Considering
real and
imaginary parts of this function, we arrive at the general
amplitude
a shorter
period.
defined,
at t = and
0. the
solution
4t
16. W (e2t , g(t)) = e2t g � (t) − 2e2ty(t)
g(t) =
= c3ecos(3t/2)
. Dividing
both
sides by e2t , we find that g must
+
c
sin(3t/2).
1
2
satisfy
ODE g � − 2g
3e2t . Therefore,
g(t) =Linear
3te2t + ce2tHomogeneous
.
4.2 theTheory
of=Second
Order
Equa2 t 0) is a stable center.
(b) W
We(t,see
that=the
critical
point
17.
g(t))
tg � (t)
− g(t)
= t(0,
e . Dividing both sides of the equation by t, we have
tionsOF SECOND
4.2.
THEORY
LINEAR HOMOGENEOUS EQUATIONS
259
g � − g/t
= tet . Therefore,
g(t) ORDER
= tet + ct.
(c)
From
our
solution
in
part
(a),
we
see
that
260
CHAPTER 4. SECOND ORDER LINEAR EQUATIONS
18. W (f, g) = f g � − f � g. Also, W (u, v) = W (2f − g, f + 2g). Upon evaluation, W (u, v) =
1.
Write
the
IVP
as
�
�
3c1 3
3c2
6.
�
5f gWrite
− 5fthe
g =IVP
5Was
(f, g).
y
+
cos(3t/2).
�
2(t) = − 1 �� sin(3t/2)
� y = 1. equation,
y2 differential
+
15. No. Substituting
y� = sin(t )y ��into
the
2
+
y
+
(tan
x)y
=
0. g � +we4fhave
�
�
t (u, v) = −4f
19. W (f, g) = f g − f g = t cos t − sin
g. Therefore, W (u, v) =
x t−and
2 W
2
2
2
2
2
Therefore,
−4t
cos
t +function
4 sin t. −4t =sin(t
Since
the
3/t
continuous
for2tfor
all
t >x�
0such
andsin(t
t�0 =)q(t)
1>
IVP
guaranteed
+
2continuous
cos(t
cos(t
)p(t)
= the
0.� +
� is)are
�
� )+
Since
the
coefficientp(t)
functions
all
that
x
�= 0,
2,
nπ
π/2isand
x0 = 3,
y
cos(3t/2)
sin(3t/2)
286have
4. for
SECOND
ORDER
LINEAR
to
all
tCHAPTER
20.
For ais
y1unique
= cossolution
2t , yto
=for−4
y1��all
++x4y
Similarly,
forEQUATIONS
y2 = sin 2t,
x1�� =have
=>2t.
c10. Therefore,
c2such
.< 3π/2.
1 =
the
IVP
guaranteed
unique
solution
that
2
<
x
3
3 0.
� acos
y must have
− 2 sin(3t/2)
cos(3t/2)
For
the
to be valid,
p(t)
= −1/t,
is they
not continuous,
or even
y2.2�� =
−4equation
sin
Therefore,
y2�� +we
4y2 = 0. Since
W (cos
2t, sin which
2t)2 = 2,
form a fundamental
Write
the2t.
IVP
as
4.3.
LINEAR
HOMOGENEOUS
EQUATIONS
WITH
CONSTANT
COEFFICIENTS
263
7.
� ∞.
defined,
at t =continues
0.
4 −3t/2 ��sin t
The solution
to oscillate
as3tt →
set.
��
� 2t
�
7
e
e
y −
y +
y= �
. t/2 2t
�4t
2t � W (e2t ,2te−3t/2t)−
=sides
1by
−2y
t� −
1−
16.
W
(e2t
, g(t))
==
3e
.y2t
Dividing
both
thatyrepeated
g mustt
t = �e g (t)
�� − 2e
t g(t)
��2 t+
t eby .te , there
31� −3t/2
�
37.
The
characteristic
equation
is
given
λ
4λ
+
4
=
0.
Therefore,
one
21.
For
y
=
e
,
y
=
y
=
e
.
Therefore,
−
+
y
=
e
= find
0.isFor
2e
−
e
2− 2e + etwe
1
1
2 = te ,
3
1
1
Therefore,
2 1 2t
�1
2t
2t
�
satisfy
the
ODE
g
−
2g
=
3e
.
Therefore,
g(t)
=
3te
+
ce
.
�
t
��
t
��
�
t
t
t
x
coefficient
are
continuous
for
1 (2
and
t0 =−−2
1, the
is
root,
λthe
=t)e
−2.and
Therefore,
general
solution
ySince
+
y2 functions
= (2 the
+ t)e
. Therefore,
y2isδu
−given
2yall +tby
y<2 =
+ t)e
2(1 <
+ t)e
+ teIVP
= 0.
2 = (1
−δx2 /2
−δx2 /2
2 2 /2 2
t
t
2t
�
2
t
y
(x)
=
c
e
e
du
+
c
e
.
8.
2g(t) solution
2 longest
x2
guaranteed
have
ae unique
all
1. −2t
The
interval
�−2tt <
Further,
(eto, te
. Therefore,
solutions
form
a �fundamental
set. of
17.
W (t,W
g(t))
= )tg=(t)
−
= t1 ethe
. for
Dividing
both
sides
of the equation
by existence
t, we haveis
1 +tc2 tesin t.�
�
t
ty(t) = c10e�� cos
(−∞,
1).
g22.−For
g/t = te
. Therefore,
g(t)
= 0.
tet,Therefore,
+sinct.t) = x2 y �� − x(x �+=2)y
�� (cos
�
= 0,
x,
y1� =
1 and
yW
�− sin1t cos t� 1.1 + (x + 2)y1 = −x(x + 2) +
1 =solution
Setting cy21 =
we
arrive
at
the
x
x
Therefore,
3.
Write
–3(u,
–2 =and
–1 (2f
22g).
3Upon evaluation,
(x
+
2)x
For
, y2� =W
(1
+ x)e
y02�� =
(21f++
x)e
. Therefore,
x2 y2�� − x(x
+
18.
W
(f,=the
g)0.=IVP
f g �yas
−
f � xe
g. xAlso,
v)
W
− g,
W+
(u,2)y
v)2� =
2 =
�
−2t
x1x −2t
�
3t
4
2
x
2
x
x
x
2 x
�
�
y
(t)
=
−2c
e
+
c
(1
−
2t)e
.
��
�
–1
1
2
(x
+
2)y
=
x
(2
+
x)e
−
x(x
+
2)(1
+
x)e
+
(x
+
2)xe
=
0.
Further,
W
(x,
xe
)
=
x
e
.
5f
g
−
5f
g
=
5W
(f,
g).
2
2
9.
2
y + y (x) =�ye−δx
+ /2
y=
δu
/2
� .
e
−2tdu.
2 − 4)� e−2tt(t − 4)
t(t
t(t
−
4)
�
te
Therefore,
form
a fundamental
set.
�
–2 0
Now
the
initial
conditions,
need
�� = 4f
19.
Wusing
(f, g)the
= fsolutions
g � − f �W
g=
t cos
−we
sin
t ��and −2t
W
(u, v) = −4f
g.. Therefore, W (u, v) =
(e−2t
, tet −2t
)=
e−4t
−2tg� +
−2e
(1
−
2t)e
� functions ��
�� t �= 0,� 4 and t0 = 3, the IVP
Since
the
coefficient
are
continuous
for
all
t
such
that
23. cos
Fort y+1 4=sin
x,t.y1 = 1 and y1 = 0. Therefore,
− 2x cot x)y1 − xy1 + y1 = −x + x = 0.
−4t
c1 e2 −–3c2 e(12 =
�
��
is guaranteed
to
have
a
unique
solution
for
all
t
such
that 0 <(1t −
< 4.
For
y
=
sin
x,
y
=
cos
x
and
y
=
−
sin
x.
Therefore,
x cot
x)y2��for
− yxy2�=+sin
y2 2t,
=
2 y
2 , y �� = −4 cos 2t.
2
2 Equations
2 �� + 4y1 =with
4.3For
Linear
Constant
Co20.
Therefore,
y
0.
Similarly,
10.
1 = cos 2t Homogeneous
2
�
�
1
1= 1.
−2c
e
+
3c
e
1
2
−
cot
x)coefficient
sin
x − x cos
x 0.= Since
0.�isx Further,
W� sin
(x,2t)
sin
x)
xand
cosform
x
xthewhich
4.2�� =
Since
the2t.
function
3=
ln(|t|)
all
t �=
t0 −
=asin
2,
IVP is
is
xex 2t,
y−(1
−4x sin
Therefore,
y2��xW
++(x,
4ysin
W (cos
2,=0they
fundamental
x
2 x=
2 xe
� continuous
� for
)
=
=
x
e
.
efficients
−2
−2
x
nonzero
for
0
<
x
<
π.
Therefore,
the
functions
form
a
fundamental
set
of
solutions.
�
�
guaranteed
to
have
a
unique
solution
for
all
t
>
0.
The
solution
of
this
system
of
equations
is
c
=
7e
and
c
=
5e
.
Therefore,
the
specific
1 1(1 + x)e
2
set.
24.
−2(1+t)
−2(1+t)
solution
is
y(t)
=
7e
+
5te
.
24.
t
�
��
t
��
�
t
t
t
t
5. Write
IVP
as
21.
For ythe
1 = e , y1 = y1 = e . Therefore, y1 − 2y
2 1 + y1 = e − 2e �+ e = 0. For y2 = te ,
1.
11.
�
(a)
The
characteristic
equation
is
given
by
9λ
+
9λ
−
4
=
0.
Therefore,
the
roots
are
x
ln
|x|
�
t
��
t
�y �� − 2y � + ��
t= �(2 + t)e�t − 2(1
�= e−te.t sin
y(a)
(1 + t)e
and
+
t)et .yy��Therefore,
y2e0.
t)e−
+2ete−tt =
−t y�2 = (2 −t
��+
−t + −t
cos t
+
2 =For
2t
2 yy=
t and
2t e
,t(e
y1t sin
= 2t−e
Therefore,
e−e
+
= 0.
0.
� 2tx − 3 ysolution
1 = te W
λ =y−4/3,
1/3.
Therefore,
the
is
1 general
1 − y1 − 2y1 �=
t,
e
cos
t)
=
=
.
x
−
3
2
t
�
�
Further,
W
(e
,
te
)
=
e
.
Therefore,
the
solutions
form
a
fundamental
set.
2t
�
2t
��
2t
��
�
2t
2t
2t
(a) The
equation
is
given
by
λ
+
2λ
−
3
=
0.
Therefore,
the
two
distinct
roots
e
(sin
t
+
cos
t)
e
(cos
t
−
sin
t)
For ycharacteristic
=
e
,
y
=
2e
and
y
=
4e
.
Therefore,
y
−
y
−
2y
=
4e
−
2e
−
2e
=
0.
2
2
2
2
2
2
−4t/3
t/3bythat
Since
the
coefficient
functions
are
continuous
for
x
such
x
=
�
0,
3
and
x
=
1,
the
2t
�−tTherefore,
��t the
2 ��all
�
are
λ
=
−3,
1.
general
solution
is
given
0
y(t) =the
c1 exfunctions
+x(x
c2 eform
Further,
) = 3e
. Therefore,
a 1fundamental
of solutions.
22. For
y1 = W
x, (e
y1 =, e1 and
y1 =
0. Therefore,
y1 −
+.2)y
+ (x + 2)y1 set
= −x(x
+ 2) +
1.5
12.
IVP
is
guaranteed
to
have
for
that
0
<
x
<
3.
x a � unique solution
x �
�� all x such x
2 ��
�
(x + 2)x = 0. For y2 = xe , y2 = (1 + x)e and
x)e . Therefore, x y2 − x(x + 2)y2� +
2 (2 +
2t =
�= c ycos
−3t 1 + cos 2θ �
θ
2 linear and yy(t)
e
+
c
e
2equation
x
x
2 x
(b)
Since
the
is
,
y
,
y
are
all
linear
solutions,
1
2
�
� = 0. Wof(x,
4 5+ (x + 2)xex. =combinations
(cos
θ,
1 ++
cos
2θ)
=
(x
2)y2 =our
x (2
+Wx)e
x(x
2)(1
+3see
x)e
0. Further,
xex ) = xthey
e .
(b)+ From
solution
in−part
(a),
y we
�that
�
1
−2
sin
θ
cos
θ
−2
sin
2θ
are
also
solutions
of
the
differential
equation.
Therefore, the solutions form a fundamental set.
4ctherefore,
cunstable.
1 −4t/3
2 t/3 �� 2
(b) The critical
(0, 0) is ��a saddle
point,
�
2 point
t+
(t)W
=(y−
eWx)y
. 14 ,2t
23.
For
y1� =t y,11��Wand
Therefore,
(1
x cot
−
+ yTherefore,
x, 1=
0.5
13.
If (y
y1y1 ,1=y=
, then
=(y2.
t12,yy1��4 )−e=
2y6e
t2 (2)
=
If
y2 −x
= t+−1{y
then
(c) W
=
−6e
)==y0.0,
and
(y−
y5 )xy
=1� 0.
, y30.}
1 =
1−=
3 )t x,
2 ,yy13Therefore,
3
3
�
��
��
�
�
−3
2 ��
2
−3
−1
For
y22t ={y. sin
y2 = fundamental
− solutions,
sin
(13 }−the
x cot
=
y(c)
tcos
y2x−and
2y2 y=2sets
t=(2t
) −x.2t Therefore,
= 0.{ySince
equation
isxynot.
linear,
and
yx,
of
but
and
{y4x)y
, y52}−do
2 + y2the
2 =For
1 ,Therefore,
4 } form
2, y
y above,
we see−1that
2
Therefore,
function
y3 =
t
+
c
t
will
also
be
a
solution.
−(1
−
x cot
x)csin
x
−
x
cos
x
+
sin
x
=
0.
Further,
W
(x,
sin
x)
=
x
cos
x
−
sin
x
which
is
1
2
−3t
–1
2 . � 3 �
y � =�c1 e11t −�3c
2e
t
1
nonzero
for 0 < x < π. Therefore,
functions form
−4t/3
t/3
25. We compute:
14.
x = cthe
+ ca2 efundamental
. set of solutions.
1e
−4/3
1/3
solution as the two parameter family
24.
TheTherefore,
solution ywe
→ can
0 asrewrite
t → ∞.our
�
K[c
x
+
c
x
]
=
(c
x
+
c
x
)
−
P(t)((c
c2 x2 ) = c1 (x��1 − P(t)x1 ) + c2 (x�2 − P(t)x2 ) =
1
2 2the equation
1 1
2by
(a) 1Dividing
have
� 2 y,�we �
� 1 x1�+
2
t +3 =
38. For
Theycharacteristic
equation
λ c+1 e5λ
0. yTherefore,
the
−t
−t x1 is ��given
�
−t two
+ c2ye��−3t
(a)
, y1� = −e
and =
y1 =ye−tby
. Therefore,
−
+ e−tdistinct
− 2e−t roots
= 0.
1 = e √
1−3t
1 − 2y1 = e
=
�
t
are For
λ =y(−5
±
Therefore,
solution
by
2t 13)/2.
�
2t
�� the
2tgeneral
��e is given
�
2t
2t
2t
x
y
c
e
−
3c
1
2
1
2
y2 −�2y2 = 4e − 2e − 2e = 0.
2 = e , y2 = 2e and y2 = 4e y.�� Therefore,
2 −�
+ �
(y ��
)2 =y0.
√
√
y
Further, W (e−t , e2t ) = 3et . Therefore,
the
functions
form
set of solutions.
1
1a fundamental
(−5− 13)t/2
y(t) = c1 e(−5+= 13)t/2
.e−3t .
c1 + ec2t e+ c2
1
−3
This
equation
will
not
have
the
form
required.
(b)
Since
the
equation
is
linear
and
y
,
y
,
y
are
all
linear
combinations of solutions, they
3
4
5
Therefore,
√
√
are also solutions of(−5
the+differential
equation.
√
√
13)c
(−5 − 13)c
�� 2 ��
�
� 2
��
13)t/2
(b) Consider T [y
+=
y2 ]. We see that1 eT(−5+
[y1 +13)t/2
y2 ] =
y �1(t)
+ (y1 + y2 )(y1 +ey(−5−
2 ) + (y1 +. y2 ) �= y1 y1 +
2
2
� 2
��
� 2
t
t
(y1(y
) 1+
(y2 ), W=(yT2[y
[y2W
]. (y1 , y4 )3 = 6e and W (y4 , y5 ) = 0. Therefore, {y1 , y3 }
(c) W
, y3y)2 y=2 +
−6e
, y13])+=T0,
Nowand
using
the
initial
conditions,
we
need
{y1 , y4 } form fundamental sets of solutions,
but {y2 , y3 } and {y4 , y5 } do not.
2
(c) If y1 = 1, then y1� = 0 = y1�� . Therefore,x2 T [y1 ] = 0. If y2 = t1/2 , then y2� = t−1/2 /2 and
c1 + c2 = 1
1
y2�� =compute:
−t−3/2 /4. Therefore, T [y2 ]√= −t1/2 (t−3/2 /4)+(t
√ −1/2 /2)2 = −t−1 /4+t−1 /4 = 0. But
25. We
1/2 +
−3/21
2
−3/2
13)c
(−5 −−1/213)c
T [c1 y1 + c2 y2 ] = (c1 + c2 t(−5
)(−c
/4)
/2)
/4 �= 0 in general.
–3 2 t–2
–1
22 =
+ +0 (c2 t 1 x1
=3−c
0. 1 c2 t
�
�
2
2
K[c1 x1 + c2 x2 ] = (c1 x1 + c2 x2 ) − P(t)((c1 x1 +–1c2 x2 ) = c1 (x1 − P(t)x1 ) + c2 (x�2 − P(t)x2 ) =
√
√
The solution of this system of equations is c–21 = (1 + 5/ 13)/2 and c2 = (1 − 5/ 13)/2.
Therefore, the specific solution is
–3
√ = 0.
y√
+ (v0 + (γy0 /2m))t
(−5 − 13)c2
(−5 +0 13)c
1
+
= 0.
2
The solution of this equation is 2
1
t
√ 3
√
–2
–1
y01 (1 + 25/ 13)/2
The solution of this system of equations
is
c
=
and
c
=
(1
−
5/
13)/2.
1
2
0
t=−
.
v
+
(γy
/2m)
Therefore, the specific solution is
0
–1 0
√
√
–2√
In order for there to be a positive
t satisfying
this
equation,
we√13)t/2
need v0 + (γy0 /2m) < 0.
1 + 5/ time
13 (−5+
1 − 5/
13 (−5−
13)t/2
y(t)
=
e
+
e
.
4.3.
LINEAR
HOMOGENEOUS
EQUATIONS
WITH
CONSTANT
COEFFICIENTS
287
y –3
That
is, we need
v0 < −γy0 /2m.
2
2
21.
–4
–5
1
(a) Let y = Re−γt/2m cos(νt−δ).
Then y attains a maximum when νtk −δ = 2kπ. Therefore,
–6
1
the time between successive
maxima
is Td = tk+1 − tk = (2(k + 1)π − 2kπ) = 2π/nu.
0.8
ν
(b)
The solution y → −∞ as 0.6
t → ∞.
y
k /2m two with roots λ = 2, −3. We take
44. We need to find a characteristic equation
ofe−γt
degree
y(tk )
=
2
0.4
p(λ) = (λ − 2)(λ + 3) = λ + λ − 6. y(t
Therefore,
thek+1
differential
equation is y �� + y � − 6y = 0.
/2m
e−γt
k+1 )
)/2m with one repeated root λ = −2.
k+1 −γtktwo
45. We need to find a characteristic
equation
= of
e(γtdegree
0.2
We take p(λ) = (λ+2)2 = λ20+4λ+4.
Therefore,
differential
equation is y �� +4y � +4y = 0.
γTthe
0.5
1
1.5
2d /2m .2.5
3
=
e
296
CHAPTERt 4. SECOND ORDER LINEAR EQUATIONS
46. We need to find a characteristic equation of degree two with roots λ = −3 ± 4i. We take
The
y(a)
→
0 as(b),
t−
→(−3
∞.
p(λ)
=solution
(λ parts
− (−3
+ 4i))(λ
− 4i)) ==ln[y(t
λ2 + k6λ
+ 25.
Therefore,
differential=equation
(c)
From
and
we
have∆
)/y(t
= γTd /2mthe
= 2πγ/2mν
πγ/mν.
k+1 )]
2.0
��
�
is y + 6y + 25y = 0.
1
1.5
2
2
39.
characteristic
equation
+2 λ + 1.25
which The
has roots
λm= =
− 1/2
± i. Therefore,
22.The
The
spring constant
is k1.0is=λ 16/(1/4)
= =640,lb/ft.
mass √
/ft. The
2 thislb-s
47. The characteristic equation is
λ
−
(2α
−
1)λ
+
α(α
−
1)
=
0.
Solving
equation,
we
−t/2
−t/2
damping
coefficient
γ ==
2 lb-sec/ft.
The+quasifrequency
isThe
ν =derivative
2 31 rad/sec.
the
general
solution
isis y(t)
c0.51−
e 1.�Therefore,
cos(t)
c2 ethe general
sin(t).
of y isTherefore,
y � (t) =
�
�
�
see
that
the
roots
are
λ
=
α,
α
solution
is
1
c1 e−t/2 − 12 cos(t) − sin(t) + c2 e−t/2 −2π
sin(t) +2 cos(t)
the initial conditions, we
4
2π 3 . Using
21
!0.5
√ c1αt=
√and
≈c21.1285.
(α−1)t
have c1 = 3 and −c1 /2 + c2 = ∆
1. =
Therefore,
=
3
=
5/2,
and we conclude that the
y(t) = c131
e + c231
e
.
(1/2)2
!1.0
solution is
!1.5
5
2 e−t/2
In
order
for
the
solution
to
tend
to−t/2
we
need
α,
αsin(t).
−
1 < 0.that
Therefore,
the solutions
y(t)
=m
3e
cos(t)
23. For problem 17, the mass is
=zero,
1/4
lb-s+
/ft.
We
suppose
∆ = 3 and
Td = 0.3
2
will
all
tend
to
zero
as
long
as
α
<
0.
All
solutions
will
become
unbounded
if
α
> 0 and
seconds.
Fromy problem
weaknow
thatamplitude
∆ = γTd /2m.
Therefore,
we have
The
solution
oscillates21,
with
growing
as x →
0.
α − 1 > 0. This will occur exactly when α > 1.
0.3γ
3
=(3 − α)λ=−0.6γ.
48. The characteristic equation is λ23 +
2(α − 1) = 0. Solving this equation, we
2(1/4)
2.5
see
that Mechanical
the roots are λ = α and
− 1, −2.Electrical
Therefore, the Vibrations
general solution is
4.4
2
Therefore, the damping coefficient
is γ = 5 lb-sec/ft.
1.5
y(t) = c1ise(α−1)t√+ c2 e−2t .
24. The general solution of 1this equation
1. R cos δ = 3 and R sin δ = 4 implies R = 25 = 5 and δ = arctan(4/3) ≈ 0.9273.
�
�
Therefore,
0.5
In
order for all solutions to
tend
zero,2k/3t)
we need
− 1 <2k/3t).
0. Therefore, the solutions will
y(t)
= Atocos(
+ Bαsin(
−2t
all tend to zero as long as 0α < 1. 2yDue
the
term
c
e
,
we
can never guarantee that all
=�5to
cos(2t
−
0.9273).
2
4
6
8
10
The
period
of
this
system
is
T
=
2π/
2k/3
sec.
Therefore,
if
the
period is T = π seconds,
t
solutions will become unbounded
as t → ∞.
�
–0.5
√
√
then k must satisfy 2π/ 2k/3 = π. The solution of this equation
is k = 6. The initial
2. RSuppose
cos δ = the
−1 roots
and Rare
sindistinct,
δ = 3 λimplies
R = 4 = 2 and�δ = π + arctan(−3) = 2π/3.
49.
1 , λ2 . Then the solution is
�
conditions y(0) = 2 and y (0) = v imply A = 2 and√B = 2k/3�= v. Since k = 6, we
Therefore,
+ 2 =
The
solution
y oscillates
as amplitude
decays toy(t)
zero
asc tsystem
λ→
know
that 2B
= v. The
of the
22 + (v/2)2 . Since
1 t ∞. is λ2 t A2 B
+2π/3).
c2 e .
√ the
1 e �−
y ==
2 cos(t
2 + (v/2)2 = 3 which implies v = 2 5.
amplitude
is assumedequation
to be 3,iswe
mustbyhave
40.
The characteristic
given
2λ2 + λ2−
4 = 0. Therefore, the two distinct roots
√
√
√
1t
Solving
the
equation
y(t)
=
0,
we
see
that
we
must
c1δby
eλ=
= −c2 eλ2 t which
implies
are
(−1
±4 and
33)/4.
Therefore,
the general
solution
ishave
given
25.λR=cos
3.
δ
=
R
sin
δ
=
−2
implies
R
=
20
=
2
5
and
arctan(−1/2)
≈ −0.4636.
(λ1 −λ2 )t
√
eTherefore,
= −c2 /c1 . First, in order(−1+
to √
guarantee
any
solution
of this equation, we would
33)t/4
(−1− 33)t/4
y(t)
=
c
e
+
c
e
.
√
1
2
need c2 /c1 < 0. Then, applying the natural logarithm function to the equation, we see that
y = 2 5 cos(3t + 0.46362).
t = ln(−c2 /c1 )/(λ1 − λ2 ).
Therefore,
√
√ − √33)c2
√
√
(−1
+
33)c
(−1
1
�
(−1+
33)t/4
33)t/4
4. R cos δ = y−2
π + arctan(3/2)
≈ 4.1244.
(t) and
= R sin δ = −3e implies R += 13 and δ =e(−1−
.
4
4
Therefore,
√ need
Now using the initial conditions, we
y = 13 cos(πt + π + arctan(3/2)).
c1 + c2 = 0
√
√
5.
(−1 + 33)c1 (−1 − 33)c2
+
= 1.
4
4
(a) The spring constant is k = 2/(1/2) = 4 lb/ft. The mass m = 2/32 = 1/16 lb-s2 /ft.
Therefore, the equation of motion is
e
− identities
cos(2t)
− ±10β) =
sin(2t)
> π/2.
4. Consider the trigonometric
sin(α
sin α cos β ±tcos
α sin β. Adding these
5
2
two identities we obtain sin(α − β) + sin(α + β) = 2 sin α cos β. Here, our expression is
sin(3t) + sin(4t). Therefore,
we let α + β = 4t and α − β = 3t. Solving this system of
phi 1.5
4.6
Forced
Vibrations,
Response,
and
Resequations, we have α = 7t/2 and
β = Frequency
t/2. Therefore, we
can write sin(3t)
+ sin(4t)
=
1
4.6.
FORCED
VIBRATIONS,
FREQUENCY
RESPONSE,
AND
RESONANCE
323
2 sin(7t/2)
cos(t/2).
onance
0.5
5. The spring constant is k = 4/(1.5/12) = 32 lb/ft. The mass is m = 4/32 = 1/8 lb-s2 /ft.
The initial conditions are y(0) = 2/12 = 1/6 ft. and y � (0) = 0.
1. Consider
trigonometric
identities
cos(α
β) 20
=2cos
α cos30the
β ∓equation
sin α sin β.
Subtracting
Assuming
no the
damping,
but an 0external
F±
cos(3t),
describing
the
5 force,
10
15(t) =
25
w
6.
The
spring
constant
is
k
=
5(9.8)/.1
=
490
N/m.
The
mass
is
m
=
5
kg.
The
damping
these
two
identities
we
obtain
cos(α
−
β)
−
cos(α
+
β)
=
2
sin
α
sin
β.
Here,
our
expression
motion is
1 let
force
is γ =
= 50
N-sec/m.weThe
external
is Fα(t)
Therefore,
the
is cos(9t)
− 2/.04
cos(7t).
Therefore,
α32y
− β==2force
9t and
+=
β 10
= sin(t/2).
7t. Solving
this system
y �� +
cos(3t)
8
equation
describing
the
motion
is
of The
equations,
have αis =
8t 12
andlb/ft
β =and
−t.theTherefore,
we can
cos(9t)
− cos(7t)
=
9.
spring we
constant
k =
mass is 6/32
lb-s2write
/ft. The
forcing
term is
which
cansin(−t)
be rewritten
as
2
sin(8t)
=
−2
sin(8t)
sin(t).
�
4 cos(7t). Therefore, the equation
of��50y
motion
is = 10 sin(t/2)
5y �� +
+ 490y
y
+
256y
= 16
cos(3t).
2.
Consider
the
trigonometric
identities
sin(α
± β)
= sin α cos
β ±RESONANCE
cos α sin β. Subtracting
4.6. FORCED VIBRATIONS, FREQUENCY
RESPONSE,
AND
323
6
��
which
can
be
rewritten
as
these two identities we obtain sin(α +
β)
− sin(α
− β) = 2 cos α sin β. Here, our expression
y
+
12y
=
4
cos(7t)
�� 32
10yα� + 98y
sin(t/2).
is sin(7t) − sin(6t). Therefore, ywe+
let
β ==7t2 and
α − β = 6t. Solving this system of
The initial conditions are y(0) = 2/12 = 1/6 ft. and y � (0) = 0.
equations,
wesimplified
have α are
=to13t/2
= t/2.
which
can be
The
initial
conditions
y(0) =and
0 mβand
y � (0) Therefore,
= 0.03 m/s.we can write sin(7t) − sin(6t) =
6.2 sin(t/2)
The spring
constant
is
k
=
5(9.8)/.1
=
490
N/m.
The mass is m = 5 kg. The damping
64
cos(13t/2).
y �� + 64y =
cos(7t).
7. is γ = 2/.04 = 50 N-sec/m. The
force
external3 force is F (t) = 10 sin(t/2). Therefore, the
3. Consider the trigonometric identities cos(α ± β) = cos α cos β ∓ sin α sin β. Adding these
equation
describing
the
is
The
solution
of we
theofobtain
homogeneous
is yh+(t)
c(t)
cos(8t)
+ cβ.
Then
we look
foris
(a)
solution
themotion
homogeneous
isβ)=
yh=
=
+ c2 sin(16t).
To find
12
2 sin(8t).
1 cos(16t)
twoThe
identities
cos(α −problem
β) +problem
cos(α
coscα
cos
Here,
our expression
acos(πt)
particular
solution
of
the
form
y
(t)
=
A
cos(7t).
(We
do
not
need
to
include
a
term
of
the
a solution
of theTherefore,
inhomogeneous
we10
look
of the this
formsystem
Y (t) =
�� p let problem,
− cos(2πt).
− 490y
β = πt
and
α for
+ βa =solution
2πt. Solving
of
5ywe
+ 50yα� +
=
sin(t/2)
form
sin(7t)
since
there
are
no
first
derivative
terms
in
the
equation.)
Substituting
yp into
A
cos(3t)
(since
there
is
no
first
derivative
term,
we
may
exclude
the
sin(3t)
function).
equations, we have α = 3πt/2 and β = πt/2. Therefore, we can write cos(πt) + cos(2πt)
=
the
ODE,
we
conclude
that
A
=
64/45.
Therefore,
the
general
solution
of
this
differential
Looking
for
a
solution
of
this
form,
we
arrive
at
equation
which
can
be
rewritten
as
2 cos(3πt/2) cos(πt/2).
��
+ 10y �++ 64
98ycos(7t).
= 2 sin(t/2).
equation
is y(t)
c1 cos(8t)−9A
+ cyidentities
sin(8t)
initial
conditions
y(0)
= 0 ftthese
and
2 cos(3t)
4. Consider
the=trigonometric
±cos(3t)
β) = The
sin
cos
β±
cos α sin β.
Adding
+sin(α
256A
=α
16
cos(3t).
45
�
�
The
conditions
are
y(0)
m and
y (0)c+2=β)
m/s.
ytwo
(0)initial
= 0 ft/sec
imply
that
c1 =−0−64/45
and
=0.03
0.= Therefore,
the Here,
solution
this IVP isis
identities
we
obtain
sin(α
β)
+ sin(α
2 sin α cos β.
ourofexpression
Therefore,
we
need
A
to
satisfy
247A
=
16
or
A
=
16/247.
Therefore,
the
64
64
sin(3t)
+ sin(4t).
Therefore,
we let α + β = 4t and α − β = 3t. Solving thissolution
system of
of
7.
y(t)
=
−
cos(8t)
+
cos(7t).
the
inhomogeneous
problem
is
45
45
equations, we have α = 7t/2 and β = t/2. Therefore, we can write sin(3t) + sin(4t) =
(a)
The solution
of the homogeneous problem is yh (t) = c116cos(16t) + c2 sin(16t). To find
2 sin(7t/2)
cos(t/2).
y(t) = c1 cos(16t)
+ c2 we
sin(16t)
cos(3t). of the form Y (t) =
a solution of the inhomogeneous
problem,
look +
for a solution
5. The spring constant is k = 4/(1.5/12) = 32 lb/ft. The 247
mass is m = 4/32 = 1/8 lb-s2 /ft.
A cos(3t) (since there is no
first derivative term, we may exclude the sin(3t) function).
2
� (t) = 2 cos(3t), the equation describing the
Assuming
no damping,
but
external
force,yF
The initial
arean
y(0)
= 1/6
0. equation
Therefore, c1 , c2 must satisfy
Looking
for conditions
a solution of
this
form,
weand
arrive(0)
at=the
motion is
1
1 ��
16
1
−9A cos(3t)
y +
+c256A
32y
2 cos(3t)
= = 16 cos(3t).
1 + =cos(3t)
82
0
2476
68
4
10
t
Therefore,
we
need
A
to
satisfy
247A
=
16
or
A
=
16/247.
Therefore, the solution of
16c
=
0.
which can be rewritten as –1
2
��
the inhomogeneous problem is y + 256y = 16 cos(3t).
Therefore, we have c1 =–2151/1482 and c2 = 0. Therefore, the solution is
16
y(t) = c1 cos(16t)
+
cos(3t).
151 + c2 sin(16t)16
247
y(t) =
cos(16t) +
cos(3t).
1482
247
The initial conditions are y(0) = 1/6 and y � (0) = 0. Therefore, c1 , c2 must satisfy
10.
(b) The spring constant is k = 8/(1/2) = 16 lb/ft and the mass is 8/32 = 1/4 lb-s2 /ft. The
16 of motion
1
forcing term is 8 sin(8t). Therefore, the equation
is
c1 +
=
247 6
4.6. FORCED VIBRATIONS, FREQUENCY
RESPONSE,
AND RESONANCE
327
1 ��
0.
0.15
y + 16y16c
= 28 =
sin(8t)
4
0.1
Therefore,
we have c1to=y 151/1482
and c2 = 0. Therefore, the solution is
which
can be simplified
0.05
y �� 151
+ 64y = 32 sin(8t).
16
y(t) =
cos(16t) +
cos(3t).
0
1482
1
2
3 247 4
The solution of the homogeneous problem is y (t) = c cos(8t)+c5 sin(8t). Then we look for a
–0.05
h
t
1
2
particular solution of the form yp (t) = At sin(8t) + Bt cos(8t). Substituting yp into the ODE,
(b)
we conclude that A = 0 and–0.1
B = −2. Therefore, the general solution of this differential
–0.15
equation is y(t) = c1 cos(8t) + c2 sin(8t) − 2t cos(8t). The initial conditions y(0) = 1/4 ft and
y � (0) = 0 ft/sec imply that c0.15
1 = 1/4 and c2 = 1/4. Therefore, the solution of this IVP is
1
1
y(t) = 4 cos(8t) + 4 sin(8t) − 2t cos(8t). Solving for y � , we see that y � (t) = (−2 + 16t) sin(8t).
0.1
We see that y � (t) = 0 whenyt = 1/8 or sin(8t) = 0. Therefore, the first four times the velocity
is zero are t = 1/8, π/8, π/4, 0.05
3π/8.
3
4 mass5 is 8/32 = 1/4 lb-s2 /ft. The
11. The spring constant is k =08/(1/2)1 = 162lb/ft and
the
when k = mω 2 ; that is, when m = k/ω 2 = 16/4 = 4 slugs.
1.5
Then
� t
�
te et
−1
X (t)The
= mass
.
t
12.
spring VIBRATIONS,
constant is k =
3 N/m.
The damping
equals y � . The
1 FREQUENCY
−e
0is 2 kg. AND
4.6. The
FORCED
RESPONSE,
RESONANCE
329
external
force
is
3
cos(3t)
−
2
sin(3t)
N.
Therefore,
the
equation
of
motion
is
Therefore,
0.5
�� �
this time interval, the forcing ��term� is F (t) =� 0.
the general solution is given by
t
tet Therefore,
e−
−1
2y +−1
y +0.83y =1 3 cos(3t)
2 sin(3t).
0.6
1.2
1.4
1.6
1.8
2
X
(t)g(t)
=
the solution of the homogeneous
problem, y(t)−e
=
x tc1 cos(t) + c2 sin(t). Considering our initial
(b)
0
t
conditions,
we need
c1 = 0 and
cCHAPTER
solution
the IVPEQUATIONS
for 2π < We
t is
�
2 = −4A.
328
4.0 �
SECOND
ORDER
LINEAR
Since
the system
is damped,
the steady
stateTherefore,
response
isthe
equal
to theof
particular
solution.
given
byaamplitude
y(t)
= −4A
sin(t).
Toy summarize,
conclude
thatisthe
solution
= we
. spring
(c)
The
a form
similar
system
a tlinear
given
by is given by
look
for
solution
offor
the
= A with
cos(3t)
p (t)
e + B sin(3t). Substituting an equation of this

form into the ODE, we conclude
−1/6 and B = 1/6.
the steady-state
sin(t)A+=At
0 ≤ tTherefore,
≤π
 −Athat
5
Therefore,
response is
= √+ 2πA − At� � π. < t ≤ 2π
−3A R
sin(t)
y(t) =
 �
1 25 − 49ω�2 + 25ω 4
−4A
sin(t)
0 2π < t.
y(t)
=
(sin(3t)
−1
X (t)g(t)
dt =− cos(3t)).
dt
6
et
� �
15. The
Theamplitude
inductanceofisthe
L steady-state
= 1 henry. The
resistance
= 5 × 103 ohms. The capacitance
13.
response
is given
0 R by
=
.
−6
Amplitude
C = 0.25 × 10 farads. The5 forcing term is due to
et the 12−volt battery. Therefore, the
1
equation for charge Q is
R= � 2
Therefore,
(ω0 −� +
ω 2(4
)2 ×
+ 10
4δ62)Q
ω 2 = 12.
Q4 �� + 5000Q
� m
�
= (t)
−1
The initial
conditions
Q(0)3 x
0, =
Q� (0)2 = 0.X
The
solution2 of the homogeneous problem
4.7.
VARIATION
OF are
PARAMETERS
343
p
m2X(t)
(ω0 − ω 2 )2 +(t)g(t)
4m2 δ 2dt
ω
−1000t
−4000t
is Qh (t) = c1 e
+ c2 e
. Looking
for
a
solution
of
the inhomogeneous problem, we
�
�
�
�
m −t
2
0 =
−e3 × 100−6 .. Therefore, the general solution
� Qp (t)
= form
find a particular solution of the
2
=
2
2 2 this
2
Therefore, c1 − c2 = −1 −1000t
and c2 − 1 −4000t
=
1.(ωSolving
of equations, we have c1 = 1
−t
m
−
ω
+−6γ. 2system
0 3 ×
e+
te)−t
eωtConsidering our initial conditions,
is given by Q(t) = c1 e
+
c
e
10
2
1
and c2 = 2. Therefore, the solution
of
the
IVP
is
�
�
−6
we conclude
that
c1 be
= a−4maximum
× 10−6 and
c−1
= 10denominator
. Therefore,
solution of
the IVPthe
is
2 the
The
amplitude
will
when
is athe
minimum.
Consider
�
�
=
.
−6
−1000t
−4000t 0.6 0.8
−6
∼
−t
t
−t
1t= 0.001,
1.6 we
1.8 have
2 Q(0.001) = 1.5468
Q(t) = 10f (z)(−4e
3).This
At
0.01,
10 2 .
2
e tfunction
−
2e1.2w+ 1.4
tehas
+a 2t
function
= m2 (ω02 +
−ez)
+ γ+2 z.
minimum
when z = ω02 − γ 2×/2m
−6 x(t) =
−6
.
∼
t our
and
Q(0.01)
2.9998×10
,
respectively.
From
function
Q,
we
see
that
Q(t)
→
3×10
=
2 1
2
2
2
2
2e at
−ω
t−
Therefore,
the
amplitude
reaches
a
maximum
=
ω
−
γ
/2m
.
Since
ω
=
k/m,
we
max
0
0
Therefore,
� �
as
t conclude
→ ∞. that
can
� −1 2 �
γ4 .is
(t)2 =
2 xpin
8.
solution ofof
theParameters
equation
16.The general
ωmax
= ωproblem
.
4.7
Variation
0 1 −t
� �
� 2km
�
� �
−t
(a)
As X(t)
the system
is2 damped, the
homogeneous
solution
will
0
−e
−1be the transient solution, and
2 the fundamental
5.
Let
be
matrix,
x(t)
= expression
c1be −t
+for
c2 the
+ solution
Substituting
ω = ωmax
into
the
amplitude,
we. haveof the inhomogeneous
−t
the
steady-state
solution
will
given
by
a
particular
1.
e �
te
� t
cosofm
tthesin
t A cos(ωt)+B sin(ωt). Substituting
problem. We look for a particular solution
form
X(t)
.
� =�
�t =
R
−
sin
t
cos
t satisfy
this
function
into
the
ODE,
we
see
that
A,
B
must
4 /4m2 + γ 2 (ω 2 − γ
2 /2m2 )
The initial condition x(0) = 1 0 γimplies
(a)
�
��
� 0�
�
m
�
�
�
�
�
�
�
�
x
(t)
x
(t)
u
(t)
x
(t)u
(t)
+
x
(t)u
(t)
Then
1
11
12
1
11
1
12
2
�
�
= �(2
Xu =
.
0 2−2ωu2 )A
−1
1
+ t=ωB
cos
− sin
t2 =
4−1
x21 (t)= cx1X
(t)
x=
ω(t)
/4m
22−1
21 (t)u
1 (t) + x.22 (t)u2 (t)
x(0)
c−22γ(t)
4 2+
0 γ+=
.
1 1 m sin
0 t cos t0
0
2
�
= − ωA + (2 − ω. )B = 0.
Therefore,
Therefore, −c − 1 = 1 and c =γω
41 − γ 2this
� we have c1 = 0 and
/4mk
Therefore,
0.�0 Solving
of equations,
2
1
�
� system
�
x
u�2 + x12 u��2
�
�
1 + x11 u1 + x�
11 u
12
2�
4
(Xu) is
=
. and B = 8ω/[16ω 4 −
c2 =The
−2. solution
Therefore,
the solution
ofAthe
IVP
ist ω−
� 32(2
of this
system
63ωt2 +� 64]
sin
cos
−1
x=
u1cos
+−
x21 u)/[16ω
+ xt�22 u−
2 + x22 u2
21
1
X
(t)g(t)
=
14. 63ω
First,
problem
yh (t) = 2c)1cos(ωt)+
cos(t) +
2 for 0 ≤ t ≤ π, the solution of�the homogeneous
�− sin
t −tof�cos
+64]. Therefore, the steady-statesin
part
thet�solution
istY (t) =is [32(2−ω
−e
−1
�
�
c2 sin(t).
Then
we
look
for
a
particular
solution
of
the
form
y
(t)
=
Bt.
We
see
that a
2
Now
8ω sin(ωt)]/[16ω4 − 63ω +x(t)
64]. =
+
. �p
� −2
�
�
−t
1
te
t
�Therefore,
�
particular solution is given by yp� (t) =
the ugeneral
solution for 0 ≤ t ≤ π
=
.
x�11At.
1
0 x12
= The
u� = y(0)
.= 0 implies c1 = 0. Then
� initial
�
�
(b)y(t)
Here
= 2 and
1/8.+X
Therefore,
is
= ωc012 cos(t)
+ cδ2 =
sin(t)
At.
condition
x21 x22
u2
The
general
solution
of the
equation
in problem
y9.� (t)
= −c
+ c2 cos(t)
+ A.
Therefore,
y � (0) =5c2is+ A = 0 implies c2 = −A. Therefore,
1 sin(t)
1
the solution of this IVP is y(t)|G(iω)|
sin(t)
� +
�= −A=
�
� At for
� 0 �≤ t ≤ π.
� Then at time t = π,
2 + 4ω 2t/64
cos ttime interval,
cos tterm is F (t) = A(2π − t).
(2 −sin
ω 2t)the
y(π) = Aπ and y � (π) =
2A.
In
this
forcing
x(t) = c1
+ c2
+
.
− sin
t
cos
sin=t C + Dt. Substituting a
4tform y−t
Therefore, we look for a particular
solution
of
the
(t)
p
=√
.
4−
function of this form into the
that
C63ω
= 22πA
+ 64and D = −A. Therefore, the
� ODE,
�t we see 16ω
1 is1 y(t)
The initial
condition
implies
general
solution
for πx(0)
< t=
− 2π
= c1 cos(t) + c2 sin(t) + 2πA − At. Now considering
Also,
�
�
�
� � �
√�
the initial conditions y(π) = Aπ and� y�(π) = �
2A,�we
need
0 4 − 63ω
1 2 + 64 .
φ(ω) = arccos1 4(2 − ω02 )/ 16ω
x(0) = c1
+c
+
=
.
0− c1 +2 πA
1 = Aπ 0
1
− cthe
= 2A. of the IVP is
2−A
Therefore, c1 = 1 and c2 = 1. Therefore,
solution
� Therefore,
� � the�solution
�
�
Therefore, c1 = 0 and c2 = −3A.
π < t ≤ 2π is
cos t
sin t
t cosoft the IVP for
�
x(t)
=
+
+
.
y(t) = −3A sin(t) + 2πA − At. Then
t t= 2π, y(2π)
− sin tat timecos
−t sin t= 0 and y (2π) = −4A. In
1
Therefore, a particular solution is Y (t) = − t cos(2t)
+ (sin(2t)) ln | sin(2t)|. Therefore,
y(t) = c1 cos(2t) + c2 sin(2t)2 +
sin(2t
4 − 2s)g(s) ds.
the general solution is y(t) = c1 cos(2t) + c2 sin(2t)2 − 32 t cos(2t) + 34 sin(2t) ln | sin(2t)|.
346
4. SECOND
ORDER
EQUATIONS
22. By
direct
substitution,
it canCHAPTER
be verified
that
y1is(t)yh=(t)t and
(t) =LINEAR
tet +
arec2solutions
ofThe
the
18.
The
solution
of the homogeneous
equation
=
cy12cos(t/2)
sin(t/2).
2 t
homogeneous
of theseform
functions
is W (y1 , yset
)
=
t
e
.
Rewriting
functions
y1 (t)equations.
= cos(t/2)The
andWronskian
y2 (t) = sin(t/2)
a fundamental
of
solutions.
The
2
the equation
standard
form,
have
Wronskian
of in
these
functions
is we
W (y
1 , y2 ) = 1/2. Writing the ODE in standard form, we see
that g(t) = sec(t/2)/2. Using the method of variation of parameters, a particular solution
t(t + 2) � t + 2
is given by Y (t) = u1 (t)y1 (t) + yu��2 (t)y
− 2 (t)2 where
y + 2 y = 2t.
t
t
�
cos(t/2)(sec(t/2))
Therefore, g(t) = 2t. uUsing
a particular solution is
− method of variationdtof=parameters,
2 ln[cos(t/2)]
1 (t) = the
2W (t)
given by Y (t) = u1 (t)y1 (t) + u2 (t)y2 (t) where
�
�
sin(t/2)(sec(t/2))
tet (2t)
u2 (t) =
dt = t.
u1 (t) = − 2W (t) dt = −2t
W (t)
Therefore, a particular solution is Y (t) �= 2 cos(t/2) ln[cos(t/2)] + t sin(t/2). Therefore, the
t(2t)
general solution is y(t) = c1 cos(t/2)
u2 (t)+=c2 sin(t/2)dt+=2 cos(t/2)
−2e−t . ln[cos(t/2)] + t sin(t/2).
W
(t)
19. The solution of the homogeneous equation is yh (t) = c1 et +c2 tet . The functions y1 (t) = et
2
t
Therefore,
particular
is Y (t)
− 2t2 =The
−4t2Wronskian
.
and
y2 (t) =a te
form a solution
fundamental
set=of−2t
solutions.
of these functions is
2t
W
. Using the method
variation
a particular tsolution is given
1 , ydirect
2 ) = e substitution,
23.(yBy
it can beofverified
thatofyparameters,
1 (t) = 1 + t and y2 (t) = e are solutions of
by
(t) = u1 (t)y1 (t)
+ u2 (t)yThe
2 (t) where
theYhomogeneous
equations.
Wronskian of these functions is W (y1 , y2 ) = tet . Rewriting
the equation in standard form, we� have tet (et )
1
u1 (t) = −
dt = − ln(1 + t2 )
2
W1(t)(1
2
+ t + t )1
� y �� − t t y � + y = te2t .
e (et )
t
u2 (t) =
dt = arctan(t).
2)
W
(t)(1
+
t
Therefore, g(t) = te2t . Using the method of variation of parameters, a particular solution is
given by Y (t) = u1 (t)y1 (t) + u2 (t)y2 (t) where 1
Therefore, a particular solution is Y (t) = − et ln(1 + t2 ) + tet arctan(t). Therefore, the
� t 2 2t
1 t e (te ) 2
t
t
t 1 2t
general solution is y(t) = c1 e +
c2 te=−
u1 (t)
− 2 e ln(1 + tdt) +
= te
− arctan(t).
e
W (t)
2
20. The solution of the homogeneous
� equation 2tis yh (t) = c1 e3t + c2 e2t . The functions
(1 + t)(te ) of solutions.
y1 (t) = e3t and y2 (t) = e2t form
The Wronskian of these
u2 (t)a =fundamental set dt
= tet .
5t
W
(t)
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) + u2 (t)y
1 2 (t) where
1
2t
Therefore, a particular solution is Y�(t) = − (1 + t)e�
+ te2t = (t − 1)e2t .
2
2
e2s (g(s))
−3s
t
u
(t)
=
−
ds
=
e
g(s)ds
1
24. By direct substitution, it can be verified
W (s) that y1 (t) = e and y2 (t) = t are solutions
t
of the homogeneous equations. The
Wronskian of these
� 3s
� functions is W (y1 , y2 ) = (1 − t)e .
(g(s))
Rewriting the equation in standard eform,
we have
u2 (t) =
ds = − e−2s g(s)ds.
W (s)
t �
� 1 3(t−s)
y �� − is Y (t)
y −
y =g(s)ds
2(1 − −
t)e�−te.2(t−s) g(s)ds. Therefore, the
Therefore, a particular solution
=
e
1−t � 1−t
�
general solution is y(t) = c1 e3t + c2 e2t + e3(t−s) g(s)ds − e2(t−s) g(s)ds.
21. The solution of the homogeneous equation is yh (t) = c1 cos(2t)+c2 sin(2t). The functions
y1 (t) = cos(2t) and y2 (t) = sin(2t) form a fundamental set of solutions. The Wronskian of
these functions is W (y1 , y2 ) = 2. Using the method of variation of parameters, a particular
solution is given by Y (t) = u1 (t)y1 (t) + u2 (t)y2 (t) where
�
�
sin(2s)(g(s))
1
u1 (t) = −
ds = −
sin(2s)g(s) ds
W (s)
2
�
�
1
cos(2s)(g(s))
u2 (t) =
ds =
cos(2s)g(s) ds.
W (s)
2