——————————————————————————
——
CHAPTER 3.
________________________________________________________________________
page 83
Chapter Three
Section 3.1
1.
Let
, so that
and
.
Direct substitution into the differential
C œ /
C
œ < /
C
œ < /
<>
w
<>
ww
<>
equation yields
.
Canceling the exponential, the characteristic
a
b
< #< $ /
œ !
#
<>
equation is
The roots of the equation are
,
.
Hence the
< #< $ œ ! Þ
< œ $ "
#
general solution is
.
C œ  /  /
"
#
>
$>
2.
Let
.
Substitution of the assumed solution results in the characteristic equation
C œ /
<>
< $< # œ ! Þ
< œ #
"
#
The roots of the equation are
,
.
Hence the general
solution is
.
C œ  /
 /
"
#
>
#>
4.
Substitution of the assumed solution
results in the characteristic equation
C œ /
<>
#< $< " œ ! Þ
< œ "Î# "
#
The roots of the equation are
,
.
Hence the general
solution is
.
C œ  /
 /
"
#
>Î#
>
6.
The characteristic equation is
, with roots
.
Therefore the
%< * œ !
< œ „$Î#
#
general solution is
.
C œ  /
 /
"
#
$>Î#
$>Î#
8.
The characteristic equation is
, with roots
.
Hence the
< #< # œ !
< œ "„
$
#
È
general solution is
.
C œ  /B: "
$ >  /B: "
$ >
"
#
Š
‹
Š
‹
È
È
9.
Substitution of the assumed solution
results in the characteristic equation
C œ /
<>
< < # œ ! Þ
< œ # "
#
The roots of the equation are
,
.
Hence the general
solution is
.
Its derivative is
.
Based on the
C œ  /
 /
C
œ # /
 /
"
#
"
#
#>
>
w
#>
>
first condition,
, we require that
.
In order to satisfy
,
C !
œ "
  œ "
C
!
œ "
a b
a b
"
#
w
we find that
.
Solving for the constants,
and
.
Hence the
#  œ "
 œ !
 œ "
"
#
"
#
specific solution is
.
C >
œ /
a b
>
11.
Substitution of the assumed solution
results in the characteristic equation
C œ /
<>
'< &< " œ ! Þ
< œ "Î$ "Î#
#
The roots of the equation are
,
.
Hence the general
solution is
.
Its derivative is
.
Based
C œ  /
 /
C
œ  /
Î$  /
Î#
"
#
"
#
>Î$
>Î#
w
>Î$
>Î#
on the first condition,
, we require that
.
In order to satisfy the
C !
œ "
  œ %
a b
"
#
condition
, we find that
.
Solving for the constants,
C
!
œ "
 Î$  Î# œ !
 œ "#
w
a b
"
#
"
and
.
Hence the specific solution is
.
 œ )
C >
œ "# /
) /
#
a b
>Î$
>Î#
12.
The characteristic equation is
, with roots
,
.
Therefore the
< $< œ !
< œ $
!
#
general solution is
, with derivative
.
In order to
C œ   /
C
œ $  /
"
#
#
$>
w
$>
satisfy the initial conditions, we find that
, and
.
Hence the
  œ #
$  œ $
"
#
#
specific solution is
.
C >
œ " /
a b
$>
13.
The characteristic equation is
, with roots
< &< $ œ !
#
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——————————————————————————
——
CHAPTER 3.
________________________________________________________________________
page 84
<
œ
„
&
"$
#
#
"ß#
È
.
The general solution is
, with
C œ  /B:
&
"$ >Î#  /B:
&
"$ >Î#
"
#
Š
‹
Š
‹
È
È
derivative
C
œ
 /B:
&
"$ >Î#
 /B:
&
"$ >Î#
&
"$
&
"$
#
#
w
È
È
Š
‹
Š
‹
È
È
"
#
.
In order to satisfy the initial conditions, we require that
, and
  œ "
"
#
&
"$
&
"$
#
#
È
È

 œ !
 œ
" &Î
"$
Î#
"
#
"
.
Solving for the coefficients,
and
Š
‹
È
 œ
" &Î
"$
Î# Þ
#
Š
‹
È
14.
The characteristic equation is
, with roots
#< < % œ !
#
<
œ
„
"
$$
%
%
"ß#
È
.
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