NOT FOR SALE

722 ¤ CHAPTER 17 SECOND-ORDER DIFFERENTIAL EQUATIONS

solution is = 1
( 5+ 15 )
+ 2
( 5 15 )
. Entering the initial conditions gives 1 = 1
and 2 = 1
, so
2 15 2 15

= 1 ( 5+ 15 ) 1 ( 5 15 )
.
2 15 2 15

= 20: = 5± 5 and the solution is = 1

( 5+ 5)
+ 2
( 5 5)
so again the motion is overdamped.

The initial conditions give 1 = 1

and 2 = 1
, so = 1 ( 5+ 5) 1 ( 5 5)
.
2 5 2 5 2 5 2 5

5
= 25: we have equal roots 1 = 2 = 5, so the motion is critically damped and the solution is =( 1 + 2 ) .
5
The initial conditions give 1 = 0 and 2 = 1, so = .
5
= 30: = 5± 5 so the motion is underdamped and the solution is = 1 cos 5 + 2 sin 5 .
1 1 5
The initial conditions give 1 = 0 and 2 = 5
, so = 5
sin 5 .

= 40: = 5± 15 so we again have underdamping.

5
The solution is = 1 cos 15 + 2 sin 15 ,
1
and the initial conditions give 1 = 0 and 2 = 15
.
1 5
Thus = 15
sin 15 .

00 2
9. The differential equation is + = 0 cos 0 and 0 6= = . Here the auxiliary equation is + =0

with roots ± =± so ( )= 1 cos + 2 sin . Since 0 6= , try ( )= cos 0 + sin 0 .

2 2
Then we need ( ) 0 ( cos 0 + sin 0 ) + ( cos 0 + sin 0 )= 0 cos 0 or 0 = 0 and

2 0 0 2
0 = 0. Hence = 0 and = 2
= 2 2
since = . Thus the motion of the mass is given
0 ( 0)

0
by ( ) = 1 cos + 2 sin + 2 2 cos 0 .
( 0)

10. As in Exercise 9, ()= 1 cos + 2 sin . But the natural frequency of the system equals the frequency of the

external force, so try ( ) = ( cos + sin ). Then we need

2 2
(2 ) cos (2 + ) sin + cos + sin = 0 cos or 2 = 0 and
2 2 2
2 = 0 [noting + = 0 and + = 0 since = ]. Hence the general solution is

()= 1 cos + 2 sin +[ 0 (2 )] sin .

0
11. From Equation 6, ( ) = ( ) + ( ) where ( ) = 1 cos + 2 sin and ( ) = 2 2
cos 0 . Then
( 0)

2 2
is periodic, with period , and if 6= 0, is periodic with period 0
. If 0
is a rational number, then we can say

0
= = 0
where and are non-zero integers. Then

2 2 2 2 2
+ · = + · + + · = ( )+ + 0
· = ( )+ + · 0
= ( )+ ( ) = ( )

so ( ) is periodic.

c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
 NOT FOR SALE
SECTION 17.3 APPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS ¤ 723

12. (a) The graph of = 1 + 2 has a -intercept when 1 + 2 =0 ( 1 + 2 )=0 1 = 2 .

Since 0, has a -intercept if and only if 1 and 2 have opposite signs.

(b) For 0, the graph of crosses the -axis when 1

1
+ 2
2
=0 2
2
= 1
1

1
( 1 2) ( 1 2)
2 = 1 = 1 . But 1 2 1 2 0 and since 0, 1. Thus
2

( 1 2)
| 2| = | 1| | 1 |, and the graph of can cross the -axis only if | 2 | | 1 |.

00 0 0
13. Here the initial-value problem for the charge is + 20 + 500 = 12, (0) = (0) = 0. Then
10 3
()= ( 1 cos 20 + 2 sin 20 ) and try ()= 500 = 12 or = 125
.
10 3 3
The general solution is ()= ( 1 cos 20 + 2 sin 20 ) + 125
. But 0 = (0) = 1 + 125
and
0 10 0
( )= ( )= [( 10 1 + 20 2 ) cos 20 + ( 10 2 20 1 ) sin 20 ] but 0 = (0) = 10 1 + 20 2 . Thus the charge
1 10 3 10 3
is ()= 250
(6 cos 20 + 3 sin 20 ) + 125
and the current is ( ) = 5
sin 20 .

00 0 0
14. (a) Here the initial-value problem for the charge is 2 + 24 + 200 = 12 with (0) = 0 001 and (0) = 0.
6 3
Then ()= ( 1 cos 8 + 2 sin 8 ) and try ( )= = 50 and the general solution is
6 3 3
()= ( 1 cos 8 + 2 sin 8 ) + 50
. But 0 001 = (0) = + 50
so 1 = 0 059. Also
0 6 0
()= ()= [( 6 1 + 8 2 ) cos 8 + ( 6 2 8 1 ) sin 8 ] and 0 = (0) = 6 1 +8 2 so
6 3
2 = 0 04425. Hence the charge is ()= (0 059 cos 8 + 0 04425 sin 8 ) + 50
and the current is
6
()= (0 7375) sin 8 .

(b)

10
15. As in Exercise 13, ()= ( 1 cos 20 + 2 sin 20 ) but ( ) = 12 sin 10 so try

()= cos 10 + sin 10 . Substituting into the differential equation gives

( 100 + 200 + 500 ) cos 10 + ( 100 200 + 500 ) sin 10 = 12 sin 10

3 3
400 + 200 = 0 and 400 200 = 12. Thus = 250
, = 125
and the general solution is
10 3 3 3 3
( )= ( 1 cos 20 + 2 sin 20 ) 250
cos 10 + 125
sin 10 . But 0 = (0) = 1 250
so 1 = 250
.
0 3 6 10
Also ()= 25
sin 10 + 25
cos 10 + [( 10 1 + 20 2 ) cos 20 + ( 10 2 20 1 ) sin 20 ] and
0 6 3
0= (0) = 25 10 1 + 20 2 so 2 = 500 . Hence the charge is given by
10 3 3 3 3
()= 250
cos 20 500
sin 20 250
cos 10 + 125
sin 10 .

c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated,
° licated, or posted to a publicly accessible website, in whole or in part.
par
 NOT FOR SALE
724 ¤ CHAPTER 17 SECOND-ORDER DIFFERENTIAL EQUATIONS

6
16. (a) As in Exercise 14, ()= ( 1 cos 8 + 2 sin 8 ) but try ()= cos 10 + sin 10 . Substituting into the

differential equation gives ( 200 + 240 + 200 ) cos 10 + ( 200 240 + 200 ) sin 10 = 12 sin 10 ,
1 6 1
so = 0 and = 20
. Hence, the general solution is ()= ( 1 cos 8 + 2 sin 8 ) 20
cos 10 . But
1 0 6 1
0 001 = (0) = 1 20
, ()= [( 6 1 + 8 2 ) cos 8 + ( 6 2 8 1 ) sin 8 ] 2
sin 10 and
0
0= (0) = 6 1 + 8 2 , so 1 = 0 051 and 2 = 0 03825. Thus the charge is given by
6 1
( )= (0 051 cos 8 + 0 03825 sin 8 ) 20
cos 10 .

(b)

1 2
17. ()= cos( + ) ()= [cos cos sin sin ] ()= cos + sin where

cos = 1 and sin = 2 ()= 1 cos + 2 sin . [Note that cos2 + sin2 = 1 2
1 + 2
2 = 2
.]

2
18. (a) We approximate sin by and, with = 1 and = 9 8, the differential equation becomes 2
+ 9 8 = 0. The auxiliary

2
equation is +98=0 = ± 9 8 , so the general solution is ( ) = 1 cos 98 + 2 sin 98 .
0 1
Then 0 2 = (0) = 1 and 1 = (0) = 98 2 2 = 98
, so the equation is

1
( ) = 0 2 cos 98 + 98
sin 98 .

0 5
(b) ()= 02 9 8 sin 98 + cos 98 = 0 or tan 98 = 98
, so the critical numbers are

1 1 5
= 98
tan 98
+ 98
( any integer). The maximum angle from the vertical is

1 1 5
98
tan 98
0 377 radians (or about 21 7 ).

(c) From part (b), the critical numbers of ( ) are spaced 98

apart, and the time between successive maximum values

2
is 2 98
. Thus the period of the pendulum is 98
2 007 seconds.

1
(d) ( ) = 0 0 2 cos 98 + 98
sin 98 =0 tan 98 = 02 98

1 1
= 98
tan 02 98 + 0 825 seconds.

0
(e) (0 825) 1 180 rad s.

c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
 NOT FOR SALE
SECTION 17.4 SERIES SOLUTIONS ¤ 725

17.4 Series Solutions

0 1 0
1. Let ( ) = . Then ( )= and the given equation, = 0, becomes
=0 =1

1
= 0. Replacing by + 1 in the first sum gives ( + 1) +1 = 0, so
=1 =0 =0 =0

[( + 1) +1 ] = 0. Equating coefficients gives ( + 1) +1 = 0, so the recursion relation is

=0

1 0 1 1 1 0 1 0
+1 = , =0 1 2 . Then 1 = 0, 2 = 1 = , 3 = 2 = · 0 = , 4 = 3 = , and
+1 2 2 3 3 2 3! 4 4!
0 0
in general, = . Thus, the solution is ( ) = = = 0 = 0 .
! =0 =0 ! =0 !

0 0 1
2. Let ( ) = . Then = =0 = 0 or
=0 =1 =0

1 +1
= 0. Replacing with + 1 in the first sum and with 1 in the second
=1 =0

gives ( + 1) +1 1 = 0 or 1 + ( + 1) +1 1 = 0. Thus,
=0 =1 =1 =1

1 + [( + 1) +1 1] = 0. Equating coefficients gives 1 = 0 and ( + 1) +1 1 = 0. Thus, the

=1

1
recursion relation is +1 = , = 1 2, . But 1 = 0, so 3 = 0 and 5 = 0 and in general 2 +1 = 0. Also,
+1
0 2 0 0 4 0 0 0
2 = , 4 = = = , 6 = = = and in general 2 = . Thus, the solution
2 4 4·2 22 · 2! 6 6·4·2 23 · 3! 2 · !
2
2 0 2 2 2 2
is ( ) = = 2 = = 0 = 0 .
=0 =0 =0 2 · ! =0 !

0 1
3. Assuming ( ) = , we have ( )= = ( + 1) +1 and
=0 =1 =0

2 +2 0 2
= = 2 . Hence, the equation = becomes ( + 1) +1 2 =0
=0 =2 =0 =2

2
or 1 +2 2 + [( + 1) +1 2] = 0. Equating coefficients gives 1 = 2 = 0 and +1 =
=2 +1
for = 2 3, . But 1 = 0, so 4 = 0 and 7 = 0 and in general 3 +1 = 0. Similarly 2 = 0 so 3 +2 = 0. Finally
0 3 0 0 6 0 0 0
3 = , 6 = = = , 9 = = = , , and 3 = . Thus, the solution
3 6 6·3 32 · 2! 9 9·6·3 33 · 3! 3 · !
3 3
3 0 3 3 3 3
is ( ) = = 3 = = 0 = 0 = 0 .
=0 =0 =0 3 · ! =0 3 ! =0 !

0 1
4. Let ( )= ( )= = ( + 1) +1 . Then the differential equation becomes
=0 =1 =0

+1
( 3) ( + 1) +1 +2 =0 ( + 1) +1 3 ( + 1) +1 +2 =0
=0 =0 =0 =0 =0

c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated,
° licated, or posted to a publicly accessible website, in whole or in part.
par
 NOT FOR SALE
726 ¤ CHAPTER 17 SECOND-ORDER DIFFERENTIAL EQUATIONS

3( + 1) +1 + 2 =0 [( + 2) 3( + 1) +1 ] =0
=1 =0 =0 =0

since = . Equating coefficients gives ( + 2) 3( + 1) +1 = 0, thus the recursion relation is

=1 =0

( + 2) 2 0 3 1 3 0 4 2 4 0 5 3 5 0
+1 = , =0 1 2 . Then 1 = , 2 = = 2 , 3 = = 3 , 4 = = 4 , and
3( + 1) 3 3(2) 3 3(3) 3 3(4) 3
( + 1) 0 +1
in general, = . Thus the solution is ( ) = = 0 .
3 =0 =0 3

+1 9 0
Note that 0 = for | | 3.
=0 3 (3 )2

0 1 00
5. Let ( )= ( )= and ( )= ( + 2)( + 1) +2 . The differential equation
=0 =1 =0

1
becomes ( + 2)( + 1) +2 + + = 0 or [( + 2)( + 1) +2 + + ] =0
=0 =1 =0 =0

since = . Equating coefficients gives ( + 2)( + 1) +2 + ( + 1) = 0, thus the

=1 =0

( + 1)
recursion relation is +2 = = , =0 1 2 . Then the even
( + 2)( + 1) +2
0 2 0 4 0
coefficients are given by 2 = , 4 = = , 6 = = , and in general,
2 4 2·4 6 2·4·6
0 ( 1) 0 1 3 1 5 1
2 = ( 1) = . The odd coefficients are 3 = , 5 = = , 7 = = ,
2 · 4 · ··· · 2 2 ! 3 5 3·5 7 3·5·7
1 ( 2) ! 1
and in general, 2 +1 = ( 1) = . The solution is
3 · 5 · 7 · · · · · (2 + 1) (2 + 1)!

( 1) 2 ( 2) ! 2 +1
( )= 0 + 1 .
=0 2 ! =0 (2 + 1)!

00 2 00
6. Let ( ) = . Then ( )= ( 1) = ( + 2)( + 1) +2 . Hence, the equation =
=0 =2 =0

becomes ( + 2)( + 1) +2 = 0 or [( + 2)( + 1) +2 ] = 0. So the recursion relation

=0 =0 =0

0 2 0 4 0
is +2 = , =0 1 . Given 0 and 1, 2 = , 4 = = , 6 = = , ,
( + 2)( + 1) 2·1 4·3 4! 6·5 6!
0 1 3 1 1 5 1 1
2 = and 3 = , 5 = = = , 7 = = , , 2 +1 = . Thus, the solution
(2 )! 3·2 5·4 5·4·3·2 5! 7·6 7! (2 + 1)!
2 2 +1
2 2 +1
is ( ) = = 2 + 2 +1 = 0 + 1 . The solution can be written
=0 =0 =0 =0 (2 )! =0 (2 + 1)!

+ 0 + 1 0 1
as ( ) = 0 cosh + 1 sinh or ( ) = 0 + 1 = + .
2 2 2 2

c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
 SECTION 17.4 SERIES SOLUTIONS ¤ 727

0 1 00
7. Let ( )= ( )= = ( + 1) +1 and ( )= ( + 2)( + 1) +2 . Then
=0 =1 =0 =0

00 +1
( 1) ( )= ( +2)( +1) +2 ( +2)( +1) +2 = ( +1) +1 ( +2)( +1) +2 .
=0 =0 =1 =0

Since ( + 1) +1 = ( + 1) +1 , the differential equation becomes

=1 =0

( + 1) +1 ( + 2)( + 1) +2 + ( + 1) +1 =0
=0 =0 =0

[ ( + 1) +1 ( + 2)( + 1) +2 + ( + 1) +1 ] = 0 or [( + 1) 2 +1 ( + 2)( + 1) +2 ] = 0.
=0 =0

Equating coefficients gives ( + 1)2 +1 ( + 2)( + 1) +2 = 0 for = 0 1 2, . Then the recursion relation is

( + 1) 2 +1 1 2 1 3 1
+2 = +1 = +1 , so given 0 and 1, we have 2 = 2 1
, 3 = 3 2
= 3 1
, 4 = 4 3
= 4 1
, and
( + 2)( + 1) +2

1
in general = , = 1 2 3, . Thus the solution is ( ) = 0 + 1 . Note that the solution can be expressed as
=1

0 1 ln(1 ) for | | 1.

00 2
8. Assuming ( ) = , ( )= ( 1) = ( + 2)( + 1) +2 and
=0 =2 =0

+1 00
( )= = 1 . The equation = becomes
=0 =1

( + 2)( + 1) +2 1 = 0 or 2 2 + [( + 2)( + 1) +2 1] = 0. Equating coefficients

=0 =1 =1

1
gives 2 = 0 and +2 = for = 1 2, . Since 2 = 0, 3 +2 = 0 for =0 1 2 . Given 0,
( + 2)( + 1)
0 3 0 0 1
3 = , 6 = = , , 3 = . Given 1, 4 = ,
3·2 6·5 6·5·3·2 3 (3 1)(3 3)(3 4) · · · · · 6 · 5 · 3 · 2 4·3
4 1 1
7 = = , , 3 +1 = . The solution can be written
7·6 7·6·4·3 (3 + 1)3 (3 2)(3 3) 7·6·4·3

(3 2)(3 5) · · · · · 7 · 4 · 1 3 (3 1)(3 4) · · · · · 8 · 5 · 2 3 +1
as ( ) = 0 + 1 .
=0 (3 )! =0 (3 + 1)!

0 1
9. Let ( ) = . Then ( )= = = ,
=0 =1 =1 =0

00 00 0
( )= ( + 2)( + 1) +2 , and the equation = 0 becomes
=0

[( + 2)( + 1) +2 ] = 0. Thus, the recursion relation is

=0

+ ( + 1)
+2 = = = for = 0 1 2, . One of the given conditions is (0) = 1. But
( + 2)( + 1) ( + 2)( + 1) +2

0 1 2 1 4 1
(0) = (0) = 0 + 0 + 0 + ··· = 0, so 0 = 1. Hence, 2 = = , 4 = = , 6 = = , ,
=0 2 2 4 2·4 6 2·4·6

c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated,
° licated, or posted to a publicly accessible website, in whole or in part.
par
 NOT FOR SALE
728 ¤ CHAPTER 17 SECOND-ORDER DIFFERENTIAL EQUATIONS

1 0 0 1
2 = . The other given condition is (0) = 0. But (0) = (0) = 1 + 0 + 0 + ··· = 1, so 1 = 0.
2 ! =1

1
By the recursion relation, 3 = = 0, 5 = 0, , 2 +1 = 0 for = 0, 1, 2, . Thus, the solution to the initial-value
3
2 2
2 ( 2) 2 2
problem is ( ) = = 2 = = = .
=0 =0 =0 2 ! =0 !

2 +2
10. Assuming that ( ) = , we have = and
=0 =0

00 2 +2 +2
( )= ( 1) = ( + 4)( + 3) +4 =2 2 +6 3 + ( + 4)( + 3) +4 .
=2 = 2 =0

00 2 +2
Thus, the equation + = 0 becomes 2 2 +6 3 + [( + 4)( + 3) +4 + ] = 0. So 2 = 3 = 0 and
=0

0
the recursion relation is +4 = , = 0 1 2, . But 1 = (0) = 0 = 2 = 3 and by the recursion
( + 4)( + 3)

0 1
relation, 4 +1 = 4 +2 = 4 +3 = 0 for = 0 1 2, . Also, 0 = (0) = 1, so 4 = = ,
4·3 4·3

4 ( 1)2 ( 1)
8 = = , , 4 = . Thus, the solution to the initial-value
8·7 8·7·4·3 4 (4 1)(4 4)(4 5) · · · · · 4 · 3
4
4
problem is ( ) = = 0 + 4 =1+ ( 1) .
=0 =0 =1 4 (4 1)(4 4)(4 5) · · · · · 4 · 3

+1 2 0 2 1 +1
11. Assuming that ( ) = , we have = = , = = ,
=0 =0 =0 =1 =0

00 2 +1
( )= ( 1) = ( + 3)( + 2) +3 [replace with + 3]
=2 = 1

+1
=2 2 + ( + 3)( + 2) +3 ,
=0

00 2 0 +1
and the equation + + = 0 becomes 2 2 + [( + 3)( + 2) +3 + + ] = 0. So 2 = 0 and the
=0

( + 1)
recursion relation is +3 = = , = 0 1 2, . But 0 = (0) = 0 = 2 and by the
( + 3)( + 2) ( + 3)( + 2)

0 2 1 2
recursion relation, 3 = 3 +2 = 0 for = 0, 1, 2, . Also, 1 = (0) = 1, so 4 = = ,
4·3 4·3
5 4 2·5 22 52 22 52 · · · · · (3 1)2
7 = = ( 1)2 = ( 1)2 , , 3 +1 = ( 1) . Thus, the solution is
7·6 7·6·4·3 7! (3 + 1)!

22 52 · · · · · (3 1)2 3 +1
( )= = + ( 1) .
=0 =1 (3 + 1)!

2 00 +2
12. (a) Let ( ) = . Then ( )= ( 1) = ( + 2)( + 1) +2 ,
=0 =2 =0

0 +2 +2
( )= = ( + 2) +2 = 1 + ( + 2) +2 , and the equation
=1 = 1 =0

c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
 NOT FOR SALE
CHAPTER 17 REVIEW ¤ 729

2 00 0 2 +2
+ + = 0 becomes 1 + {[( + 2)( + 1) + ( + 2)] +2 + } = 0. So 1 = 0 and the
=0

0
recursion relation is +2 = , = 0 1 2, . But 1 = (0) = 0 so 2 +1 = 0 for =0 1 2 .
( + 2)2
1 2 1 1 4 1
Also, 0 = (0) = 1, so 2 = , 4 = = ( 1)2 = ( 1)2 4 , 6 = = ( 1)3 , ,
22 42 42 22 2 (2!)2 62 26 (3!)2
2
1
2 = ( 1) . The solution is ( ) = = ( 1) .
22 ( !)2 =0 =0 22 ( !)2

(b) The Taylor polynomials 0 to 12 are shown in the graph.

Because 10 and 12 are close together throughout the

interval [ 5 5], it is reasonable to assume that 12 is a good

approximation to the Bessel function on that interval.

17 Review

00 0
1. (a) + + = 0 where , , and are constants.
2
(b) + + =0

(c) If the auxiliary equation has two distinct real roots 1 and 2, the solution is = 1
1
+ 2
2
. If the roots are real and
equal, the solution is = 1 + 2 where is the common root. If the roots are complex, we can write 1 = +
and 2 = , and the solution is = ( 1 cos + 2 sin ).

2. (a) An initial-value problem consists of finding a solution of a second-order differential equation that also satisfies given
0
conditions ( 0) = 0 and ( 0) = 1, where 0 and 1 are constants.

(b) A boundary-value problem consists of finding a solution of a second-order differential equation that also satisfies given
boundary conditions ( 0) = 0 and ( 1) = 1.

00 0
3. (a) + + = ( ) where , , and are constants and is a continuous function.
00 0
(b) The complementary equation is the related homogeneous equation + + = 0. If we find the general solution
of the complementary equation and is any particular solution of the original differential equation, then the general
solution of the original differential equation is ( ) = ( )+ ( ).

(c) See Examples 1–5 and the associated discussion in Section 17.2.

(d) See the discussion on pages 1177–1179 [ ET 1153–1155].

4. Second-order linear differential equations can be used to describe the motion of a vibrating spring or to analyze an electric
circuit; see the discussion in Section 17.3.

5. See Example 1 and the preceding discussion in Section 17.4.

c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated,
° licated, or posted to a publicly accessible website, in whole or in part.
par