(t)/(200 + t) (since
overflowing is 121/125 lbs/gallon. The
of salt
t gallons
of water are added every t minutes). Using the equation for Q above, we see that
Therefore,
� approach 1 lb/gal as t → ∞. 53
2.2.
WITH
FIRSTthe
ORDER
EQUATIONS
if theMODELING
tank had infinite
capacity,
concentration
would
dV
dh
= A(h)
= −αa 2gh.
dt
dt
5.
(c) The cross-sectional area of the cylinder is A(h) = π(1)2 = π. The outflow cross-sectional
2
(a) area
Let Q(t)
beπ(.1)
the quantity
salt inpart
the (a),
tank.
is a =
= 0.01π.of From
weWe
takeknow
α = that
0.6 for water. Then by part (b),
we have
�
dQ dh
= rate
in − rate
out.
π
=
−0.006π
2gh.
dt dt
�
This is a separable equation with solution h(t) = 0.000018gt2 − 0.006 2gh(0)t + h(0).
Setting h(0) = 3 and g = 9.8, we have h(t) = 0.0001764t2 − 0.046t + 3. Then h(t) = 0
implies t ≈ 130.4 seconds.
7.
(a) The equation for S is
dS
solution of this equation is t = − ln(80/130)/.08 ∼
= 6.07 minutes.
17.
(a) The solution of this separable equation is given by
u30
u =
.
3αu30 t + 1
3
Since u0 = 2000, the specific solution is
2000
.
(6t/125 + 1)1/3
2.2. MODELING WITH FIRST ORDER EQUATIONS
(b)
u(t) =
63
The units on the left must match the units on the right. Since the units for tm g/v0 =
2000 cancel. As a result, we can conclude that kv0 /mg is dimen(s · m/s2 )/(m/s), the units
sionless.
1800
25.
1600
(a) The equation of motion1400
is given by mdv/dt = −kv − mg. The solution of this equation
is given by v(t) = −mg/k
+ (v0 + mg/k)e−kt/m .
1200
(b) Applying L’Hospital’s rule,
1000 as k → 0, we have
−kt/m
0 20 40 60 80 100 120 140
160 180 200
lim −mg/k
+ (v0 + mg/k)e
= v0 − gt.
t
k→0
(c)
(c) We look for τ so that u(τ ) = 600. The solution of this equation is t ≈ 750.77 seconds.
lim −mg/k + (v0 + mg/k)e−kt/m = 0.
m→0
18.
(a)
26. The differential equation for Q is
(a) The equation of motion is given dQ
by = kr + P − Q(t) r.
dt
V
dv
4
�4
3
m = −6πµav + ρ πa g − ρ πa3 g.
Therefore,
dt
3
3
dc
V
= kr + P − c(t)r.
We can rewrite this equation as dt
−rt/V
The solution of this equation is c(t)
P/r
. Therefore
3 + (c0 − k − P/r)e
6πµa= k +
4 πa
g �
�
v +
v=
(ρ − ρ).
limt→∞ c(t) = k + P/r.
m
3 m
Multiplying by the integrating factor e6πµat/m , we have
(e6πµat/m v)� =
4 πa3 g �
(ρ − ρ)e6πµat/m .
3 m
Integrating this equation, we have
�
2a2 g(ρ� − ρ) 6πµat/m
v=e
e
+C
9µ
2a2 g(ρ� − ρ)
=
+ Ce−6πµat/m .
9µ
−6πµat/m
�
Therefore, we conclude that the limiting velocity is vL = (2a2 g(ρ� − ρ))/9µ.
(b) By the equation above, we see that the force exerted on the droplet of oil is given by
4
4
Ee = −6πµav + ρ� πa3 g − ρ πa3 g.
3
3
74
CHAPTER 2. FIRST ORDER DIFFERENTIAL
EQUATIONS
2
�
2 conclude
2.3
Differences
between
Nonlinear
EquaTherefore,
that
the limitingLinear
velocity is vand
L = (2a g(ρ − ρ))/9µ.
Therefore,
2kwe
sin2 tdt = dx.
Also,
rt
�
tions
2 v)see
which
be
as
kertthe
. =The
of
equation
is
given
by by
v= =
(b)
By can
the
that
force
exerted
on this
the
droplet
oilhave
is given
(c)
Letting
θequation
= written
2t, weabove,
have
sin2=
(θ/2)dθ
dx.
Integrating
both
sides,ofwe
x(θ)
� (ekwe
� solution
t
t
�t
�t
−rt
2 dr 2
(kk+2 (θ
cre−−rt
)/r.
Then,
using the
factdrthat
y = 1/v,y we
conclude
y = r/(k that
+ crey(θ)
). =
sin
θ)/2.
Further,
fact
t, that
wedsconclude
s p(r) the
e−using
g(s)
ds that
≤ � 4 e−=3 s kp(r)sin
|g(s)|
4
3
1. kRewriting
the
equation
as
2
2
2
Ee = −6πµav + ρ t0 πa g − ρ πa g.
t0 v satisfies
sin (θ/2)
= Therefore,
k (1 − cos(θ))/2.
30. Here
n = 3.
3
ln t � 3t 2t
y� +
y = −p0 (t−s)
≤
et sin
M
t+−2�v
3 for= e,
−
If v =
0, (c),
thenwe
solving
they/x
above
equation
we3θ).
have
(d) From
part
see that
= (1
−v �cos
θ)/(θ
−
If xds= 1 and y = 2, the solution
2σ.
t0
of using
the equation
is 2.3.1,
θ ≈ 1.401.
Substituting
θ�t into either
the in
equations
in
and
Theorem
we conclude
that athat
solution
isofguaranteed
toofexist
the interval
3 value
� −p0 (t−s)
4πa
g(ρ
−
ρ)
�
2�t
−2�t
e
linear with
part
thatintegrating
k ≈ 2.193.efactor
� is given by v = (σ +c�e )/�.
0This
<
t equation
<(c),
3. we isconclude
= ≤e M. Its solution
√. √
p0 �/ �t0σ + c�e−2�t .
Then, using the fact that y 2 = 1/v, we see that 3E
y=±
2. Rewriting the equation as
�
�
1
e−p0 (t−t0 )
1
31. Here n = 3. Therefore, v satisfies �
= M y = 0−
y + Linear
2.3 Differences between
andp0 Nonlinear Equat(t − 4) p0
�
v + 2(Γ cos t +
MT )v =is2.guaranteed to exist in the interval
tions
and using
Theorem 2.3.1, we conclude that a≤solution
p0 sin t+T t)
0This
< t equation
< 4.
is linear with integrating factor e2(Γ
. Therefore,
1. Rewriting the equation as
3. By Theorem 2.3.1, we conclude
guaranteed to exist in the interval
� 2(Γ sinthat
�
lnt)tav �solution
2t2(Γis
t+T
sin t+T t)
�
e
=
2e
y
+
y
=
π/2 < t < 3π/2.
t −e−t
3 2 . Therefore, |g(t)| ≤ 1 for all t ≥ 0.
(c) Let p(t) = 2t + 1 ≥ 1 for all t ≥ 0 andt −
let3 g(t) =
4.
Rewriting
the equation
asconclude that a solution is guaranteed to exist in the interval
which
implies
and
using
Theorem
we
By
the
answer
to2.3.1,
partEQUATIONS
(a),
2.4.
AUTONOMOUS
AND POPULATION
DYNAMICS
77
2t AND NONLINEAR
3t2
2.3.
DIFFERENCES
BETWEEN
EQUATIONS
69
�
0 < t < 3.
� ytLINEAR
+
y
=
4 − t2 � 4 − s))
t2 ds + ce−2(Γ sin t+T t) .
sin t+T t)
� t exp(2(Γ sints +�T
v = 2e−2(Γ
t
2. Rewriting the equation
as
ds
dr −s2
and using Theorem 2.3.1,
we=conclude
that
is guaranteed
y(t)
e− t00(2s+1)
e− s (2r+1)
e ds to exist in the interval
1+a solution
�
3
0
y
+
y
=
0
−∞ < t < −2.
�
t(t −2 4) � t
Then v = y −2 implies y = ± 1/v.
−(t2 +t)
−t
2 s
=e
+ ea−tsolution
e is
dsguaranteed to exist in the interval
5. using
Rewriting
the equation
as
y(t)
and
Theorem
2.3.1,
we
conclude
that
� 0
2 1 is y = 1/2 + ce−2t . For y(0) = 0,
32. The solution of the initial value �problem
y + 12y3t
=
2t
2
0<
< 4.that c = −1/2. Therefore,−ty(t)
y=
−2t
wet see
(1t−
fort20 ≤ t ≤ 1. Then y(1) = 12 (1 − e−2 ).
2 e
= e .y +=412−
4) −
0
�
−2
0
–2 by
1
2to exist y(1)
3. Next,
By Theorem
2.3.1,
conclude
that
a solution
is. The
guaranteed
in the 12interval
the solution
of 2.3.1,
ywe
+2y
= conclude
0 is given
y–1a=solution
ce−2t
initial
condition
(1−e
)
t
and using
Theorem
we
that
is guaranteed
to exist in1=the
interval
1
1 2
−2
−2
2
−2t
–1
π/2
< t <ce3π/2.
implies
=
(1
−
e
).
Therefore,
c
=
(e
−
1)
and
we
conclude
that
y(t)
=
(e
−
1)e
2
−2
< see
t <that
2. y2 satisfies the property that2 y is bounded for all time t ≥ 0.
We
for
t
>
1.
–2
-a/b
4. 6.
Rewriting
thethe
equation
as
Rewriting
equation
as
2
�
−2t
2t =
3t
33. The solution of y + 2y = 0 with
y(0)
for 0 ≤ t ≤ 1. Then
�
1y 1=is –3given
cot t by y(t) = e
�
y
+
−2
�
y
+
y
=
2
2
y(1) = e . Then, for t > 1, the solution
of
the
equation
y
+
y
=
0
is
y = ce−t . Since we
4 − tln t 4 −lntt
−1
−1
−1 −t
want
y(1)
=
e−2 , we 2.3.1,
need ce
= e−2 . Therefore,
c = ePopulation
Therefore, to
y(t)
= ein
ethe =
e−1−t
2.4
Autonomous
and
Dynamics
and using
Theorem
weEquations
conclude
that a solution
is .guaranteed
exist
interval
for
1 <t t><1.π.
3.
1. 34.
7. Using the fact that
t
t�t−
y ds
7t
p(s)
0
(a) Multiplying the equationf by
e
, we have
=
=⇒
fy = −
,
2t + 5y
(2t + 5y)2
� �t
��
�t
6
p(s) ds are satisfied
p(s) ds as long as 2t + 5y �= 0.
we see that the hypothesis of Theorem
e t0 2.3.2
y = e t0
g(t).
5
8. Using the fact that
4
Integrating we have
y
2
2 1/2
3 t −y )
f = (1f −
=⇒ f�y =
−
,
t � s (1 − t2 − y 2 )1/2
�t
p(r)
dr
p(s)
ds
e2t0
y(t) = y0 +
e t0
g(s) ds,
t0
we see that the hypothesis of1Theorem 2.3.2 are satisfied as long as t2 + y 2 < 1.
whichthe
implies
9. Using
fact that
�
0
0.5
−
t2
1.5
�t 1
p(s) ds y
� t2.5
3
dr
2
y(t)
= y0 e t0
+1 − et2−+s yp(r)
ln |ty|
− g(s)
2y 2 lnds.
|ty|
t
0
f=
=⇒
f
=
,
y
2
2
2
2
2
1−t +y
y(1 − t + y )
The equilibrium points are y ∗ = 0, 1, 2. Since f � (0), f � (2) > 0, those equilibrium point are
we
see thatSince
the hypothesis
2.3.2 are satisfied
as long as y, t �= 0 and 1−t2 +y 2 �= 0.
unstable.
f � (1) < 0, of
y ∗Theorem
= 1 is asymptotically
stable.
10. Using the fact that
f = (t2 + y 2 )3/2 =⇒ fy3 = 3y(t2 + y 2 )1/2 ,
we see that the hypothesis2of Theorem 2.3.2 are2.5satisfied for all t ∈ R.
2
The equilibrium points are y ∗∗ = 0, 1, 2. Since
f � (0), f � (2) > 0, those equilibrium point are
�
The only equilibrium
point
is
y
=
0.
Since
f
(0)
< 0, the equilibrium point is asymptotically
unstable. Since f � (1) < 0, y ∗ = 1 is asymptotically stable.
stable.
3
3
2
2.4. AUTONOMOUS EQUATIONS AND POPULATION
DYNAMICS
y(t) 2.5
2
81
1
2
01
y(t) 0
–1 12 1.5
–1
1
10
–2
0.5
8
–3
0
f 6
–1
–2
0
–2
1
t
2
1
t
2
4
7.
2
4.
2
(a) The function f (y) = k(1 −
0 =⇒0 y = 1. 1Therefore,
y ∗ = 1 is the only critical
–2 y) = –1
2
y
80 point.
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
(b)
The equilibrium points are y ∗ = 0, 1, −1. Since f � (−1) < 0, y = −1 is asymptotically stable.
Since f � (1) > 0, y = 1 is unstable. The equilibrium point y = 0 is semistable.
8
6
3
f
4
1
2
y(t)
2
1
0
–2
–2 –1
0–1
10
2
y
1 3
t
42
–1
-1
(c) This is a separable equation with solution –2
y(t) = [y0 + (1 − y0 )kt]/[1 + (1 − y0 )kt]. If
y0 < 1, then y → 1 as t → ∞. If y0 > 1, then the denominator will go to zero at some
finite time T = 1/(y0 − 1). Therefore, the solution will go towards infinity at that time
T.
10.
8.
–2
–1
6 1
y
2
3
4
4
–2
f
2
–2
–4
f
–1
–6
1
y
2
–2
–4
–8
–6
∗
The equilibrium
only equilibrium
SinceSince
f � (1) f<� (−1),
0. The
equilibrium
is semistable.
The
pointspoint
are is
y ∗ y= =0,1.
1, −1.
f � (1)
< 0, thepoint
equilibrium
points
�
y = 1, −1 are asymptotically stable. Since f (0) > 0, the equilibrium point y = 0 is
unstable.
3
82
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
3
2
y(t)
1
3 2
2
0
y(t)
–1
–2
0
1
–1
0
–2
-2
1
1
t
0–2
–1
2
1
t
2
–1–3
-1
–2
13.
11.
4
0.6
3
0.4
f
f 2
0.2
1
0
–0.2
0.2 0.4 0.6 0.8
–1
–0.5
1
y
1.2 1.4 1.6 1.8
0.5
1
y
1.5
2
2
y ∗ = 0, 1.2.They
are both
semistable.
84 The equilibrium points are
CHAPTER
FIRST
ORDER
DIFFERENTIAL EQUATIONS
The equilibrium points are y ∗ = 0, b2 /a2 . Since f � (0) < 0, the equilibrium point y = 0 is
asymptotically stable. Since f � (b2 /a2 ) > 0, the equilibrium point y = b2 /a2 is unstable.
3
y(t)
1
2
3
1
2
b /a
0
2.5
2
–2
0
–1
2
–1
y(t) 1.5
–2
0
1
t
2
1
–3
0.5
–2
14.
–1
0
1
t
2
(a) The equation is separable. Using partial fractions, it can be written as
12.
�
�
1
1/K
+
dy = rdt.
y 1 − y/K
Integrating both sides and using the initial condition y0 = K/3, we know the solution y
satisfies
�
�
� �
�
�
�K �
y
�
�
ln �
= rt + ln �� �� .
�
1 − y/K
2
To find the time τ such that y = 2y0 = 2K/3, we substitute y = 2K/3 and t = τ into
the equation above. Using the properties of logarithmic functions, we conclude that
that t ≈ 7.97. Similarly, if y0 ==6,k we
see that y ≤ 4.05
− απ
a hfor t ≈ 7.97. For all initial data
3
2 < y0 < 6, the conclusion also holds. � πh �
2/3
3a
= k − απ
V 2/3 .
18.
πh
(a) The surface area of the cone is given by
(b) The equilibrium volume can be found by setting dV /dt = 0. We see that the equilibrium
�
√
volume is
S = πa h2 + a2 +�πa2 �=3/2πa�2 ( �
(h/a)2 + 1 + 1)
k
πh
� .
=
πa2 h 3V��
2
απ
3a
=
·
(h/a) + 1 + 1
3
h
� we
�
� fact
�2/3
To find the equilibrium height,
use
that the height and radius of the conical
2/3 the
πa2 h
3a
pond maintain a constant
if he , ae represent the equilibrium values for
= cπratio. Therefore,
·
πh
the h and a, we must have h�e /a3e�= h/a. Further,
we notice that the equilibrium volume
2/3
3a
can be written as
2/3 �
1/2 � �
= cπ π � k �
V�
k.
h
π
πh
V =
= a2e h2e ,
3 απ
απ
a
3
Therefore, if the rate of evaporation is proportional to the surface area, then rate out =
1/2
where
he =2/3
(k/απ)
(h/a) and ae = (kαπ)1/2 . Since f � (V ) = − 23 απ(3a/πh)2/3 V −1/3 <
απ(3a/πh)
V 2/3 . Therefore,
0, the equilibrium is asymptotically stable.
dV
= rate in − rate out
(c) In order to guarantee that the
dt pond does not overflow, we need the rate of water in to
2
�2/3 �
be less than or equal to the rate of water�out.
2/3 need k − απa ≤ 0.
3a Therefore,
π 2 �we
= k − απ
ah
πh
3
� �2/3
3a
= k − απ
V 2/3 .
πh
(b) The equilibrium volume can be found by setting dV /dt = 0. We see that the equilibrium
volume is
� �3/2 � �
k
πh
V =
.
απ
3a
To find the equilibrium height, we use the fact that the height and radius of the conical
pond maintain a constant ratio. Therefore, if he , ae represent the equilibrium values for
the h and a, we must have he /ae = h/a. Further, we notice that the equilibrium volume
can be written as
� � � �1/2 � �
π k
k
h
π
V =
= a2e h2e ,
3 απ
απ
a
3
where he = (k/απ)1/2 (h/a) and ae = (kαπ)1/2 . Since f � (V ) = − 23 απ(3a/πh)2/3 V −1/3 <
0, the equilibrium is asymptotically stable.
(c) In order to guarantee that the pond does not overflow, we need the rate of water in to
be less than or equal to the rate of water out. Therefore, we need k − απa2 ≤ 0.