which simplifies to
g
y = 0.
L
�
�
Therefore,
the
frequency
is
ω
=
g/L,
and,
therefore,
the
period
is
T
=
2π/
g/L =
4.4
Mechanical
and
Electrical
Vibrations
0
�
2π
302 L/g.
CHAPTER 4. √
SECOND ORDER LINEAR EQUATIONS
1.
cos δ =y 3is and
R sin δof=the
4CHAPTER
implies
R 4.
=
25 = 5 and
δ = LINEAR
arctan(4/3)
≈ 0.9273.
15.
Suppose
a solution
differential
equation
302 R
SECOND
ORDER
EQUATIONS
Therefore,
Let w = y − v. Then w satisfies the same
differential equation, since the equation is linear,
my �� + γy � + ky = 0
�
and,
w(tThen
0, satisfies
w (t0 )EQUATIONS;
=the
yy1 .=
Therefore,
y = vequation,
+
w UNDETERMINED
where
v, w
the
specified
5 cos(2t
− 0.9273).
4.5. wfurther,
NONHOMOGENEOUS
METHOD
OF
COEFFICIENTS311
0) = w
Let
= y − v.
same
differential
since
thesatisfy
equation
is
linear,
�
�
conditions.
and
y(t
)
=
y
,
y
(t
)
=
y
.
Then
suppose
v
is
a
solution
of
and, satisfies
further, the
w(t0initial
) = 0,conditions
w (t0 ) =
y
.
Therefore,
y
=
v
+
w
where
v,
w
satisfy
the
specified
0
1
√ 1 0
√0
�
302
CHAPTER
4.
SECOND
ORDER
LINEAR
EQUATIONS
2
2.
coscharacteristic
δ differential
= −1 and equation,
R
sin
δ = but
implies
R
4 problem
=
2 and
π 00+
arctan(−3)
=00)2π/3.
the
same
satisfies
the=rinitial
conditions
v0θ−r
andcos(ω
v has
(t
=sin
0.θ.
conditions.
5. R
The
equation
for3we
the
homogeneous
λ=v(t
+
9)t)=
0,
which
16.
Using
trigonometric
identities,
note
that
sin(ω
= isrδ sin(ω
cos
t)roots
0 t−θ)
Therefore,
λ=
±3i.toTherefore,
theidentities,
solution
ofwe
the
yh (t)
c1 cos(3t)+c
In
order
write A cos(ω
t)homogeneous
inthat
the form
rproblem
sin(ω
t −isθ),
we =
need
A = −r sin
θ and
2 sin(3t).
0 t) + B sin(ω
0note
16.
Using
trigonometric
r sin(ω
0 t−θ)0 = r sin(ω0 t) cos θ−r cos(ω0 t) sin θ.
2 problem,
2
2 equation,
Let
w
=
y
−
v.
Then
w
satisfies
the
same
differential
since
the
equation
is
linear,
To
find
a
solution
of
the
inhomogeneous
we
look
for
a
solution
of
the
form
y
=
B
=
r
cos
θ.
Therefore,
we
must
have
A
+
B
=
r
and
tan
θ
=
−A/B.
From
the
text,
we
y 0=
− 2π/3).
In order
to 3twrite A
cos(ω
t) 2incos(t
the form
r sin(ω0 t − θ), we need A = −r√sin pθ(t)
and
0 t)� + B sin(ω
3t
2
3t
2
2
and,
further,
w(t
=+
0,D.
w Substituting
(t
) =have
y01t)
. Therefore,
=and
v cos(ω
+tan
w where
v,
w differential
satisfy
the
specified
2 function
2 0 t)
2
Ae= +
Bteθ.in+
Ct0 )eto
this
form
into
the
equation,
know
order
write
A0cos(ω
+
=
R
δ),
we
need
R=
A
+B
0θt −
B
rthat
cos
Therefore,
we
must
AaB
+sin(ω
B√
=yrof
=
−A/B.
From
the
text,
we
√
√
conditions.
and
equating
like Therefore,
terms,
we
have
18A
+Rsin(ω
6B
+
= cos(ω
0,and
+δ),
12C
=we0,need
=
1+and
and
and
δδ =
in
order
forR
cos(ω
t==
−2R
δ)
r18B
sin(ω
−we
θ),need
02C
0 tarctan(−1/2)
3.
Rtan
cos
=inB/A.
4order
and
R
sin
δ=
implies
=
20
5=
=
know
that
to
write
A−2
cos(ω
R18C
=≈r −0.4636.
A2R
B2
0 t) + B
0 t)
0 tδ−
9D
=
6.
The
solution
of
these
equations
is
A
=
1/162,
B
=
−1/27,
C
=
1/18
and
D
=
2/3.
tan
θ
=
−1/
tan
δ.
16.
trigonometric
identities,
we note
r sin(ω
r sin(ω
t) cos
sin θ.
andUsing
tan δ =
B/A. Therefore,
in order
for Rthat
cos(ω
= r=
sin(ω
weθ−r
needcos(ω
r = 0Rt) and
Therefore,
0 t−θ)
0 t − δ)
0 t −0θ),
√
Therefore,
the
solution
of
the
inhomogeneous
problem
is
2
In
tospring
write
A cos(ω0 t)
the=form
r sin(ωThe
θ), weisneed
= 1/4
−r sin
θ and
tanorder
θThe
= −1/
tan δ.
17.
constant
is+kB=sin(ω
64
lb/ft.
8/32A =
lb-s
/ft.
0 t − mass
y8/(1.5/12)
= 20 t) 5incos(3t
+ 0.46362).
2
2
2
B
=
r
cos
θ.
Therefore,
we
must
have
A
+
B
=
r
and
tan
θ
=
−A/B.
From
the
text,
we
Therefore,
the equation
ofismotion
is
1 The
1 2is3t8/322 = 1/4
√ lb-s2 /ft.
17. The spring
constant
k =+8/(1.5/12)
= 164√
lb/ft.
3t
3t mass
2
y(t)
=
c
cos(3t)
c
sin(3t)
+
e
−
te
+
t
e
+
.
1
know
thatδ in
to write
A
δ),+we
need
+ B2
02t) + B sin(ω
0 t) = R cos(ω
162
27 δ0 t=− 18
3 R = ≈A4.1244.
Therefore,
theorder
equation
4.
R cos
=
−2
and
R of
sinmotion
δ cos(ω
= −3is
R
13 and
π
arctan(3/2)
1implies
��
� =
+ 64y
=δ)0.= r sin(ω0 t − θ), we need r = R and
and tan δ = B/A. Therefore, in orderyfor+Rγycos(ω
0t −
Therefore,
4
1
�� homogeneous
tan
= −1/
tan δ.
6. θThe
characteristic
equation √
for the
is λ2 + 2λ + 1 = 0, which
√
ycos(πt
+ γy+� +
64y
= 0.problem
y
=
13
π
+
arctan(3/2)).
4.5. the
NONHOMOGENEOUS
EQUATIONS;
METHOD
UNDETERMINED
COEFFICIENTS311
4damping when
The
motion
will constant
experience
γ = 2OF
km
=
8 homogeneous
lb-s/ft.
has
repeated
root λ is=critical
Therefore,
of
the
problem
is
17.
The
spring
k−1.
= 8/(1.5/12)
=the
64 solution
lb/ft.
The
mass
is 8/32 = 1/4
lb-s2 /ft.
√
−t
−t
−6
yh (t)The
= cinductance
e will
+
c2 te L .=ofTo
find
aisdamping
solution
of
theγinhomogeneous
problem,
we look for
The
motion
experience
critical
when
=C
2 =km
= 8 lb-s/ft.
1the
18.
0.2
henry.
The capacitance
0.810
farads. Therefore,
thea
Therefore,
equation
motion
5.
2 for
−t the homogeneous problem is λ2 + 9 = 0, which has roots
5.
The
characteristic
equation
solution
of
the
form
y
(t)
=
At
e
.
Substituting
a
function
of
this
form
into
the
differential
equation
for charge qLpis = 0.2 henry.1 The capacitance C = 0.810−6 farads. Therefore, the
18.
λ =The
±3i.inductance
Therefore,
thelike
solution
of the
homogeneous
is yhA(t)==1 cand
1 cos(3t)+c
2 sin(3t).
equation,
and
equating
terms,
we
= 72.=problem
Therefore,
the solution
of
10
y���� have
+ γy ��2A
+ 64y
0.
2
equation
for
charge
q
is
0.2q
+
Rq
+
q
=
0.
4
(a)
The
spring
constant
is
k
=
2/(1/2)
=
4
lb/ft.
The
mass
m
=
2/32
=
1/16
lb-s
/ft.
To
find
a
solution
of
the
inhomogeneous
problem,
we
look
for
a
solution
of
the
form
y
(t)
=
p
the inhomogeneous problem is
87
10
√
3t
3t the2 equation
3t
��is
�
�
Therefore,
of
motion
Ae + Bte will
+ Ct e + D. Substituting
function
this
into
the differential equation,
0.2q
+a Rq
+ γof
q=
0. form
The
damping
when
km
=
8 7lb-s/ft.
The motion
motion will experience
experience critical
critical
= 222 −t
0.2
· 10
/8 = 103 ohms.
−t when
8−tR =
y(t) damping
=
c1 e+
+ c+
.
2 te2C +=t e
and equating like terms, we have
18A
6B
0,
18B
+
12C
1 and
�
−6 √ = 30, 18C = √
7 /8
18.
The
inductance
L
=
0.2
henry.
The
capacitance
C
=
0.810
farads.
Therefore,
the
1
The
motion
will
experience
critical
damping
when
R
=
2
0.2
·
10
=
10
ohms.
310
CHAPTER
4.
SECOND
ORDER
LINEAR
EQUATIONS
19.
If 6.
theThe
system
is critically
damped oris yoverdamped,
γ ≥ 2C =
km.
γ =D 2= 2/3.
km
9D =
solution
of these equations
A�� +
= 4y
1/162,
Bthen
= −1/27,
1/18Ifand
=
0
√
equation
for
charge qthen
isequation
16 overdamped,
7.
The
characteristic
for theis homogeneous
problem
2λ22 √
+km.
3λ +If1 γ= =0, 2which
(critically
damped),
given
by
Therefore,
the solution
ofthe
thesolution
inhomogeneous
problem
is then γis ≥
7
19.
If the
system
is critically
damped
or
km
10
��the solution
�
has
roots
λ
=
−1,
−1/2.
Therefore,
of
the
homogeneous
problem
is
y
(t)
=
4.5
Nonhomogeneous
Equations;
Method
of
Undeter0.2q
+
Rq
+
q
=
0.
h
(critically
damped),
then the solution
is given
by
y(t) =
(A + Bt)e
1 8 −γt/2m
1. 3t
1 2 3t 2
which
−t/2be simplified to
3t
c1 e−t
+ c2 ecan
. y(t)
To =find
a
solution
of
the
inhomogeneous
problem,
we
will
first
look
for
c1 cos(3t) + c2 sin(3t) +
e−γt/2m
− �te + t e + .
mined
Coefficients
7 /8 = 1033 ohms.
162
2
The
motion
critical
damping
227
0.2
1018
= (A
+when
Bt)e
. is,
yA�� Ct
+
64y
a solution
Y1 (t)
=
Ay(t)
+
Bt
+
to==R
account
for√
the
inhomogeneous
t2 .
In
this
case,ofwill
ifthe
y experience
=form
0, then
we
must
have
+
Bt
0,0.=that
t·=
−A/B
(assuming term
B �= 0).
√
√
Substituting
a function
ofnever
this damped
form (unless
intoorthe
differential
and
equating
terms,
If
0,characteristic
the
isequation
zero
ABt
==
0).0, Ifthat
γ equation,
>is,2 t γ=
km
(overdamped),
the
2 km.
19.
If= the
system
is then
critically
overdamped,
then
If1 γ=like
=then
In
this
case,
if solution
yconditions
= 0,
we must
have
A
+
−A/B
(assuming
B2ofwhich
�=km
0).
2 λ2
6.B
The
forthe
the
homogeneous
is≥
+ 2λ
+solution
0,
The
initial
are
y(0)
=
1/4
ft,
y � (0)
= problem
0 problem
ft/sec.
The
general
the
√
1.
The
characteristic
equation
for
homogeneous
is
λ
−
2λ
−
3
=
0,
which
has
we
have
A
+
3B
+
4C
=
0,
B
+
6C
=
0
and
C
=
1.
The
solution
of
these
equations
is
solution
is
given
by
(critically
damped),
then
the
solution
is
given
by
Ifroots
Bdifferential
=
0,=repeated
the
solution
never
zero
(unless
Aλ+1the
=
0).
If
2problem
km homogeneous
(overdamped),
then
the
3t
−tis
has
the
rootisλis
=y(t)
−1.
Therefore,
solution
of
the
problem
equation
=
A
cos(8t)
B
sin(8t)
ft.
The
initial
condition
implies
22 tγ >
λ
3,
−1.
Therefore,
the
solution
of
the
homogeneous
is
y
(t)
=
c
e
+
c
e
t
λ
h a solution
1
A = 14, B =−t−6, C =
Y1 ==Ae
14 − +
6tBe
+ t . Next, we look for
of2 the.
y(t)
−t 1. Therefore,
solution
is
given
by
1
y
(t)
=
c
e
+
c
te
.
To
find
a
solution
of
the
inhomogeneous
problem,
we
look
for =a
−γt/2m
h
1
2
To A
find
solution
inhomogeneous
problem,
we
look
for
a solution
the function
form yp (t)
y(t)
(A
. t.
inhomogeneous
the 2form
Y2=(t)
=+D
+ λE
sin
Substituting
into
= a1/4
andproblem
Bof=the
0. of
Therefore,
the
solution
cos(8t)
ft. ofthis
λBt)e
t is ty(t)
1cos
2 t=
−t(19)
where
λ
are
given
by
equation
in
the
text.
Assume
for
the
moment,
that
A,
B
=
�
0.
y(t)
=
Ae
+
Be
2t λ1 ,of
solution
the
form
y
(t)
=
At
e
.
Substituting
a
function
of
this
form
into
the
differential
2
4 andequation,
p
Ae ODE
. Substituting
a function
of
this
form
into
the
differential
we
have
the
and equating
like
terms,
we
find
that
D
=
−9/10
E
=
−3/10.
Therefore,
the
λ1 t
λ2 t
(λ1 −λ2 )t
In
thisyλcase,
yequating
=
0, Ae
then
we
must
have
A + 2A
Bt
is,for
t==
−A/B
(assuming
BB�=�=one
0).
Then
=λand
0 ifare
implies
=terms,
−Be(19)
implies
ethat
−B/A.
There
only
equation,
wewhich
have
= 0,
2.
Therefore,
A moment,
= 1 and
theissolution
√
where
given
bylike
equation
in
text.
Assume
the
that
A,
0.of
1 ,of
2the inhomogeneous
solution
problem
is the
2t
2t
2t
2t
If
B
=
0,
the
solution
is
never
zero
(unless
A
=
0).
If
γ
>
2
km
(overdamped),
then
the
solution
to
this
equation.
If
A
=
0
or
B
=
0,
then
there
are
no
solutions
to
the
equation
the inhomogeneous
4Aeλ2 t −which
4Ae implies
− 3Ae e(λ
=1 −λ
3e2 )t. = −B/A. There is only one
Then
y = 0 implies problem
Aeλ1 t = is−Be
solution
is
given
by
ysolution
= 0 (unless
they
are
both
zero,
in
which
case,
the
solution
identically
zero).
9 areis no
3
to this equation.
If−tA+ =
0−t/2
or +
B 14
=−
0,
then
solutions
to the equation
−t
−tt2there
2 −t
y(t)
=
c
c
6t
+
t
−
sin
t.
λ1c
t 2 te
λ−
1e
2 e y(t)
2 tt e cos
y(t)
=
c
e
+
+
.
1
=
Ae
+
Be
Therefore,
we
need
−3A
=
3,
or
A
=
−1.
Therefore,
the
solution
of
the
inhomogeneous
10
10
20.
the system
critically
damped,
thencase,
the general
solution
is
y = If
0 (unless
theyisare
both zero,
in which
the solution
is identically
zero).
problem
is
where
λ1 , λsystem
by equation
(19)then
in the
text.
Assume
for the
moment,
that A, B �= 0.
2 are given
2
−γt/2m
3t
−t problem
2t
20.
If
the
is critically
damped,
the
general
solution
is2 2λ
7. The
The
characteristic
equation
for
homogeneous
+
3λ
+ 1 =has
0, which
y(t)
(A
+
Bt)e
. .2 )t is is
8.
characteristic
equation
for
the
homogeneous
problem
λ
+
1
=
0,
which
c
e
+
c
e
−
λ1 t
λ2the
t =
(λ1e−λ
1
2
Then y = 0 implies Ae = −Be which implies e
= −B/A. There is onlyroots
one
has
rootsTherefore,
λ = −1, −1/2.
Therefore,
the
solution−γt/2m
ofproblem
the homogeneous
problem
isc2ysin(t).
λ
=
±i.
the
solution
of
the
homogeneous
is
y
(t)
=
c
cos(t)
+
h (t) =
h
1
y(t)
=
(A
+
Bt)e
.
solution
to conditions
this equation.
If A
A=
= y00 or
BB==0,v0then
there
are If
nov0solutions
to the
The
initial
imply
+ (γy
0,we2λ
then
= equation
γy
/2m.
−t
0 /2m).
c1 efind
+accharacteristic
e−t/2 . To
find
a solution
of the
inhomogeneous
problem,
will
2.
The
equation
for and
the
homogeneous
problem
is λ=2 +
+B
5first
=
0,look
To
of
the
inhomogeneous
problem,
wesolution
look
foris
aidentically
solution
ofzero).
the
form
y0pwhich
(t) for
=
2solution
yIn
=this
0 (unless
they
are
both
zero,
in
which
case,
the
case,
the
specific
solution
is
2
2
The
initial
imply
A==Ay+
B
= v0to
+Substituting
(γy
vfunction
0, problem
then
B =is
γy
a cos(2t)
solution
of
Bt
+solution
Ct
account
forIfathe
inhomogeneous
term
t=
.
has
roots+λconditions
−1form
±+
2i.Y
Therefore,
thesin(2t).
of
the
homogeneous
yh0 /2m.
(t)
0+and
0 /2m).
0 =
A
B=the
sin(2t)
Ct
cos(2t)
Dt
of this
form
into
1 (t)
−t If the system is critically damped, then the general−γt/2m
20.
solution
is
In
this
case, the
solution
is=a(y
a+function
ofand
this
form
into
differential
equation,
and
like
eSubstituting
(cdifferential
cspecific
Toy(t)
find
solution
of theweinhomogeneous
problem,
we−
look
for3,a
the
equation,
equating
terms,
have
4D
= equating
0, −3B
4Cterms,
=
1 cos(2t)
2 sin(2t)).
+ the
(γy
. − 3A
0 like
0 /2m)t)e
solution
of
the
form
y
(t)
=
A
cos(2t)
+
B
sin(2t).
Substituting
a
function
of
this
form
into
we
have
A
+
3B
+
4C
=
0,
B
+
6C
=
0
and
C
=
1.
The
solution
of
these
equations
is
−γt/2m
−3C = 1, and −3D =p 0. The solution
of
these
equations
= 0, B = −5/9, C = −1/3
−γt/2m
y(t)
(Awould
+ 0Bt)e
. is. A/2m)t
y(t)
=0,(y=
(γy
/2m)t)e
0+
20 + (γy
As
t
→
∞,
y
→
0.
In
order
for
y
=
we
need
y
=
0,
but
there
are
no
0
the
equation,
equating
wet .have
B−
3 and
A + 4Bof=the
0.
A =differential
−6,
C = the
1. and
Therefore,
=inhomogeneous
14terms,
− 6t +
Next,
we4A
look
a solution
and
D14,
=B
0. =
Therefore,
solution
of Ythe
problem
is = for
1 like
positive
times
t
satisfying
this
equation.
If
v
=
�
0,
the
specific
solution
is
0
The
initial
conditions
imply
A
=
y
and
B
=
v
+
(γy
/2m).
If
v
=
0,
then
B
=
γy
/2m.
The
solution
of
these
equations
is
A
=
−12/17
and
B
=
3/17.
Therefore,
the
solution
of
the
inhomogeneous
problem
of
the
form
Y
(t)
=
D
cos
t
+
E
sin
t.
Substituting
this
function
into
As t → ∞, y → 0. In order for y =0 0, 2we would0 need y00 + (γy0 /2m)t
= 0, but there are
no
0
0
5 = −9/10
1 E = −3/10. Therefore, the
−γt/2m
inhomogeneous
problem
is
In
this
case,
the
specific
solution
is
the
ODE
and
like
terms,
we
find
that
D
and
positive
times
tequating
satisfying
this
equation.
If
v
=
�
0,
the
specific
solution
is
y(t) =y(t)
c1 cos(t)
−0 /2m))t)e
sin(2t) − t cos(2t).
= (y0+
+c(v
.
2 sin(t)
0 +0(γy
9
3
solution of the inhomogeneous problem is
−γt/2m
=0 (y
. . 3
12 −γt/2m
= (y
+0(v+0 (γy
+ (γy
0 /2m)t)e
0 /2m))t)e
−ty(t)y(t)
y(t)
=
e
(c
cos(2t)
+
c
sin(2t))
−
cos(2t)
sin(2t).
2
1
2
9
9. The characteristic equation
for
the
homogeneous
problem
is +
λ2 +3ω
= 0, which has roots
−t/2
2
= c1 e−t
+ would
14 − 6tneed
+ t17
cos0 /2m)t
t −17 0=
sin0,t. but there are no
As t → ∞, y → 0.y(t)
In order
for+yc=
y−0 + (γy
2 e 0, we
λ = ±ω0 i. Therefore, the solution of the homogeneous10problem 10
is yh (t) = c1 cos(ω0 t) +
positive times t satisfying this equation. If v0 �= 0, the specific solution
is
2
y �� +
y(t) = e
(c1 cos( √15t/2) + c2 sin( √15t/2)) + 1e − 1e .
6 t 4 −t
−t/2
y(t)
=
e
The initial conditions imply(c1 cos( 15t/2) + c2 sin( 15t/2)) + e − e .
6
4
2
12. The characteristic equation for cthe+ homogeneous
problem
is
λ
−
λ
− 2 = 0, which has
c2 + 2 = 1
1
2
−t
2t
12.
The
characteristic
equation
for
the
homogeneous
problem
is
λ
−
λ
2 = c0,1 ewhich
roots λ = −1, 2. Therefore, the solution of the homogeneous problem is yh−(t)
+ c2 ehas
.
−t
3c
−
c
+
4
+
3
=
0.
1
2
roots
λ=
−1, 2. Therefore,
the solution of
the homogeneous
is yhof(t)the
= form
c1 e y+p (t)
c2 e2t
To
find
a solution
of the inhomogeneous
problem,
we look forproblem
a solution
=.
2t
−2t
To find
a solution
of the inhomogeneous
problem,
we look
for a solution
of the and
formequating
yp (t) =
Ate
+Be
. Substituting
a function of this
form into
the differential
equation,
Therefore,
c1 = −2 and c2 = 1 which implies the particular solution of the IVP is
2t
−2t
Ate
+Be
.
Substituting
a
function
of
this
form
into
the
differential
equation,
and
equating
like terms, we have A = 1/6 and B = 1/8. Therefore, the solution of the inhomogeneous
like terms,
and=B−2e
= 1/8.
of the inhomogeneous
3t
−t
problem
is we have A = 1/6 y(t)
+ eTherefore,
+ 2e2t + the
3te2tsolution
.
1
1
problem is
y(t) = c1 e−t + c2 e2t + 1te2t + 1e−2t .
−t
8 e−2t .
y(t) =for
c1 ethe
+homogeneous
c2 e2t + 6 te2t +
17. The characteristic equation
problem
is λ2 + 4 = 0, which has
6
8
13.
equation
for the homogeneous
problem
is λ2 +isλ y−h (t)
2 ==0,c1which
rootsThe
λ =characteristic
±2i. Therefore,
the solution
of the homogeneous
problem
cos(2t)has
+
2
t
−2t
The
characteristic
equation
for
the
homogeneous
problem
is
λ
+
λ
−
2
0,
which
has
roots
λ
=
1,
−2.
Therefore,
the
solution
of
the
homogeneous
problem
is
y
(t)
=
c
e
+
c
e
.
c13.
sin(2t).
To
find
a
solution
of
the
inhomogeneous
problem,
we
look
for
a
solution
of
the
h
1
2
2
t
−2t
roots
λ
=
1,
−2.
Therefore,
the
solution
of
the
homogeneous
problem
is
y
(t)
=
c
e
+
c
e
h
1differential
To
find
a solution
of the+inhomogeneous
problem,awe
look forofathis
solution
of the
yp2(t) =.
form
yp (t)
= At cos(2t)
Bt sin(2t). Substituting
function
form into
theform
To +
find
solution of the
inhomogeneous
problem,
wedifferential
look
the equating
form yp (t)
=
3 for a solution
At
B. aSubstituting
a function
of this form
into the
equation,ofand
like
equation,
and equating
like terms,
we have
yinto
p (t) = − t cos(2t). Therefore, the solution of
At
+
B.
Substituting
a
function
of
this
form
the
differential
equation,
and
equating
like
4 of these equations is A = −1 and
terms, we have −2A = 2 and A − 2B = 0. The solution
the
inhomogeneous
problem
is
terms,
we
have
−2A
=
2
and
A
−
2B
=
0.
The
solution
of these equations
is A = −1 and
B = −1/2. Therefore, the solution of the inhomogeneous problem
is
B = −1/2. Therefore, the solution of the inhomogeneous problem is
31
y(t) = y(t)
c1 cos(2t)
−
t+ c2 sin(2t)
−2t
= c1 e + c2 e − t −4 t1cos(2t).
.
4.5. NONHOMOGENEOUS EQUATIONS;
OF
2 .UNDETERMINED COEFFICIENTS313
y(t) = c1 et +METHOD
c2 e−2t − t −
2
The initial conditions imply
The initial conditions imply
c1 = 2
1
3 =0
c1 + c2 −
2c2 − =
2 −1.
4
c1 − 2c2 − 1 = 1.
Therefore, c1 = 2 and c2 = −1/8 which implies the particular solution of the IVP is
Therefore, c1 = 1 and c2 = −1/2 which implies the particular solution of the IVP is
11
31
y(t) = 2y(t)
cos(2t)
= et −
− 8 sin(2t)
e−2t − t−−4 t .cos(2t).
2
2
18.
The characteristic
characteristicequation
equationfor
forthe
thehomogeneous
homogeneousproblem
problem
is2 λ
5 = has
0, which
14. The
is λ
+2 4+=2λ
0, +
which
roots
has
roots
λ
=
−1
±
2i.
Therefore,
the
solution
of
the
homogeneous
problem
is
y
(t)
=
λ = ±2i. Therefore, the solution of the homogeneous problem is yh (t) = c1 cos(2t)+c2 sin(2t).
h
−t
eFirst,
(c1NONHOMOGENEOUS
cos(2t)
find
a solutionMETHOD
ofproblem,
the inhomogeneous
a
to find +
a csolution
of To
the
inhomogeneous
weUNDETERMINED
look for problem,
a solutionwe
oflook
the for
form
4.5.
EQUATIONS;
OF
COEFFICIENTS315
2 sin(2t)).
−t
−t 2
2
solution
(t)correspond
= Ate cos(2t)
+ Bte
Substituting
a function
of form
this
Y1 (t) = Aof+the
Bt form
+ Ct ypto
with the
term sin(2t).
t . Substituting
a function
of this
−t
into the
equation,
and equating
like terms,
we have
A =yp−1/8,
0 and
form
intodifferential
the differential
equation,
and equating
like terms,
we have
(t) = B
te =sin(2t).
t
C
=
1/4.
Next,
considering
the
inhomogeneous
term
3e
,
we
look
for
a
solution
of
the
form
Therefore, the solution of the inhomogeneous problem is
t
Y2 (t) = De . Substituting this function into the equation and equating like terms, we have
D = 3/5. Therefore, the
the inhomogeneous
is
y(t)solution
= e−t (cof
+ c2 sin(2t)) +problem
te−t sin(2t).
1 cos(2t)
The initial conditions imply
y(t) = c1 cos(2t) + c2 sin(2t) −
1 1 2 3 t
+ t + e.
8 4
5
c1 = 1
−19
c1 + 2c2 = 0.
+ c1 = 0
40
Therefore, c1 = 1 and c2 = 1/2 which implies
the particular solution of the IVP is
3
+
2c
2 = 2. �
�
5 1
−t
−t
cos(2t)
+ implies
sin(2t) the
+ te
sin(2t).solution of the IVP is
Therefore, c1 = −19/40y(t)
and=ce2 = 7/10
which
particular
2
19
7
1 1
3
y(t) = − cos(2t) +
sin(2t) − + t2 + et .
19.
40
10
8 4
5
The initial conditions imply
2
(a)
has roots
15. The
The characteristic
characteristic equation
equation for
for the
the homogeneous
homogeneousproblem
problemisisλλ2+3λ
− 2λ=+0,1which
= 0, which
has
−3t
λ
=
0,
−3.
Therefore,
the
solution
of
the
homogeneous
problem
is
y
(t)
=
c
+
c
e
h
1
2
the repeated root λ = 1. Therefore, the solution of the homogeneous problem is yh (t) =.
to find
solution
of theofinhomogeneous
problem,
we look
a solution
of the
c1 et Therefore,
+ c2 tet . First,
to afind
a solution
the inhomogeneous
problem,
wefor
look
for a solution
42 t
3 3 t 2
2
−3t
form
Y (t)
t +A
+A
+B
+D sin 3t+E
cos 3t.
3 t+A4 )+t(B
1 t+B
of the
form
Y1 =
(t)t(A
= 0At
e +1 tBt
e 2 tto+A
correspond
with0 tthe
term
te2t )e
. Substituting
a function
3 t
4.7
Variation of Parameters
4.7. VARIATION OF PARAMETERS
341
1. Let X(t) be the fundamental matrix,
4.
�
�
−t
0
−e
(a)
X(t)
= −t� �
.
�
��
�
e
te−t
x11 (t) x12 (t)
u1 (t)
x11 (t)u1 (t) + x12 (t)u2 (t)
Xu =
=
.
x21 (t) x22 (t)
u2 (t)
Then
21 (t)u1 (t) + x22 (t)u2 (t)
� t xt �
te e
X−1 (t) =
.
t
Therefore,
−e
0
� �
�
�
�
�
x
u
+
x
u
+
x
u
+
x
u
�
1
11
2
12
11
1
12
2
Therefore,
(Xu) =
.
x�21 u1 +�x21 u�1 + �
x�22�u2 +�x22 u�2
t
t
te e
−1
X−1 (t)g(t) =
t
Now
−e
t� �
� �
� 0
�
�
�
x
x
u�1
�
11
120
X� =
u
=
.
x�21 =x�22 t .
u�2
e
Therefore,
�
Therefore,
� � �
0
X (t)g(t) dt =
dt
et
� �
0
= t .
e
−1
�
xp (t) = X(t) X−1 (t)g(t) dt
�
�� �
−t
0
−e
0
4.7. VARIATION OF PARAMETERS
343
= −t
e
te−t
et
� �
−1
Therefore, c1 − c2 = −1 and c2 − 1 =
this system of equations, we have c1 = 1
= 1. Solving
.
and c2 = 2. Therefore, the solution of the tIVP is
� −t
�
Therefore,
�−t + 2t
e − 2e�t + te
−1
x(t) =
.
xp (t) 2e
= t − t − .1
t
8. Let
The X(t)
general
solution
of the equation
5.
be the
fundamental
matrix, in problem 4 is
� ��
� −t �� � �
cos t −esin
t
0
= + c2
. −1 .
x(t) = cX(t)
1
−t
−t t +
−
sin
t
cos
e
te
t
�
�t
Then
�
�
The initial condition x(0) = 1 0 −1 implies cos t − sin t
X (t) =
.
� �
�sin t� cos
�t � � �
0
−1
−1
1
x(0) = c1
+ c2
+
=
.
Therefore,
1 �
0
0�
�0�
cos t − sin t
cos t
X−1c1(t)g(t)
Therefore, −c2 − 1 = 1 and
= 0. =
Solving
this
system
of
equations,
we have c1 = 0 and
sin t cos t
− sin
t
� is
c2 = −2. Therefore, the solution of the �
IVP
1
= � . −t � � �
0 −e
−1
x(t) = −2
+
.
te−t
t
9. The general solution of the equation in problem 5 is
�
�
�
� �
�
cos t
sin t
t cos t
x(t) = c
+c
+
.
�
−t
te−t (3e
3
t
t)
t
Therefore, the particular solution
is
Y
(t)
u1 (t) = − = 2e − e =
dte=. − t2
W (t)
2
11. The solution of the homogeneous equation is yh (t) = c1 e2t + c2 e−t . The functions
� −t −t
y1 (t) = e2t and y2 (t) = e−t form a fundamental
e (3e set
) of solutions. The Wronskian of these
dt = 3t.
t u2 (t) =
functions is W (y1 , y2 ) = −3e . Using the method
W (t) of variation of parameters, a particular
solution is given by Y (t) = u1 (t)y1 (t) + u2 (t)y
2 (t) where
Therefore, a particular solution is Y (t) = − 32 t2 e−t + 3t2 e−t = 32 t2 e−t .
� −t −t
2 −3t
e (2e )is yh (t) =
13. The solution of the homogeneous
equation
c1 et/2 + c2 tet/2 . The functions
u
(t)
=
−
dt
=
−
e
1
W (t) set of solutions.
9
y1 (t) = et/2 and y2 (t) = tet/2 form a fundamental
The Wronskian of these
� 2t −t
t
functions is W (y1 , y2 ) = e . Using the
method
of parameters, a particular
e (2e
) of variation
2t
(t) =(t) + u (t)y (t)
dt where
=− .
solution is given by Y (t) = uu12(t)y
1
2 (t) 2
W
3
�
2 (4et/2 )2 −t
tet/2
Therefore, a particular solution
(t) = − e−t − dt
te= .−2t
As2 −2e−t /9 is a solution of the
u1 (t)is =Y −
9 (t) 3
W
2
� t/2 t/2
homogeneous equation, we can omit it eand
that yp (t) = − te−t is a particular
(4econclude
)
3
u2 (t) =
dt = 4t.
solution of the inhomogeneous problem. W (t)
12. The solution of the homogeneous equation
is yh (t)
= c e−t + c te−t . The functions
2 t/2
Therefore,
a particular solution
is Y (t) = −2t2 et/2 + 4t
e = 12t2 et/2 . 2
−t
−t
y1 (t) = e and y2 (t) = te form a fundamental set of solutions. The Wronskian of these
−2t
14. The solution
homogeneous
is yhof
(t)variation
= c1 cos(t)
c2 sin(t). The
functions
functions
is W (y1of
, ythe
. Using equation
the method
of +parameters,
a particular
2) = e
y1 (t) = cos(t)
= usin(t)
form a fundamental set of solutions. The Wronskian of
solution
is givenand
by yY2 (t)
(t) =
1 (t)y1 (t) + u2 (t)y2 (t) where
these functions is W (y1 , y2 ) = 1. Using�the method of variation of parameters, a particular
−t
−t
346
CHAPTER
4.(3e
SECOND
ORDER
LINEAR EQUATIONS
)
3
solution is given by Y (t) = u1 (t)y
(t)y
1 (t) + u2te
2 (t) where
u1 (t) = −
dt = − t2
W (t)
2
�
sin(t)
tan(t)
Wronskian of theseu functions
is
W
(y
,
y
)
=
1/2.
Writing
the
ODE+in
standard form, we see
1
2
� dt
= sin(t)
− ln(sec(t)
tan(t))
1 (t) = −
−t
eof−t (3e
)
(t)
that g(t) = sec(t/2)/2. Using the W
method
variation
of parameters, a particular solution
u2 (t) =
dt = 3t.
4.7.
VARIATION
PARAMETERS
345
is
given
by Y (t) = uOF
W (t)
1 (t)y
1 (t) + u2 (t)y2 (t) where
�
Therefore, a particular solution is cos(t/2)(sec(t/2))
Y�(t) = − 32 t2 e−t + 3t2 e−t = 32 t2 e−t .
cos(t) tan(t) dt = 2 ln[cos(t/2)]
u1 (t) =u −
t/2
dt = − cos(t).t/2
2 (t) =
2W
13. The solution of the homogeneous
equation
W(t)
(t) is yh (t) = c1 e + c2 te . The functions
t/2
t/2
y1 (t) = e and y2 (t) = te form a� fundamental set of solutions. The Wronskian of these
sin(t/2)(sec(t/2))
Therefore,
is=Y (t)the
= (sin(t)
− ln(sec(t)
cos(t) − cos(t)
sin(t) =
functions isa particular
W (y1 , y2 ) solution
= etu. 2 (t)
Using
method
of variation
a particular
dt +
= tan(t)))
t. of parameters,
2W
(t)
−
cos(t)
ln(sec(t)
+
tan(t)).
Therefore,
the
general
solution
is
y(t)
=
c
cos(t)
+
c
sin(t)
−
1
2
solution is given by Y (t) = u1 (t)y1 (t) + u2 (t)y2 (t) where
cos(t) ln(sec(t)
+ tan(t)).
Therefore,
a particular
solution is Y (t)�= 2 cos(t/2) ln[cos(t/2)] + t sin(t/2). Therefore, the
t/2
tet/2 is
(4ey+
general
solution
is
y(t)
=
c1 cos(t/2)
c2 sin(t/2)
2) cos(t/2)
ln[cos(t/2)]
+ t sin(t/2).
2
15. The solution of the homogeneous
equation
==
c1 −2t
cos(3t)+c
The functions
h (t)
2 sin(3t).
u1 (t) = +
−
dt
W
(t)
t
t
y
(t)
=
cos(3t)
and
y
(t)
=
sin(3t)
form
a
fundamental
set
of
solutions.
The
Wronskian
1 The solution of the
2 homogeneous equation is yh (t) = c1 e +c2 te . The functions
19.
y1 (t) = eoft
� t/2 t/2
thesey2functions
W (ya1 ,fundamental
y2 ) = 3. Using
method
variation of parameters, a particular
ethe(4e
) of The
and
(t) = tet isform
u2 (t) = set of solutions.
dt = 4t. Wronskian of these functions is
2t
solution
by Y (t)
u1 (t)y1 (t)
u2W
(t)y
where
2 (t)
(t)
W
(y1 , y2 )is=given
e . Using
the=method
of +
variation
of parameters,
a particular solution is given
by Y (t) = u1 (t)y1 (t) + u2 (t)y�2 (t) where
2 2 t/2
2 t/2
sin(3t)(9
sec−2t
(3t))
Therefore, a particular
is�Y (t) =
e dt+=4t−
ecsc(3t)
= 2t2 et/2 .
u1 (t) solution
=−
1
Wte
(t)t (et )
14. The solution of theuhomogeneous
equation 2is dt
yh (t)
= cln(1
=−
+ t2+) c2 sin(t). The functions
1 cos(t)
� =−
1 (t)
2
W
(t)(1
+
t
)
2
cos(3t)(9
sec
(3t))
y1 (t) = cos(t) andu2y(t)
set of solutions.
2 (t)
� formt a t fundamental
== sin(t)
dt = ln(sec(3t)
+ tan(3t)).The Wronskian of
e
(e
)
W
(t)
these functions is W (y1u, y(t)
of variation of parameters, a particular
2) =
= 1. Using the method
dt = arctan(t).
2
t2 )2 (t) where
solution is given by Y (t) = u1 (t)y1W
(t)(t)(1
+ u2+
(t)y
Therefore, a particular solution is Y (t) = −1 + sin(3t) ln(sec(3t) + tan(3t)). Therefore, the
�
1 t sin(3t) ln(sec(3t)
general solution
is y(t) =solution
c1 cos(3t)
c2 sin(3t)
+ tan(3t)) Therefore,
− 1.
sin(t)
Therefore,
a particular
is +
Ytan(t)
(t)
= dt−=+
esin(t)
ln(1 −
+ ln(sec(t)
t2 ) + tet +
arctan(t).
the
u1 (t) = −
tan(t))
2
−2t
−2t
W
(t)
1 t
16. Thesolution
solutionis of
the= homogeneous
c1 e + c2 te . The functions
general
y(t)
c1−2t
et + c2 tet − equation
e ln(1 +ist2 )yh+(t)tet=arctan(t).
2
−2t
y1 (t) = e
and y2 (t) = te
form a fundamental set of solutions.
The Wronskian of these
20.
The solution
of
the
homogeneous
equation
isofyhvariation
(t) = c1 eof3t parameters,
+ c2 e2t . The
functions
−4t
functions
is
W
(y
,
y
)
=
e
.
Using
the
method
a particular
1 2
3t
2t
ysolution
(t)
=
e
and
y
(t)
=
e
form
a
fundamental
set
of
solutions.
The
Wronskian
of these
1
2 Y (t) = u (t)y (t) + u (t)y (t) where
is given by
1
1
2
2
5t
functions is W (y1 , y2 ) = −e . Using the method of variation of parameters, a particular
�
solution is given by Y (t) = u1 (t)y1 (t) + te
u2−2t
(t)y
(t)e−2t
where
(t2−2
)
u1 (t) = −�
�dt = − ln(t)
W (t)
e2s (g(s))
u1 (t) = − �
ds = e−3s g(s)ds
−2t
−2
W
e (s)
(t e−2t )
1
u2 (t)
=
dt = − .
� 3s
�
t
e (g(s))W (t)
1
1
Therefore, a particular solution is Y (t) = − (1 + t)e2t + te2t = (t − 1)e2t .
2
2
t
24. By direct substitution, it can be verified that y1 (t) = e and y2 (t) = t are solutions
of the homogeneous equations. The Wronskian of these functions is W (y1 , y2 ) = (1 − t)et .
Rewriting the equation in standard form, we have
t �
1
y �� −
y −
y = 2(1 − t)e−t .
1
−
t
1
−
t
CHAPTER 4. SECOND ORDER LINEAR EQUATIONS
348
Therefore, g(t) = 2(1 − t)e−t . Using the method of variation of parameters, a particular
solution is given by Y (t) = u1 (t)y1 (t) + u2 (t)y2 (t) where
�
t(2(1 − t)e−t )
1
u1 (t) = −
dt = te−2t + e−2t
W (t)
2
�
2(1 − t)
u2 (t) =
dt = −2e−t .
W (t)
�
�
Therefore, a particular solution is Y (t) = te−t + 12 e−t − 2te−t = 12 − t e−t .
25. By direct substitution, it can be verified that y1 (x) = x−1/2 sin x and y2 (x) = x−1/2 cos x
are solutions of the homogeneous equations. The Wronskian of these functions is W (y1 , y2 ) =
−1/x. Rewriting the equation in standard form, we have
1
x2 − 0.25
sin x
y �� + y � +
y = 3 1/2 .
2
x
x
x
Therefore, g(x) = 3 sin(x)x−1/2 . Using the method of variation of parameters, a particular
solution is given by Y (x) = u1 (x)y1 (x) + u2 (x)y2 (x) where
�
x−1/2 cos(x)(3 sin(x)x−1/2 )
3
u1 (x) = −
dx = − cos2 (x)
W (x)
2
� −1/2
−1/2
x
sin(x)(3 sin(x)x
)
3
3x
u2 (x) =
dx = cos(x) sin(x) − .
W (x)
2
2
Therefore, a particular solution is
3
Y (x) = − cos2 (x) sin(x)x−1/2 +
2
3
= − x1/2 cos(x).
2
�
3
3x
cos(x) sin(x) −
2
2
�
cos(x)x−1/2
26. By direct substitution, it can be verified that y1 (x) = ex and y2 (x) = x are solutions
of the homogeneous equations. The Wronskian of these functions is W (y1 , y2 ) = (1 − x)ex .
Rewriting the equation in standard form, we have
y �� +
x �
1
g(x)
y −
y=
.
1−x
1−x
1−x
Therefore, the inhomogeneous term is g(x)/(1 − x). Using the method of variation of parameters, a particular solution is given by Y (x) = u1 (x)y1 (x) + u2 (x)y2 (x) where
�
sg(s)
u1 (x) = −
ds
(1 − s)W (s)
�
es g(s)
u2 (x) =
ds.
(1 − s)W (s)