or
258
CHAPTER
4. QSECOND
ORDER
LINEAR
8.
spring
constant
is k =
=
2/(1/2)
= 4 lb/ft.
m = 2/32
= 1/16EQUATIONS
lb-s2 /ft. The
−6
� The mass
1. The
The
differential
equation
is 10
linear.
with
initial
conditions
Q(0)
coulombs,
(0)
y �� + 128y
= 0= 0 coulombs/sec.
equation of motion is
� 3.92 N/m. The mass m = .02 kg. The damping
2. The
differential
equation
nonlinear.
12.
The
spring
constant
is k=is=
.196/.05
with
initial
conditions
y(0)
−1/12
ft, 1y=
(0) = 2 ft/sec.
y �� + 4y =
0
4.1
Definitions
and
Examples
constant
is
γ
=
400
dyne-sec/cm
=
.4
N-sec/cm.
Therefore,
the equation of motion is
3. The
is linear.
11.
Thedifferential
inductanceequation
L = 1 henry.
The 16
resistance R = 0. The capacitance C = 0.25 × 10−6
��
or
farads.
Therefore, the
equation1linear.
for.02y
charge
Q �is+ 3.92y = 0
+ .4y
4.
1. The
The differential
differential equation
equation is
is linear.
��
y + 64y = 0
0.5
5.
Q�� + (4 × 106 )Q = 0
2. The
The differential
differential equation
equation is
is nonlinear.
nonlinear.
or
�
with initial conditions y(0) = 1/4
ft,y ��y2+
(0)
=�4 0+ ft/sec.
0
10 then mg = 8. If the mass
20y
= 0 8is 8 lb,
6.
We
will
use
the
equation
mg
−
kL
=
0.
If 196y
the6 mass
−6
3.
The
differential
equation
is
linear.
with initial conditions Q(0) = 10 coulombs, Qt � (0) = 0 coulombs/sec.
–0.5
9.
The
spring
constant
is k=
=.02
.98/.05
=
N/m.
kg. The
equation
of
stretches
the
spring
6 y(0)
inches,
then
= 19.6
6.
− 6km==0 .1which
means
k = 4/3
with
initial
conditions
m, Ly � (0)
= 0Therefore,
m/sec.The8mass
4.
The
differential
equation
is
linear.
12.
The
spring
constant
is
k
=
.196/.05
=
3.92
N/m.
The
mass
m
=
.02
kg.
The
damping
motion
lb/inch.is
13.
Thedifferential
constant
is is
k–1nonlinear.
= =16/(1/4)
lb/ft. Thethe
mass
m = 1/2
lb·s2 /ft.
�� = 64 Therefore,
constant
isspring
γ = 400
dyne-sec/cm
.4 .1y
N-sec/cm.
equation
of motion
is The
5. The
equation
+
19.6y
=
0
7.
Again,
we
use
the
equation
mg
−
kL
=
0.
Here
the
mass
is
10
kg.
The
force
due to
damping coefficient is γ = 2 lb-sec/ft. Therefore, the equation of motion is
��
�
2
6. We will
the m/s
equation
mg −
kLmg
= .4y
0. 98
If+ the
mass
8 lb,mass
thenstretches
mg = 8.the
If spring
the mass
gravity
is g use
= 9.8
. Therefore,
=
Newtons.
.7
.02y
+
3.92y
= 0isThe
or
1
stretches
the
spring
6
inches,
then
L
=
6.
Therefore,
8
−
6k
=
0
which
means
k
=
4/3
��
�
meters. Therefore, k = 98/.7 = 140 N/m.
y y ��++2y196y
+ 64y
= 0= 0
4.2.
THEORY OF SECOND ORDER
259
or
2 LINEAR HOMOGENEOUS EQUATIONS2
lb/inch.
��
�
8. The
constant
k = 2/(1/2)
=
4
lb/ft.
The
mass
m
=
2/32
=
1/16
lb-s
/ft.
The
The spring
soft spring
has aishigher
amplitude
and
a
longer
period.
The
stiff
spring
has
a
smaller
y� (0)
+ 20y
196y = 0
with
initial we
conditions
y(0)
= 0 m,
==
.1 +m/sec.
or
7. Again,
use the
equation
mgy −
kL
0. Here the mass is 10 kg. The force due to
equation
of motion
amplitude
and
aisshorter
period.
��
�
� + 4y + 128y = 0
6.
Write
IVP
2
with
initial
= .02 m,yymg
(0)
= 98
0 m/sec.
gravity
isthe
gconditions
=
9.8asm/sy(0)
. Therefore,
Newtons. The mass stretches the spring .7
11 y=���249
��
+
4y
= x)y
0
258
CHAPTER
SECOND
ORDER
LINEAR
+
+64
(tan
0.
meters.
Therefore,
k =y(0)
98/.7
140
with
initial
conditions
0=yft,
y �N/m.
(0)
=2 y1/4
ft/sec.
16
13.
The
spring
constant
is=k=
16/(1/4)
=4.
lb/ft. =The
mass m
= 1/2 EQUATIONS
lb·s2 /ft. The
x−
4.2
Theory
Order
Homogeneous
2 2Equadamping
coefficient
isof
γ is
=Second
lb-sec/ft.
Therefore,
the
equation
of
motion
or
8. The
constant
k2 =
2/(1/2)
=4=
lb/ft.
The
mass
mmass
=
2/32
= is
1/16
lb-s
/ft.
The
14.
The
spring
constant
is
k are
= continuous
16/(1/4)
64Linear
lb/ft.
The
m
=
1/2
lb-s
/ft.
The
Since
thespring
coefficient
functions
for
all
x
such
that
x
=
�
2,
nπ
+
π/2
and
x0 =
3,
��
equation
of
motion
is
y
+
64y
=
0
damping
coefficient
γ
=
2
lb-sec/ft.
Therefore,
the
equation
of
motion
is
the IVP tions
is guaranteed to have a unique
for all x such that 2 < x < 3π/2.
1 �� solution
�
y 1+ 2y
�� + 64y = 0
y=+0 4y
=0
1
with initial conditions y(0) = 1/4
ft,21 y���16
(0)
7.
� ft/sec.
� 2y
y
+
+
64y
= 0 ��
2t
�
1.
Write the IVP as
7 t/2
e
e−3t/2
or
2t
−3t/22
�
� =mass
0.5
or The spring constant W
)�� ==��� 19.6
− em
. .1 kg. The equation of
9.
is k(e=, e.98/.05
N/m.
The
=
3
3 −3t/2
2t
�
�
�
−
e
2
y +
128y
yy4y
+++
y
=
��2e
2 1.= 0
or
motion is
64y = 0
0
��2
� 4t
6 = 0 8
10
��
y� (0)
+
4y
+
128y
with
initial
conditions
y(0)
=
0
ft,
y
=
1/4
ft/sec.
t
.1y + �19.6y
8.
Since
function
p(t) =y(0)
3/t=
is 1/4
continuous
t >=00and �t0 = 1 > 0, the IVP is guaranteed
with the
initial
conditions
ft,� y � (0)for
=�0allft/sec.
–0.5
cos
t sin The
t �� mass m = 1/2 lb-s2 /ft. The
with
initial
conditions
y(0)for
0(cos
y 0.
(0)
==
1/4
ft/sec.
14.
The
spring
constant
is=
kall
=ft,
16/(1/4)
=� 64
lb/ft.
to
have
a
unique
solution
t
>
W
t,
sin
t)
= 1. = .1 kg. The equation of
or
9. The spring
constant
is 2k lb-sec/ft.
=–1.98/.05 Therefore,
= 19.6�−N/m.
sinthe
t The
cos t�massofmmotion
damping
coefficient
isLγ =
=
equation
��
15.
The
inductance
0.2
henry.
The
resistance
R
=
3
× 102 ohms.is The capacitance
2.
Writeisthe IVP as
y
+
196y
=
0
motion
−5
3t ���for charge
4
sin t
C
equation
Q is
9. = 10 farads. Therefore, the
�� � y +
y �� − 1� y.1y
+�19.6y
y==
=0−2t
.�
+
2y
+
64y
0
−2t
with initial conditions y(0)−2t
= 0 m,
y (0)1� =e .14.tm/sec.
− 1 5te t − ORDER
1 ��
260
CHAPTER
LINEAR EQUATIONS
−2t t2−
�� �
� SECOND
W (e , te0.2Q
) =+� 300Q
= e−4t .
+
10
Q
=
0
−2t
−2t
�
or
−2e for(1all
− t2t)e
Since
the coefficient functions are continuous
< 1 and t0 = −2 < 1, the IVP is
or
��
2
249
��
�
The
soft
spring
has
a
higher
amplitude
and
a
longer
period.
Thehave
stiff
springofhas
a smaller
y
+
196y
=
0
15.
No.
Substituting
y
=
sin(t
)
into
the
differential
equation,
we
guaranteed
to
have
a
unique
solution
for
all
t
<
1.
The
longest
interval
existence
is
or
y + 4y + 128y = 0
10.
��
3
�
5
�
�
amplitude
and
a
shorter
period.
(−∞, 1).
Q2 0+m,
(1.5
)Q +xe
(5x2× 10
� ×210
�x1/4
� )q 2=x20
2y(0) =
with initial
initial conditions
conditions
(0)=
with
yy�x(0)
ft/sec.
�+.12tm/sec.
� sin(t )q(t) = 0.
−4ty(0)
sin(t=
)0(x,
+ft,2xe
cos(t
)=
cos(t
)p(t)
W
)
=
� 1 (1 + x)ex � = x e . 2
3. Write the IVP as
15. The inductance L = 0.2
henry.
resistance
R = 32 × 10 ohms. The capacitance
3t The
4
�
249
y �� the
+weequation
yhave
+
=is which
. is not continuous,Equa−5
For
the
equation
to
be
valid,
mustOrder
p(t)
= y−1/t,
or even
4.2
Theory
of
Second
Linear
Homogeneous
C
=
10
farads.
Therefore,
for
charge
Q
t(t −� 4)
t(t − 4)
t(t −
4)
11.
�
defined,
at t = 0.
t
�
�LINEAR
260
CHAPTER
EQUATIONS
sin4.t� SECOND
etORDER
cos t
t
t
2t
� ��t + e4t
�2t=
tions
Since the2t
coefficient
continuous
for+all
t5 Q
such
that t �= 0,
4 and
t.0 = 3, the IVP
W
(e
sin
=
−e
300Q
10
=
0
2tfunctions
� t, e cosare
2tt)0.2Q
t
�
�
16. W (e , g(t)) = e g (t) − 2e g(t) =
3e t. +
Dividing
byt)e , we find that g must
e (sin
cos t) both
e (cossides
t − sin
is guaranteed to have
a unique
that 02t< t < 4.
�
2t2solution for all t such 2t
satisfy
the
ODE
g
−
2g
=
3e
.
Therefore,
g(t)
=
3te
+
ce
.
15. No. Substituting y = sin(t ) into the differential equation, we have
or
1.
as
12.Write
4.
Since the
the IVP
coefficient
function
3 ln(|t|)
is�3 continuous
for
5 all t �= 0� and t0 = 2, the IVP is
17. W (t, g(t)) = tg � (t)2 − g(t)
= (1.5
t2 et .× 2Dividing
both
sides
3 � cos
Q2�� +
10
)q
0the equation by t, we have
2 2× 10
�� � )Q
�+ (5
2 2θ �
θ
1 +=ofcos
2
guaranteed
to
have
a
unique
solution
for
all
t
>
0.
y
+
y
=
1.
�
−4t
sin(t
)
+
2
cos(t
)
+
2t
cos(t
)p(t)
sin(t
)q(t)� =
�
t
t
W (cos g(t)
θ, 1 +
2θ)ct.= � t
= 0.
0.
g − g/t = te . Therefore,
= cos
te +
−2 sin θ cos θ −2 sin 2θ �
5. Write the IVP �as �
18.
(f,equation
g) = f g to
− be
f=
g.valid,
Also,
Wmust
(u, v)xhave
=for
Wp(t)
(2f
+ 2g).
Upon
evaluation,
W (u,
=
Since
the
function
p(t)
3/t
is we
continuous
allln
t−
>
0 fand
twhich
>
thecontinuous,
IVP is guaranteed
0 = 1
For W
the
=
−1/t,
is 0,
not
or v)
even
|x|g,
��
2� +
��
20.
2
−1
� If y � = t2 , then y �� = 2. Therefore,
y
+
y
y
=
13.
t
y
−
2y
=
t
(2)
−
2t
=
0.
If
y
=
t
,
then
5f
g
−
5f
g
=
5W
(f,
g).
to
have a1atunique
1
2
1 for all t > 0.
defined,
t = 0. solution
x2 − 3−3 1 x −
�
−3
2 ��
−13
y
=
2t
.
Therefore,
t
y
−
2y
=
t
(2t
)
−
2t
=
0.
Since
the
equation
is linear, the
�
�
�
�
2
2 W (f,2tg) = f g −2tf �g =2t cos2tt − sin t and
4t W (u, v) = −4f g + 4f g.2t Therefore, W (u, v) =
19.
2.
IVP=
16.Write
Wthe
(e the
,coefficient
g(t))
e cfunctions
g t(t)
g(t)continuous
=a3esolution.
. Dividing
both
sides
byxe�= ,0,we
find xthat
g must
2as
−1 − 2e are
Since
for
all
x
such
that
3
and
0 = 1, the
function
y
=
c
t
+
will
also
be
3
1 �
2
3t
4
2t ��
2t sin t 2t
−4t
cos
t
+
4
sin
t.
�
satisfy
the ODE g to
− have
2g = a3eunique
g(t)
+ cethat
y. Therefore,
− solution
y +
yx=such
. . 0 < x < 3.
IVP
is guaranteed
for =
all3te
14.
t
−
1
t
−
1
t
−
1
��
��
2 t Therefore, y1 + 4y1 = 0. Similarly, for y2 = sin 2t,
20.
= cos
y1−=g(t)
−4 cos
17. For
W (t,y1g(t))
= 2t
tg �,(t)
= t2t.
e . Dividing both sides of the equation by t, we have
��
Since
the=
functions
are
continuous
all2t,t sin
< 2t)
1 and
= −2
< a1,fundamental
the IVP is
t Therefore,
t
yg2��� −
= g/t
−4
sincoefficient
2t.
y
+
4y
=
0.ct.
Since Wfor
(cos
= 2,t0they
form
2te +
2
te
.
Therefore,
g(t)
=
(a) Dividing the equation by y, we have
guaranteed
to have a unique solution for all t < 1. The longest interval of existence is
set.
18. W1).
(f, g) = f g � − f � g. Also, W (u, v) = W (2f − g, f + 2g). Upon evaluation, W (u, v) =
(−∞,
t
�
�
t
t
t
t
� 1 + y1 = e − 2e + e = 0. For y2 = te ,
21.
y1� g==e5W
, y1(f,
=g).
y1�� = et . Therefore,
y1��1−(y2y
5f
g �For
− 5f
y �� +
)2 =
0.
�
t
��
t
��
�
t
t
t
3.
Write
y2 =
(1 +the
t)e IVP
and�as
y2 =� (2 + t)e . Therefore, yy2 − 2y2 + y2 = (2
+ t)e
− 2(1 + t)e + te = 0.
�
�
19.
W
(f,
g)
=
g
=
t
cos
t
−
sin
t
and
W
(u,
v)
=
−4f
g
+
4f
g.
Therefore,
W (u, v) =
3t
4
2
t f gt − f 2t
Further, W (e , te ) = e . Therefore,
they �solutions
form
set.
y �� + the form
+
y = a fundamental
.
equation
required.
−4t This
cos t +
4 sin t.� will not have
t(t − 4)
t(t − 4)
t(t − 4)
22. For y1 = x, y = 1 and y �� = 0. Therefore, x2 y �� − x(x + 2)y � + (x + 2)y1 = −x(x + 2) +
–2
0
–3
4.3
10.
Linear Homogeneous Equations with Constant Coefficients
1. The characteristic equation is given by 4λ2 + 12λ + 9 = 0. Therefore, we have one
(a)
270 repeated root λ = −3/2. Therefore,
CHAPTER
4. SECOND
ORDER
EQUATIONS
the general
solution
is givenLINEAR
by
(a) The characteristic equation is given by λ2 + 2λ − 3 = 0. Therefore, the two distinct roots
−3t/2 is given
−3t/2
are λcharacteristic
= −3, 1. Therefore,
theisgeneral
by
y(t)
= 2λ
csolution
+ 0.
c2 teTherefore,
.
(a) The
equation
λ2 +
the roots are λ = −4, 2.
1 e− 8 =
Therefore, the general solution is
y(t) = c1 et + c2 e−3t .
(b) For y above, we see that
y(t) = c1 e−4t�+ c2 e2t �
3c1 point,
3t −3t/2
(b) The critical point (0, 0) isy �a=saddle
−
e−3t/2 +therefore,
c2 1 − unstable.
e
.
2
(b) We see that the critical point (0,20) is an unstable saddle
point.
(c) For y above, we see that
Therefore, we can rewrite our solution
as
parameter
family
t
y � =that
c1 ethe
− two
3c2 e−3t
.
(c) From our solution in part (a), we see
� � � � �
�
x1
ysolution
c1−4t
e−3t/2
+ �c22tte−3t/2
Therefore, we can rewrite
our
as
the
two
parameter
family
�
=
=
. 3t � e−3t/2
3c1 1−3t/2
x2 � � y � y�(t)�=−−4c
ee +
+ 2c
c2 2 e1�−
2
2
�
� �
t �
−3t �
x1
y
c
e
+
c
e
1
2
t
−3t/2
= 1
=
Therefore,
e�−3t/2
�12−
�.
2 3c
3t
x=2 c1� −�y3� e
c1+
e�t c−
e−3t
y 2
1
1
2
�
�
�
�
x=
= c1 e−4t 1 t + c2 e2t1
.
y�
−4
2 e−3t .
= c1
e + c2
1
−3
3
23
x2
–3
–3
–3
–2
–2
–2
3
1
2
x2
2
x2 01
–1
1
–10
–1
0
–1 –2
–1
–1
–3
–2
–2
–3
–3
1
2
3
2
x1 2
x1
3
3
x1
1
1
(c) The critical point (0, 0) is an asymptotically stable improper node.
12.
11.
2.
(a) The characteristic equation is given by λ2 + 2λ + 2 = 0. Therefore, the roots are
λ = −1 ± i. Therefore, one complex solution is
y(t) = e(−1+i)t = e−t (cos(t) + i sin(t)).
Considering the real and imaginary parts of this function, we arrive at the general
solution
y(t) = c1 e−t cos(t) + c2 e−t sin(t).
(b) From our solution in part (a), we see that
y � (t) = c1 e−t (− cos(t) − sin(t)) + c2 e−t (− sin(t) + cos(t)).
Therefore,
x = c1 e
−t
�
�
�
�
cos(t)
sin(t)
−t
+ c2 e
.
cos(t) − sin(t)
− sin(t) + cos(t)
4.3. LINEAR HOMOGENEOUS
WITH CONSTANT COEFFICIENTS 271
1
2
= � EQUATIONS
= 1
x2
y
−5c2 e−5t
� �
� �
1
1
= c1
+3 c2
e−5t .
0
−5
2
x2
1
282
CHAPTER 4.3 SECOND ORDER LINEAR EQUATIONS
–3
0.5
0
–2 2 –1
x2
1t
1
1.5
–2
–1
2
3
x1
0
–3
1
–1
2
–2
0
1
–3
2.5
3
2
3
x1
–1
–10
–2
y
–20
–3
(c) The critical point (0, 0) is an asymptotically
stable spiral point.
–30
13.
272
CHAPTER 4. SECOND ORDER LINEAR EQUATIONS
14. (a) The characteristic equation is given by λ2 + 5λ = 0. Therefore, the two distinct roots
(a) The
equation the
is given
by solution
4λ2 − 9 is= given
0. Therefore,
the two distinct roots
arecharacteristic
λ = 0, −5. Therefore,
general
by
are λ =
3/2,
−3/2.
the general solution is given by
The solution
y→
−∞
as t Therefore,
→ ∞.
−5t
y(t)
=3t/2
c1 + c2 e−3t/2
.
2
30. The characteristic equation is given y(t)
by λ=
+c14λ
+
3
=
0.
the two distinct roots
e
+ c2 e Therefore,
.
are λ = −1, −3. Therefore, the general solution is given by
(b) Every point of the form (x0 , 0) is a critical point. They are each nonisolated, stable
(b) For y above, we see that
points.
y(t) =
c13c
e−t1 +
c2 e−3t3c. 2 −3t/2
y� =
e3t/2 −
e
.
2
2
Therefore,
(c) For y above, we see that
�
� the two
−5t
Therefore, we can rewrite your
parameter family
(t) solution
= −c e−t
−
yas
= 3c
−5ce−3t
2e . .
� � � 1� � 2 3t/2
�
x1our
y
c1 e two+parameter
c2 e−3t/2 family
Now usingTherefore,
the initialwe
conditions,
we
need
can rewrite
solution
as
the
=
= 3c1 3t/2 3c2 −3t/2
�
e
�x2 � �y� �� �2 e �−−5t2�
c1 +
x1
y1c2 = 2 c1 + c2 e 1 �
=e3t/2 + c −5t e−3t/2 .
�
==1c1−y3c
x2 −c
−5c22 e 3
32 =
−1.
� � − 2� �
2
1
1
−5t
The solution of this system of equations is c1 ==5/2
the specific
c1 and c+2 c=2 −1/2. eTherefore,
.
0
−5
5 −t
1 −3t
solution is y(t) = 2 e − 2 e .
3
2
x2
2
1
1.6
x2
0
–1
2
–3
1.4
–2
1
1
2
3
x1
–1
1.2
y
3
1.8
–3
1
–1–2 0
–1
–3
–2
–2
0.8
0.6
1
2
3
x1
0.4
–3
0.2
0
0.5
1
1.5
t
2
2.5
3
(c) The critical point (0, 0) is a saddle point, and, therefore,
unstable.
15. solution y → 0 as t → ∞.
The
14.
2
31. (a)
TheThe
characteristic
equation
is λ2is+given
4λ + 5by=25λ
0, which
has
λ =Therefore,
−2 ± i. Therefore,
characteristic
equation
− 20λ
+ roots
4 = 0.
we have one
−2t
−2t
the general
solution
is
y(t)
=
c
e
cos(t)
+
c
e
sin(t).
The
derivative
of
y
is
y � (t) =
repeated root λ = 2/5. Therefore,
the general
solution is given by
1
2
c1 e−2t (−2 cos(t) − sin(t)) + c2 e−2t (−2 sin(t) + cos(t)).
Using2t/5the initial conditions, we have
2t/5
y(t)
=
c
e
+
. conclude that the solution
1
c1 = 1 and −2c1 + c2 = 0. Therefore, c1 = 1 and c2 = 2,c2 te
and we
is
(b) The critical point (0, 0) is an unstable
improper node.
y(t) = e−2t cos(t) + 2e−2t sin(t).
(c) For y above, we see that
�
�
2c
2t
The solution y → −∞ as t → ∞.
33. The characteristic equation is given by λ2 − 6λ + 9 = 0. Therefore, there is one repeated
root, λ = 3. Therefore, the general solution is given by
4.3. LINEAR HOMOGENEOUS EQUATIONS
WITH CONSTANT COEFFICIENTS 285
y(t) = c1 e3t + c2 te3t .
4.3. LINEAR HOMOGENEOUS EQUATIONS
WITH CONSTANT COEFFICIENTS 285
Therefore,
Therefore,
y � (t) = −3c2 e−3t .
y � (t) = −3c2 e−3t .
Now using the initial conditions, we need
Now using the initial conditions, we need
c1 + c2 = −2
c1−3c
+ c22 =
= 3.
−2
−3c2 = 3.
The solution of this system of equations is c1 = −1 and c2 = −1. Therefore, the specific
The solution
system
solution
is y(t)of=this
−1 −
e−3t . of equations is c1 = −1 and c2 = −1. Therefore, the specific
solution is y(t) = −1 − e−3t .
–1
–1
–1.2
–1.2
–1.4
y –1.4
y –1.6
–1.6
–1.8
–1.8
–2
0
0.5
1
–2
0
0.5
1
1.5
t
1.5
t
2
2.5
2
2.5
3
3
The solution y → −1 as t → ∞.
TheThe
solution
y → −1 as
t → ∞.is λ2 + 1 = 0, which has roots λ = ±i. Therefore, the general
36.
characteristic
equation
� ±i. Therefore, the general
36. The is
characteristic
equation
is λ2 + 1The
= 0,derivative
which hasofroots
solution
y(t) = c1 cos(t)
+ c2 sin(t).
y is λy=
(t) = −c1 sin(t) + c2 cos(t).
�
solution
y(t) =
c1 cos(t) +wec2have
sin(t). The derivative of y is y (t) = −c1 sin(t) + c2 cos(t).
Using
theisinitial
conditions,
Using the initial conditions, we have
√
1
√3 c2 = 2
c
1+
21 c√+ 2 3 c = 2
1
2
2 √3 2 1
−
3c1 + 1c2 = −4.
− 2 c1 + 2 c2 = −4.
√
2 √ 2
The solution of this system is c1 = 1 + 2 √3 and c2 = √3 − 2, and we conclude that the
The solution
of this system is c1 = 1√+ 2 3 and c√
3 − 2, and we conclude that the
solution
is
2 =
solution is
y(t) = (1 + 2 √3) cos(t) + ( √3 − 2) sin(t).
y(t) = (1 + 2 3) cos(t) + ( 3 − 2) sin(t).
4
4
2
2
0
0
–2
–2
–4
–4
2
4
6
8
10
t
2
4
6
t
8
10
y(t)
= c1 eαt +the
c2 e(α−1)t
.
p(λ) = (λ − 2)(λ + 3) = λ2 + λ − 6.
Therefore,
differential
equation is y �� + y � − 6y = 0.
45.
We need to find a characteristic
equation of degree
two with oneLINEAR
repeated EQUATIONS
root λ = −2.
296order
SECOND
In
for the solution
to2 tendCHAPTER
to zero, we4.need
α, α − ORDER
1 < 0. Therefore,
2
�� the� solutions
We take p(λ) = (λ+2) = λ +4λ+4. Therefore, the differential equation is y +4y +4y = 0.
will all tend to zero as long as α < 0. All solutions will become unbounded if α > 0 and
2.0
46.− We
to find
characteristic
equation
two ORDER
with roots
λ = −3 EQUATIONS
± 4i. We take
α
1 >need
0. This
will aoccur
exactly
when
α > of
1. degree
298
SECOND
LINEAR
1.5 CHAPTER 4.
2
p(λ)The
= (λcharacteristic
− (−3 + 4i))(λ
− (−3
− 4i))
6λ−+2(α
25. −Therefore,
the differential
equation
2
1.0 is λ
48.
equation
+=
(3λ− +
α)λ
1) = 0. Solving
this equation,
we
��
�
is
y
+
6y
+
25y
=
0.
0.5
see that the roots are λ = α − 1, −2. Therefore, the general solution is
47. The characteristic equation is λ2 − 1(2α − 10
1)λ
+ α(α
− 1) =
0. Solving this equation, we
2
3
4
(α−1)t
−2t
!0.5
y(t)Therefore,
= c1 e
+ general
c2 e . solution is
see that the roots are λ = α, α − 1.
the
!1.0
x2
5
order for all solutions to tend to
zero,
− 1. < 0. Therefore, the solutions will
y(t)
= c1we
eαtneed
+ c2 eα(α−1)t
−2t
tend to zero as long as α < 1.
Due
to
the
term
c
, we10 can never guarantee that all
2e
0
–10
–5
5
In
all
The
solution
ybecome
oscillates
with
a growing
xx1 →
solutions
willthe
unbounded
as
→ amplitude
∞.
In
order
for
solution
to tend
to tzero,
we need as
α, α
− 10.< 0. Therefore, the solutions
–5
will Suppose
all tend the
to zero
long
as α <λ 0., λ All
solutions will become unbounded if α > 0 and
49.
rootsasare
distinct,
1
2 . Then the solution is
α − 1 > 0. This will occur exactly when α > 1.
–10
λ1 t
λ2 t
4.4The characteristic
Mechanical
and
Electrical
2
eVibrations
. 1) = 0. Solving this equation, we
1 eα)λ+−c22(α
48.
equation
is λy(t)
+=
(3c−
−
(b)
see that the roots are λ = α − 1, −2. Therefore, √
the general solution
is λ t
Solving
the
equation
y(t)
=
0, 4weimplies
see that
we
must
have
c1 eλδ1 t =
= arctan(4/3)
−c22 e 2 which
1.
R
cos
δ
=
3
and
R
sin
δ
=
R
=
25
=
5
and
≈ implies
0.9273.
7.
lb/ft.
The
mass
is
3/32
lb-s
/ft. Therefore,
the
(λ1The
−λ2 )tspring constant is k = 3/(1/4) = 12 (α−1)t
−2t
eTherefore,
= −c2 /c1 . First, in order
to
guarantee
any
solution
of
this
equation,
we
would
y(t) = c1 e
+ c2 e .
equation
of
motion
is
need c2 /c1 < 0. Then, applying the natural
logarithm function to the equation, we see that
y =35 cos(2t
− 0.9273).
��
y
+
=α0,− 1 < 0. Therefore, the solutions will
tIn=order
ln(−cfor
/c
)/(λ
−
λ
).
need
2 1all solutions
1
2 to tend to zero, we 12y
32
√ Due to the term
√ c e−2t , we can never guarantee that all
all tend
tobe
zero as long to
as α < 1.
which
can
2. R cos
δ = simplified
−1 and R sin δ = 3 implies R = 4 2= 2 and δ = π + arctan(−3) = 2π/3.
solutions
will become unbounded as t →
∞.128y = 0.
y �� +
Therefore,
49. Suppose
the roots are
are y(0)
distinct,
λy1 ,=λ22ft,
. cos(t
Then
−the
2π/3).
The
initial
= −1/12
y � (0)
=
2solution
ft/sec. is
The general solution is y(t) =
√ conditions √
A cos(8 2t) + B sin(8 2t). Considering the initial
conditions,
we arrive at the solution
λ√
1t
2t
y(t) = c1 eλ√
+ c√
.
2e
3. R cos δ = 4 and R sin δ = −2 implies R =
√ 20 = 22 5 and
√ δ = arctan(−1/2) ≈ −0.4636.
1
Therefore,
y(t) = − cos(8 2t) +
sin(8 2t).
Solving the equation y(t) = 0, we12
see√that we must
8 have c1 eλ1 t = −c2 eλ2 t which implies
y = to
2 guarantee
5 cos(3t + 0.46362).
√ in order
√
e(λ1 −λ2 )t = −c2 /c1 . First,
any solution
of this equation, we would
The �
frequency ω0 = 8 2 rad/sec. The period is T = π 2/8 seconds. The amplitude is
need c2 /c1 < 0. Then,√applying
the
natural
logarithm
function
to
the equation,
we see that
√
√
√
2−2
2 =δ =11/12
R
=
(1/12)
+
(1/4
2)
ft.
The
phase
is
δ
=
π
−
arctan(3/
2).
4.
R
cos
δ
=
and
R
sin
−3
implies
R
=
13
and
δ
=
π
+
arctan(3/2)
≈ 4.1244.
t = ln(−c2 /c1 )/(λ1 − λ2 ).
Therefore,
8. The inductance L = 1 henry.√ The resistance R = 0. The capacitance C = 0.2510−6
y =for 13
cos(πtq +
farads. Therefore, the equation
charge
is π + arctan(3/2)).
!1.5
q �� + (4106 )q = 0.
5.
The initial conditions are q(0) = 10−6 coulombs, q � (0) = 0 coulombs/sec. The general
2
(a) The ofspring
constant isisq(t)
k ==2/(1/2)
= 4 lb/ft.
The massThe
m =initial
2/32 conditions
= 1/16 lb-s
/ft.
solution
this equation
A cos(2000t)
+ B sin(2000t).
imply
−6
Therefore,
the equation
is
the specific
solution
is q(t) =of10motion
cos(2000t)
coulombs.
9.
1 ��
y + 4y = 0
163.92 N/m. The mass m = .02 kg. The damping
(a) The spring constant is k = .196/.05 =
constant is γ = 400 dyne-sec/cm = .4 N-sec/m. Therefore, the equation of motion is
which can be simplified to
�
.02y �� +
+ 3.92y
y �� .4y
+ 64y
= 0. = 0
or
The initial conditions are y(0) = ��1/4 ft, � y � (0) = 0 ft/sec. The general solution of the
y + 20y + 196y = 0,
differential equation is y(t) = A cos(8t) + B sin(8t) ft. The initial condition implies
with initial conditions
y(0) = .02 m,
y � (0) = 0 m/sec. 1The solution of this equation is
√ solution
A = 1/4−10t
and B = 0.√Therefore, the
is y(t) = cos(8t) ft.
y(t) = e
(A cos(4 6t) + B sin(4 6t)). The initial conditions
imply
4
�
�
√
√
√
y(t) = e−10t 2 cos(4 6t) + (5/ 6) sin(4 6t) cm