17.1 Second-Order Linear Equations
1. The auxiliary equation is 2 − − 6 = 0 ⇒ ( − 3)( + 2) = 0 ⇒ = 3, = −2. Then by (8) the general solution
is = 13 + 2−2.
2. The auxiliary equation is 2 − 6 + 9 = 0
...
17.1 Second-Order Linear Equations
1. The auxiliary equation is 2 − − 6 = 0 ⇒ ( − 3)( + 2) = 0 ⇒ = 3, = −2. Then by (8) the general solution
is = 13 + 2−2.
2. The auxiliary equation is 2 − 6 + 9 = 0 ⇒ ( − 3)2 = 0 ⇒ = 3. Then by (10), the general solution is
= 13 + 23.
3. The auxiliary equation is 2 + 2 = 0 ⇒ = ±√2. Then by (11) the general solution is
= 0 1 cos√2 + 2 sin√2 = 1 cos√2 + 2 sin√2.
4. The auxiliary equation is 2 + − 12 = 0 ⇒ ( − 3)( + 4) = 0 ⇒ = 3, = −4. Then by (8) the general
solution is = 13 + 2−4.
5. The auxiliary equation is 42 + 4 + 1 = 0 ⇒ (2 + 1)2 = 0 ⇒ = − 1 2. Then by (10), the general solution is
= 1−2 + 2−2.
6. The auxiliary equation is 92 + 4 = 0 ⇒ 2 = − 4 9 ⇒ = ± 2 3 , so the general solution is
= 0 1 cos 2 3 + 2 sin 2 3 = 1 cos 2 3 + 2 sin 2 3 .
7. The auxiliary equation is 32 − 4 = (3 − 4) = 0 ⇒ = 0, = 4 3, so = 10 + 243 = 1 + 243.
8. The auxiliary equation is 2 − 1 = ( − 1)( + 1) = 0 ⇒ = 1, = −1. Then the general solution is
= 1 + 2−.
9. The auxiliary equation is 2 − 4 + 13 = 0 ⇒ = 4 ± √−36
2
= 2 ± 3, so = 2(1 cos 3 + 2 sin 3).
10. The auxiliary equation is 32 + 4 − 3 = 0 ⇒ = −4 ± √52
6
=
−2 ± √13
3
, so
= 1(−2+√13 )3 + 2(−2−√13 )3.
11. The auxiliary equation is 22 + 2 − 1 = 0 ⇒ = −2 ± √12
4 =
−1 ± √3
2
, so = 1(−1+√3)2 + 2(−1−√3)2.
12. The auxiliary equation is 2 + 6 + 34 = 0 ⇒ = −6 ± √−100
2
= −3 ± 5, so = −3(1 cos 5 + 2 sin 5).
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13. The auxiliary equation is 32 + 4 + 3 = 0 ⇒ = −4 ± √−20
6
= −
2 3
± √5
3 , so
= −231 cos √35 + 2 sin √35 .
14. The auxiliary equation is 42 − 4 + 1 = (2 − 1)2 = 0 ⇒ = 1 2, so
the general solution is = 12 + 22. We graph the basic solutions
() = 2, () = 2 as well as = 22 + 32,
= −2 − 32, and = 42 − 22. The graphs are all
asymptotic to the -axis as → −∞, and as → ∞ the solutions
approach ±∞.
15. The auxiliary equation is 2 + 2 + 2 = 0 ⇒
=
−2 ± √−4
2
= −1 ± , so the general solution is
= − (1 cos + 2 sin). We graph the basic solutions
() = − cos, () = − sin as well as
= − (−cos − 2sin) and = − (2 cos + 3 sin). All the solutions oscillate with amplitudes that become
arbitrarily large as → −∞ and the solutions are asymptotic to the -axis as → ∞.
16. The auxiliary equation is 22 + − 1 = (2 − 1)( + 1) = 0 ⇒
= 1
2, = −1, so the general solution is = 12 + 2−. We graph
the basic solutions () = 2, () = − as well as = 22 + −,
= −2 − 2−, and = 2 − −. Each solution consists of a single
continuous curve that approaches either 0 or ±∞ as → ±∞.
17. 2 + 3 = 0 ⇒ = ±√3 and the general solution is
= 01 cos√3 + 2 sin√3 = 1 cos√3 + 2 sin√3. Then (0) = 1 ⇒ 1 = 1 and, since
0 = −√31 sin√3 + √32 cos √3, 0(0) = 3 ⇒ √32 = 3 ⇒ 2 = √33 = √3, so the solution to the
initial-value problem is = cos√3 + √3 sin√3.
18. 2 − 2 − 3 = ( − 3)( + 1) = 0, so = 3, = −1 and the general solution is = 13 + 2−. Then
0 = 313 − 2−, so (0) = 2 ⇒ 1 + 2 = 2 and 0(0) = 2 ⇒ 31 − 2 = 2, giving 1 = 1 and 2 = 1. Thus
the solution to the initial-value problem is = 3 + −.
19. 92 + 12 + 4 = (3 + 2)2 = 0 ⇒ = − 2 3 and the general solution is = 1−23 + 2−23. Then (0) = 1 ⇒
1 = 1 and, since 0 = − 2 3 1−23 + 2 1 − 2 3 −23, 0(0) = 0 ⇒ − 2 3 1 + 2 = 0, so 2 = 2 3 and the solution to
the initial-value problem is = −23 + 2 3 −23.
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20. 32 − 2 − 1 = (3 + 1)( − 1) = 0 ⇒ = − 1 3 , = 1 and the general solution is = 1−3 + 2. Then
0 = − 1 3 1−3 + 2, so (0) = 0 ⇒ 1 + 2 = 0 and 0(0) = −4 ⇒ − 1 3 1 + 2 = −4, giving 1 = 3 and
2 = −3. Thus the solution to the initial-value problem is = 3−3 − 3.
21. 2 − 6 + 10 = 0 ⇒ = 3 ± and the general solution is = 3(1 cos + 2 sin). Then 2 = (0) = 1 and
3 = 0(0) = 2 + 31 ⇒ 2 = −3 and the solution to the initial-value problem is = 3(2 cos − 3sin).
22. 42 − 20 + 25 = (2 − 5)2 = 0 ⇒ = 5 2 and the general solution is = 152 + 252. Then 2 = (0) = 1 and
−3 = 0(0) = 5 2 1 + 2 ⇒ 2 = −8. The solution to the initial-value problem is = 252 − 852.
23. 2 − − 12 = ( − 4)( + 3) = 0 ⇒ = 4, = −3 and the general solution is = 14 + 2−3. Then
0 = (1) = 14 + 2−3 and 1 = 0(1) = 414 − 32−3 so 1 = 1 7 −4, 2 = − 1 7 3 and the solution to the initial-value
problem is = 1 7 −44 − 1 7 3−3 = 1 7 4−4 − 1 7 3−3.
24. 42 + 4 + 3 = 0 ⇒ = − 1
2 ± √22 and the general solution is = −2 1 cos √22 + 2 sin √22 . Then
0 = (0) = 1 and 1 = 0(0) = √22 2 − 1 2 1 ⇒ 2 = √2 and the solution to the initial-value problem is
= −2 0 + √2 sin √22 = √2−2 sin √22 .
25. 2 + 16 = 0 ⇒ = ±4 and the general solution is = 1 cos 4 + 2 sin 4. Then −3 = (0) = 1 and
2 = (8) = 2, so the solution of the boundary-value problem is = −3cos 4 + 2 sin 4.
26. 2 + 6 = ( + 6) = 0 ⇒ = 0, = −6 and the general solution is = 1 + 2−6. Then 1 = (0) = 1 + 2
and 0 = (1) = 1 + 2−6 so 1 = 1
1 − 6 , 2 = −
6
1 − 6 . The solution of the boundary-value problem is
=
1
1 − 6 −
6
1 − 6 · −6 = 1 −16 − 16−−66 .
27. 2 + 4 + 4 = ( + 2)2 = 0 ⇒ = −2 and the general solution is = 1−2 + 2−2. Then 2 = (0) = 1 and
0 = (1) = 1−2 + 2−2 so 2 = −2, and the solution of the boundary-value problem is = 2−2 − 2−2.
28. 2 − 8 + 17 = 0 ⇒ = 4 ± and the general solution is = 4(1 cos + 2 sin). But 3 = (0) = 1 and
2 = () = −14 ⇒ 1 = −24, so there is no solution.
29. 2 − = ( − 1) = 0 ⇒ = 0, = 1 and the general solution is = 1 + 2. Then 1 = (0) = 1 + 2
and 2 = (1) = 1 + 2 so 1 = − 2
− 1, 2 =
1
− 1
. The solution of the boundary-value problem is = − 2
− 1
+
− 1
.
30. 42 − 4 + 1 = (2 − 1)2 = 0 ⇒ = 1 2 and the general solution is = 12 + 22. Then 4 = (0) = 1 and
0 = (2) = 1 + 22 ⇒ 2 = −2. The solution of the boundary-value problem is = 42 − 22.
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31. 2 + 4 + 20 = 0 ⇒ = −2 ± 4 and the general solution is = −2(1 cos 4 + 2 sin 4). But 1 = (0) = 1 and
2 = () = 1−2 ⇒ 1 = 22, so there is no solution.
32. 2 + 4 + 20 = 0 ⇒ = −2 ± 4 and the general solution is = −2(1 cos 4 + 2 sin 4). But 1 = (0) = 1 and
−2 = () = 1−2 ⇒ 1 = 1, so 2 can vary and the solution of the boundary-value problem is
= −2(cos 4 + sin 4), where is any constant.
33. (a) Case 1 ( = 0): 00 + = 0 ⇒ 00 = 0 which has an auxiliary equation 2 = 0 ⇒ = 0 ⇒ = 1 + 2
where (0) = 0 and () = 0. Thus, 0 = (0) = 1 and 0 = () = 2 ⇒ 1 = 2 = 0. Thus = 0.
Case 2 ( 0): 00 + = 0 has auxiliary equation 2 = − ⇒ = ±√− [distinct and real since 0] ⇒
= 1
√−
+ 2−
√− where (0) = 0 and () = 0. Thus 0 = (0) = 1 + 2 (∗) and
0 = () = 1√− + 2−√− (†).
Multiplying (∗) by √− and subtracting (†) gives 2√− − −√− = 0 ⇒ 2 = 0 and thus 1 = 0 from (∗).
Thus = 0 for the cases = 0 and 0.
(b) 00 + = 0 has an auxiliary equation 2 + = 0 ⇒ = ± √ ⇒ = 1 cos√ + 2 sin√ where
(0) = 0 and () = 0. Thus, 0 = (0) = 1 and 0 = () = 2 sin√ since 1 = 0. Since we cannot have a trivial
solution, 2 6= 0 and thus sin√ = 0 ⇒ √ = where is an integer ⇒ = 222 and
= 2 sin() where is an integer.
34. The auxiliary equation is 2 + + = 0. If 2 − 4 0, then any solution is of the form () = 11 + 22 where
1 =
− + √2 − 4
2
and 2 = − − √2 − 4
2
. But , , and are all positive so both 1 and 2 are negative and
lim→∞ () = 0. If 2 − 4 = 0, then any solution is of the form () = 1 + 2 where = −(2) 0
since , are positive. Hence lim→∞ () = 0. Finally if 2 − 4 0, then any solution is of the form
() = (1 cos + 2 sin) where = −(2) 0 since and are positive. Thus lim→∞ () = 0.
35. (a) 2 − 2 + 2 = 0 ⇒ = 1 ± and the general solution is = (1 cos + 2 sin). If () = and () = then
(1 cos + 2 sin) = ⇒ 1 cos + 2 sin = − and (1 cos + 2 sin) = ⇒
1 cos + 2 sin = −. This gives a linear system in 1 and 2 which has a unique solution if the lines are not parallel.
If the lines are not vertical or horizontal, we have parallel lines if cos = cos and sin = sin for some nonzero
constant or cos
cos = = sin sin ⇒ cos sin = cos sin ⇒ tan = tan ⇒ − = , any integer. (Note that
none of cos , cos, sin, sin are zero.) If the lines are both horizontal then cos = cos = 0 ⇒ − = , and
similarly vertical lines means sin = sin = 0 ⇒ − = . Thus the system has a unique solution if − 6= .
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(b) The linear system has no solution if the lines are parallel but not identical. From part (a) the lines are parallel if
− = . If the lines are not horizontal, they are identical if − = − ⇒ −
− = = cos cos ⇒
= − cos
cos . (If = 0 then = 0 also.) If they are horizontal then cos = 0, but = sin sin also (and sin 6= 0) so
we require
= − sin sin . Thus the system has no solution if − = and 6= − cos cos unless cos = 0, in
which case
6= − sin sin .
(c) The linear system has infinitely many solution if the lines are identical (and necessarily parallel). From part (b) this occurs
when − = and
= − cos cos unless cos = 0, in which case = − sin sin .
17.2 Nonhomogeneous Linear Equations
1. The auxiliary equation is 2 + 2 − 8 = ( − 2)( + 4) = 0 ⇒ = 2, = −4, so the complementary solution is
() = 12 + 2−4. We try the particular solution () = 2 + + , so 0 = 2 + and 00 = 2.
Substituting into the differential equation, we have (2) + 2(2 + ) − 8(2 + + ) = 1 − 22 or
−82 + (4 − 8) + (2 + 2 − 8) = −22 + 1. Comparing coefficients gives −8 = −2 ⇒
= 1
4, 4 − 8 = 0 ⇒ = 1 8, and 2 + 2 − 8 = 1 ⇒ = − 32 1 , so the general solution is
() = () + () = 12 + 2−4 + 1 4 2 + 1 8 − 32 1 .
2. The auxiliary equation is 2 − 3 = ( − 3) = 0 ⇒ = 0, = 3, so the complementary solution
is () = 1 + 23. We try the particular solution () = cos 2 + sin 2, so
0 = −2sin 2 + 2 cos 2 and 00 = −4cos 2 − 4 sin 2. Substitution into the differential
equation gives (−4cos 2 − 4 sin 2) − 3(−2sin 2 + 2 cos 2) = sin 2 ⇒
(−4 − 6)cos 2 + (6 − 4)sin2 = sin 2. Then −4 − 6 = 0 and 6 − 4 = 1 ⇒ = 26 3 and = − 13 1 .
Thus the general solution is () = () + () = 1 + 23 + 26 3 cos 2 − 13 1 sin 2.
3. The auxiliary equation is 92 + 1 = 0 with roots = ± 1 3 , so the complementary solution is
() = 1 cos(3) + 2 sin(3). Try the particular solution () = 2, so 0 = 22 and 00 = 42.
Substitution into the differential equation gives 942 + 2 = 2 or 372 = 2. Thus 37 = 1 ⇒ = 37 1
and the general solution is () = () + () = 1 cos(3) + 2 sin(3) + 37 1 2.
4. The auxiliary equation is 2 − 2 + 2 = 0 with roots = 1 ± , so the complementary solution is
() = (1 cos + 2 sin). Try the particular solution () = + + , so 0 = + and 00 = .
Substitution into the differential equation gives () − 2( + ) + 2( + + ) = + ⇒
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2 + (−2 + 2) + = + . Comparing coefficients, we have 2 = 1 ⇒ = 1 2 , −2 + 2 = 0 ⇒
= 1
2 , and = 1, so the general solution is () = () + () = (1 cos + 2 sin) + 1 2 + 1 2 + .
5. The auxiliary equation is 2 − 4 + 5 = 0 with roots = 2 ± , so the complementary solution is
() = 2(1 cos + 2 sin). Try () = −, so 0 = −− and 00 = −. Substitution gives
− − 4(−−) + 5(−) = − ⇒ 10− = − ⇒ = 10 1 . Thus the general solution is
() = 2(1 cos + 2 sin) + 10 1 −.
6. The auxiliary equation is 2 − 4 + 4 = ( − 2)2 = 0 ⇒ = 2, so the complementary solution is
() = 12 + 22. For 00 − 40 + 4 = try 1() = + . Then 0 1 = and 001 = 0, and substitution into
the differential equation gives 0 − 4 + 4( + ) = or 4 + (4 − 4) = , so 4 = 1 ⇒ = 1 4 and
4 − 4 = 0 ⇒ = 1
4 . Thus 1() = 1 4 + 1 4 . For 00 − 40 + 4 = −sin try 2() = cos + sin.
Then 0 2 = −sin + cos and 002 = −cos − sin. Substituting, we have
(−cos − sin) − 4(−sin + cos ) + 4(cos + sin) = −sin ⇒
(3 − 4)cos + (4 + 3)sin = −sin. Thus 3 − 4 = 0 and 4 + 3 = −1,
giving = − 25 4 and = − 25 3 , so 2() = − 25 4 cos − 25 3 sin. The general solution is
() = () + 1() + 2() = 12 + 22 + 1 4 + 1 4 − 25 4 cos − 25 3 sin.
7. The auxiliary equation is 2 − 2 + 5 = 0 with roots = 1 ± 2, so the complementary solution is
() = (1 cos 2 + 2 sin 2). Try the particular solution () = cos + sin, so 0 = −sin + cos
and 00 = −cos − sin. Substituting, we have
(−cos − sin) − 2(−sin + cos) + 5(cos + sin) = sin ⇒
(4 − 2)cos + (2 + 4)sin = sin. Then 4 − 2 = 0, 2 + 4 = 1 ⇒ = 10 1 , = 1 5 and the general
solution is () = () + () = (1 cos 2 + 2 sin 2) + 10 1 cos + 1 5 sin. But 1 = (0) = 1 + 10 1 ⇒ 1 = 10 9
and 1 = 0(0) = 22 + 1 + 1 5 ⇒ 2 = − 20 1 . Thus the solution to the initial-value problem is
() = 10 9 cos 2 − 20 1 sin 2 + 10 1 cos + 1 5 sin.
8. The auxiliary equation is 2 − 1 = 0 with roots = ±1, so the complementary solution is () = 1 + 2−. Try the
particular solution () = ( + )2, so 0 = (2 + + 2)2 and 00 = (4 + 4 + 4)2. Substituting, we
have (4 + 4 + 4)2 − ( + )2 = 2 ⇒ (3 + 4 + 3)2 = 2. Then 3 = 1 ⇒ = 1 3 and
4 + 3 = 0 ⇒ = − 4
9 , and the general solution is () = () + () = 1 + 2− + 1 3 − 4 9 2. But
0 = (0) = 1 + 2 − 4 9 and 1 = 0(0) = 1 − 2 − 5 9 ⇒ 1 = 1, 2 = − 5 9 . Thus the solution to the initial-value
problem is () = − 5 9− + 1 3 − 4 9 2.
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9. The auxiliary equation is 2 − = 0 with roots = 0, = 1 so the complementary solution is () = 1 + 2.
Try () = ( + ) so that no term in is a solution of the complementary equation. Then
0 = (2 + (2 + ) + ) and 00 = (2 + (4 + ) + (2 + 2)). Substitution into the differential equation
gives (2 + (4 + ) + (2 + 2)) − (2 + (2 + ) + ) = ⇒ (2 + (2 + )) = ⇒
= 1
2, = −1. Thus () = 1 2 2 − and the general solution is () = 1 + 2 + 1 2 2 − . But
2 = (0) = 1 + 2 and 1 = 0(0) = 2 − 1, so 2 = 2 and 1 = 0. The solution to the initial-value problem is
() = 2 + 1 2 2 − = 1 2 2 − + 2.
10. () = 1 + 2−2. For 00 + 0 − 2 = try 1() = + . Then 0 1 = , 001 = 0, and substitution gives
0 + − 2( + ) = ⇒ = − 1 2, = − 1 4, so 1() = − 1 2 − 1 4. For 00 + 0 − 2 = sin 2 try
2() = cos 2 + sin 2. Then 0 2 = −2sin 2 + 2 cos 2 002 = −4cos 2 − 4 sin 2, and substitution
gives (−4cos 2 − 4 sin 2) + (−2sin 2 + 2 cos 2) − 2(cos 2 + sin 2) = sin 2 ⇒ = − 20 1 ,
= − 3
20. Thus 2() = − 20 1 cos 2 + − 20 3 sin 2 and the general solution is
() = 1 + 2−2 − 1 2 − 1 4 − 20 1 cos 2 − 20 3 sin 2. But 1 = (0) = 1 + 2 − 1 4 − 20 1 and
0 = 0(0) = 1 − 22 − 1 2 − 10 3 ⇒ 1 = 17 15 and 2 = 1 6. Thus the solution to the initial-value problem is
() = 17 15 + 1 6 −2 − 1 2 − 1 4 − 20 1 cos 2 − 20 3 sin 2.
11. The auxiliary equation is 2 + 3 + 2 = ( + 1)( + 2) = 0, so = −1, = −2 and () = 1− + 2−2.
Try = cos + sin ⇒ 0 = −sin + cos, 00 = −cos − sin. Substituting into the differential
equation gives (−cos − sin) + 3(−sin + cos) + 2(cos + sin) = cos or
( + 3)cos + (−3 + )sin = cos . Then solving the equations
+ 3 = 1, −3 + = 0 gives = 10 1 , = 10 3 and the general
solution is () = 1− + 2−2 + 10 1 cos + 10 3 sin. The graph
shows and several other solutions. Notice that all solutions are
asymptotic to as → ∞. Except for , all solutions approach either ∞
or −∞ as → −∞.
12. The auxiliary equation is 2 + 4 = 0 ⇒ = ±2, so () = 1 cos 2 + 2 sin 2. Try = − ⇒
0 = −−, 00 = −. Substituting into the differential equation gives − + 4− = − ⇒
5 = 1 ⇒ = 1
5, so = 1 5 − and the general solution is
() = 1 cos 2 + 2 sin 2 + 1 5 −. We graph along with several
other solutions. All of the solutions except oscillate around = 1 5 −,
and all solutions approach ∞ as → −∞.
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13. Here () = 12 + 2−, and a trial solution is () = ( + ) cos + ( + ) sin.
14. Here () = 1 cos 2 + 2 sin 2. For 00 + 4 = cos 4 try 1() = cos 4 + sin 4 and for 00 + 4 = cos 2 try
2() = ( cos 2 + sin 2) (so that no term of 2 is a solution of the complementary equation). Thus a trial solution
is () = 1() + 2() = cos 4 + sin 4 + cos 2 + sin 2.
15. Here () = 12 + 2. For 00 − 30 + 2 = try 1() = (since = is a solution of the complementary
equation) and for 00 − 30 + 2 = sin try 2() = cos + sin. Thus a trial solution is
() = 1() + 2() = + cos + sin.
16. Since () = 1 + 2−4 try () = (3 + 2 + + ) so that no term of () satisfies the complementary
equation.
17. Since () = −(1 cos 3 + 2 sin 3) we try () = (2 + + )− cos 3 + (2 + + )− sin 3
(so that no term of is a solution of the complementary equation).
18. Here () = 1 cos 2 + 2 sin 2. For 00 + 4 = 3 try 1() = 3 and for 00 + 4 = sin 2 try
2() = ( + )cos 2 + ( + )sin2 (so that no term of 2 is a solution of the complementary equation).
Note: Solving Equations (7) and (9) in The Method of Variation of Parameters gives
0 1 = − 2
(120 − 210 ) and 0 2 = (1 20 −1 210 )
We will use these equations rather than resolving the system in each of the remaining exercises in this section.
19. (a) Here 42 + 1 = 0 ⇒ = ± 1 2 and () = 1 cos 1 2 + 2 sin 1 2 . We try a particular solution of the form
() = cos + sin ⇒ 0 = −sin + cos and 00 = −cos − sin. Then the equation
400 + = cos becomes 4(−cos − sin) + (cos + sin) = cos or
−3cos − 3 sin = cos ⇒ = − 1
3, = 0. Thus, () = − 1 3 cos and the general solution is
() = () + () = 1 cos 1 2 + 2 sin 1 2 − 1 3 cos .
(b) From (a) we know that () = 1 cos 2 + 2 sin 2 . Setting 1 = cos 2 , 2 = sin 2 , we have
120 − 210 = 1 2 cos2 2 + 1 2 sin2 2 = 1 2. Thus 0 1 = −cos sin 2
4 · 1
2
= −
1 2
cos2 · 2 sin 2 = − 1 2 2cos2 2 − 1sin 2
and 0 2 = cos cos 2
4 · 1
2
=
1 2
cos2 · 2 cos 2 = 1 2 1 − 2sin2 2 cos 2 . Then
1() = 1 2 sin 2 − cos2 2 sin 2 = −cos 2 + 2 3 cos3 2 and
2() = 1 2 cos 2 − sin2 2 cos 2 = sin 2 − 2 3 sin3 2 . Thus
() = −cos 2 + 2 3 cos3 2 cos 2 + sin 2 − 2 3 sin3 2 sin 2 = − cos2 2 − sin2 2 + 2 3 cos4 2 − sin4 2
= −cos 2 · 2 + 2 3 cos2 2 + sin2 2 cos2 2 − sin2 2 = −cos + 2 3 cos = − 1 3 cos
and the general solution is () = () + () = 1 cos 2 + 2 sin 2 − 1 3 cos.
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20. (a) Here 2 − 2 − 3 = ( − 3)( + 1) = 0 ⇒ = 3, = −1 and the complementary solution is
() = 13 + 2−. A particular solution is of the form () = + ⇒ 0 = , 00 = 0, and
substituting into the differential equation gives 0 − 2 − 3( + ) = + 2 or −3 + (−2 − 3) = + 2,
so = − 1
3 and −2 − 3 = 2 ⇒ = − 4 9. Thus () = − 1 3 − 4 9 and the general solution is
() = () + () = 13 + 2− − 1 3 − 4 9.
(b) In (a), () = 13 + 2−, so set 1 = 3, 2 = −. Then 120 − 210 = −3− − 33− = −42 so
0 1 = −( + 2)−
−42 =
1 4
( + 2)−3 ⇒ 1() = 1 4 ( + 2)−3 = 1 4 − 1 3( + 2)−3 − 1 9 −3 [by parts]
and 0 2 = ( + 2)3
−42 = −
1 4
( + 2) ⇒ 2() = − 1 4 ( + 2) = − 1 4[( + 2) − ] [by parts].
Hence () = 1 4 − 1 3 − 7 9 −3 3 − 1 4[( + 1)]− = − 1 3 − 4 9 and
() = () + () = 13 + 2− − 1 3 − 4 9.
21. (a) 2 − 2 + 1 = ( − 1)2 = 0 ⇒ = 1, so the complementary solution is () = 1 + 2. A particular solution
is of the form () = 2. Thus 42 − 42 + 2 = 2 ⇒ 2 = 2 ⇒ = 1 ⇒ () = 2.
So a general solution is () = () + () = 1 + 2 + 2.
(b) From (a), () = 1 + 2, so set 1 = , 2 = . Then, 120 − 210 = 2(1 + ) − 2 = 2 and so
0 1 = − ⇒ 1 () = − = −( − 1) [by parts] and 0 2 = ⇒ 2() = = . Hence
() = (1 − )2 + 2 = 2 and the general solution is () = () + () = 1 + 2 + 2.
22. (a) Here 2 − = ( − 1) = 0 ⇒ = 0, 1 and () = 1 + 2 and so we try a particular solution of the form
() = . Thus, after calculating the necessary derivatives, we get 00 − 0 = ⇒
(2 + ) − (1 + ) = ⇒ = 1. Thus () = and the general solution is () = 1 + 2 + .
(b) From (a) we know that () = 1 + 2, so setting 1 = 1, 2 = , then 120 − 210 = − 0 = . Thus
0 1 = −2 = − and 0 2 = = 1. Then 1() = − = − and 2() = . Thus
() = − + and the general solution is () = 1 + 2 − + = 1 + 3 + .
23. As in Example 5, () = 1 sin + 2 cos , so set 1 = sin, 2 = cos. Then 120 − 210 = −sin2 − cos2 = −1,
so 0 1 = −sec2 cos
−1
= sec ⇒ 1() = sec = ln (sec + tan) for 0 2 ,
and 0 2 = sec2 sin
−1
= −sec tan ⇒ 2() = −sec. Hence
() = ln(sec + tan) · sin − sec · cos = sinln(sec + tan) − 1 and the general solution is
() = 1 sin + 2 cos + sin ln(sec + tan) − 1.
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24. As in Exercise 23, () = 1 sin + 2 cos, 1 = sin, 2 = cos, and 120 − 210 = −1. Then
0 1 = −sec3 cos
−1
= sec2 ⇒ 1() = tan and 0 2 = sec3 sin
−1
= −sec2 tan ⇒
2() = − tan sec2 = − 1 2 tan2 . Hence
() = tan sin − 1 2 tan2 cos = tan sin − 1 2 tan sin = 1 2 tan sin and the general solution
is () = 1 sin + 2 cos + 1 2 tan sin.
25. 1 = , 2 = 2 and 120 − 210 = 3. So 0 1 = −2
(1 + −)3 = −
−
1 + −
and
1() = −1 +−− = ln(1 + −). 0 2 = (1 + −)3 = 3 + 2 so
2() = 3 + 2 = ln+ 1 − − = ln(1 + −) − −. Hence
() = ln(1 + −) + 2[ln(1 + −) − −] and the general solution is
() = [1 + ln(1 + −)] + [2 − − + ln(1 + −)]2.
26. 1 = −, 2 = −2 and 120 − 210 = −−3. So 0 1 = −(sin)−2
−−3 = sin
and 0 2 = (sin)−
−−3 = −2 sin. Hence 1 () = sin = −cos and
2() = −2 sin = cos − sin. Then () = −− cos − −2[sin − cos]
and the general solution is () = (1 − cos)− + [2 − sin + cos]−2.
27. 2 − 2 + 1 = ( − 1)2 = 0 ⇒ = 1 so () = 1 + 2. Thus 1 = , 2 = and
120 − 210 = ( + 1) − = 2. So 0 1 = − · (1 + 2)
2 = −
1 + 2 ⇒
1 = − 1 +2 = −12 ln1 + 2, 0 2 = · 2(1 + 2) = 1 +12 ⇒ 2 = 1 +12 = tan−1 and
() = − 1 2 ln(1 + 2) + tan−1 . Hence the general solution is () = 1 + 2 − 1 2 ln(1 + 2) + tan−1 .
28. 1 = −2, 2 = −2 and 120 − 210 = −4. Then 0 1 = −−2−2
3−4 = −
1 2
so 1() = −1 and
0 2 = −2−2
3−4 =
1 3
so 2() = − 1
22 . Thus () = −2 − 2−22 = − 22 and the general solution is
() = −2[1 + 2 + 1(2)].
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17.3 Applications of Second-Order Differential Equations
1. By Hooke’s Law (025) = 25 so = 100 is the spring constant and the differential equation is 500 + 100 = 0.
The auxiliary equation is 52 + 100 = 0 with roots = ±2√5, so the general solution to the differential equation is
() = 1 cos2√5 + 2 sin2√5. We are given that (0) = 035 ⇒ 1 = 035 and 0(0) = 0 ⇒
2√52 = 0 ⇒ 2 = 0, so the position of the mass after seconds is () = 035 cos2√5.
2. By Hooke’s Law (04) = 32 so = 032 4 = 80 is the spring constant and the differential equation is 800 + 80 = 0.
The general solution is () = 1 cos√10 + 2 sin√10. But 0 = (0) = 1 and 1 = 0(0) = √102 ⇒
2 = √110, so the position of the mass after seconds is () = √110 sin√10.
3. (05) = 6 or = 12 is the spring constant, so the initial-value problem is 200 + 140 + 12 = 0, (0) = 1, 0(0) = 0.
The general solution is () = 1−6 + 2−. But 1 = (0) = 1 + 2 and 0 = 0(0) = −61 − 2. Thus the position is
given by () = − 1 5−6 + 6 5−.
4. (a) (025) = 13 ⇒ = 52, so the differential equation is
200 + 80 + 52 = 0 with general solution
() = −21 cos√22 + 2 sin√22. Then 0 = (0) = 1
and 05 = 0(0) = √222 ⇒ 2 = 2 √122, so the position is
given by () = 2 √122−2 sin√22.
(b)
5. For critical damping we need 2 − 4 = 0 or = 2(4) = 142(4 · 12) = 49 12 kg.
6. For critical damping we need 2 = 4 or = 2√ = 2√2 · 52 = 4√26.
7. We are given = 1, = 100, (0) = −01 and 0(0) = 0. From (3), the differential equation is 2
2 +
+ 100 = 0
with auxiliary equation 2 + + 100 = 0.
If = 10, we have two complex roots = −5 ± 5√3, so the motion is underdamped and the solution is
= −51 cos5√3 + 2 sin5√3. Then −01 = (0) = 1 and 0 = 0(0) = 5√32 − 51 ⇒ 2 = − 101√3,
so = −5−01cos5√3 − 101√3 sin5√3.
If = 15, we again have underdamping since the auxiliary equation has roots = − 15 2 ± 5 √2 7. The general solution is
= −1521 cos 5 √2 7 + 2 sin 5 √2 7, so −01 = (0) = 1 and 0 = 0(0) = 5 √2 72 − 15 2 1 ⇒ 2 = − 103√7.
Thus = −152−01cos 5 √2 7 − 103√7 sin 5 √2 7.
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For = 20, we have equal roots 1 = 2 = −10, so the oscillation is critically damped and the solution is
= (1 + 2)−10. Then −01 = (0) = 1 and 0 = 0(0) = −101 + 2 ⇒ 2 = −1, so = (−01 − )−10.
If = 25 the auxiliary equation has roots 1 = −5, 2 = −20, so we have overdamping and the solution is
= 1−5 + 2−20. Then −01 = (0) = 1 + 2 and 0 = 0(0) = −51 − 202 ⇒ 1 = − 15 2 and 2 = 30 1 ,
so = − 2
15 −5 + 30 1 −20.
If = 30 we have roots = −15 ± 5√5, so the motion is
overdamped and the solution is = 1(−15 + 5 √5) + 2(−15 − 5 √5).
Then −01 = (0) = 1 + 2 and
0 = 0(0) = −15 + 5√5 1 + −15 − 5√5 2 ⇒
1 = −5 − 3 √5
100 and 2 = −5 + 3 100√5, so
= −5 100 − 3 √5 (−15 + 5 √5) + −5 + 3 100√5 (−15 − 5 √5).
8. We are given = 1, = 10, (0) = 0 and 0(0) = 1. The differential equation is 2
2 + 10 + = 0 with auxiliary
equation 2 + 10 + = 0. = 10: the auxiliary equation has roots = −5 ± √15 so we have overdamping and the
solution is = 1(−5 + √15) + 2(−5 − √15). Entering the initial conditions gives 1 = 2 √115 and 2 = − 2 √115, so
= 1
2 √15 (−5 + √15) − 2 √115 (−5 − √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 = 2 √1 5 and 2 = − 2 √1 5, so = 2 √1 5 (−5 + √5) − 2 √1 5 (−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 = −51 cos√5 + 2 sin√5.
The initial conditions give 1 = 0 and 2 = √15, so = √15 −5 sin√5.
= 40: = −5 ± √15 so we again have underdamping.
The solution is = −51 cos√15 + 2 sin√15,
and the initial conditions give 1 = 0 and 2 = √115.
Thus = √115 −5 sin√15.
9. The differential equation is 00 + = 0 cos0 and 0 6= = . Here the auxiliary equation is 2 + = 0
with roots ± = ± so () = 1 cos + 2 sin. Since 0 6= , try () = cos 0 + sin0.
Then we need ()−2 0(cos0 + sin0) + (cos0 + sin0) = 0 cos0 or − 2 0 = 0 and
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− 2 0 = 0. Hence = 0 and = − 0 2
0
=
0
(2 − 2 0) since 2 = . Thus the motion of the mass is given
by () = 1 cos + 2 sin + 0
(2 − 2 0) cos0.
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)cos − (2 + 2)sin + cos + sin = 0 cos or 2 = 0 and
−2 = 0 [noting −2 + = 0 and −2 + = 0 since 2 = ]. Hence the general solution is
() = 1 cos + 2 sin + [0(2)] sin.
11. From Equation 6, () = () + () where () = 1 cos + 2 sin and () = 0
(2 − 2 0) cos0. Then
is periodic, with period 2 , and if 6= 0, is periodic with period 20 . If 0 is a rational number, then we can say
0
=
⇒ =
0
where and are non-zero integers. Then
+ · 2 = + · 2 + + · 2 = () + + 0 · 2 = () + + · 20 = () + () = ()
so () is periodic.
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 11 + 22 = 0 ⇔ 22 = −11 ⇔
2 = −1
1
2 = −1(1−2). But 1 2 ⇒ 1 − 2 0 and since 0, (1−2) 1. Thus
|2| = |1|(1−2) |1|, and the graph of can cross the -axis only if |2| |1|.
13. Here the initial-value problem for the charge is 00 + 200 + 500 = 12, (0) = 0(0) = 0. Then
() = −10(1 cos 20 + 2 sin 20) and try () = ⇒ 500 = 12 or = 125 3 .
The general solution is () = −10(1 cos 20 + 2 sin 20) + 125 3 . But 0 = (0) = 1 + 125 3 and
0() = () = −10[(−101 + 202) cos 20 + (−102 − 201)sin20] but 0 = 0(0) = −101 + 202. Thus the charge
is () = − 250 1 −10(6 cos 20 + 3 sin 20) + 125 3 and the current is () = −10 3 5 sin 20.
14. (a) Here the initial-value problem for the charge is 200 + 240 + 200 = 12 with (0) = 0001 and 0(0) = 0.
Then () = −6(1 cos 8 + 2 sin 8) and try () = ⇒ = 50 3 and the general solution is
() = −6(1 cos 8 + 2 sin 8) + 50 3 . But 0001 = (0) = + 50 3 so 1 = −0059. Also
0() = () = −6[(−61 + 82)cos 8 + (−62 − 81)sin8] and 0 = 0(0) = −61 + 82 so
2 = −004425. Hence the charge is () = −−6(0059 cos 8 + 004425 sin 8) + 50 3 and the current is
() = −6(07375) sin 8.
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© Cengage Learning. All Rights Reserved.732 ¤ CHAPTER 17 SECOND-ORDER DIFFERENTIAL EQUATIONS
(b)
15. As in Exercise 13, () = −10(1 cos20 + 2 sin20) but () = 12sin10 so try
() = cos10 + sin10. Substituting into the differential equation gives
(−100 + 200 + 500) cos 10 + (−100 − 200 + 500)sin 10 = 12 sin 10 ⇒
400 + 200 = 0 and 400 − 200 = 12. Thus = − 3
250, = 125 3 and the general solution is
() = −10(1 cos 20 + 2 sin 20) − 250 3 cos 10 + 125 3 sin 10. But 0 = (0) = 1 − 250 3 so 1 = 250 3 .
Also 0() = 25 3 sin 10 + 25 6 cos 10 + −10[(−101 + 202)cos 20 + (−102 − 201)sin 20] and
0 = 0(0) = 25 6 − 101 + 202 so 2 = − 500 3 . Hence the charge is given by
() = −10 250 3 cos 20 − 500 3 sin 20 − 250 3 cos 10 + 125 3 sin 10.
16. (a) As in Exercise 14, () = −6(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)sin10 = 12 sin 10,
so = 0 and = − 1
20. Hence, the general solution is () = −6(1 cos 8 + 2 sin 8) − 20 1 cos 10. But
0001 = (0) = 1 − 20 1 , 0() = −6[(−61 + 82)cos 8 + (−62 − 81)sin8] − 1 2 sin 10 and
0 = 0(0) = −61 + 82, so 1 = 0051 and 2 = 003825. Thus the charge is given by
() = −6(0051 cos 8 + 003825 sin 8) − 20 1 cos 10.
(b)
17. () = cos( + ) ⇔ () = [coscos − sinsin] ⇔ () = 1 cos + 2 sin where
cos = 1 and sin = −2 ⇔ () = 1 cos + 2 sin. [Note that cos2 + sin2 = 1 ⇒ 2 1 + 2 2 = 2.]
18. (a) We approximate sin by and, with = 1 and = 98, the differential equation becomes 2
2 + 98 = 0. The auxiliary
equation is 2 + 98 = 0 ⇒ = ±√98, so the general solution is () = 1 cos√98 + 2 sin√98 .
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Then 02 = (0) = 1 and 1 = 0(0) = √982 ⇒ 2 = √1 98, so the equation is
() = 02cos√98 + √198 sin√98 .
(b) 0() = −02√98sin√98 + cos√98 = 0 or tan√98 = √598, so the critical numbers are
= √198 tan−1 √5 98 + √98 ( any integer). The maximum angle from the vertical is
√198 tan−1 √5 98 ≈ 0377 radians (or about 217◦).
(c) From part (b), the critical numbers of () are spaced √98 apart, and the time between successive maximum values
is 2 √98 . Thus the period of the pendulum is √298 ≈ 2007 seconds.
(d) () = 0 ⇒ 02cos√98 + √198 sin√98 = 0 ⇒ tan√98 = −02√98 ⇒
= √1 98 tan−1−02√98 + ≈ 0825 seconds.
(e) 0(0825) ≈ −1180 rads.
17.4 Series Solutions
1. Let () =
∞
=0
. Then 0() =
∞
=1
−1 and the given equation, 0 − = 0, becomes
∞
=1
−1 −
∞
=0
= 0. Replacing by + 1 in the first sum gives
∞
=0
( + 1)+1 −
∞
=0
= 0, so
∞
=0
[( + 1)+1 − ] = 0. Equating coefficients gives ( + 1)+1 − = 0, so the recursion relation is
+1 =
+ 1, = 0 1 2 . Then 1 = 0, 2 = 121 = 20 , 3 = 132 = 13 · 120 = 3! 0 , 4 = 143 = 4! 0 , and
in general, = 0
!. Thus, the solution is () =
∞
=0
=
∞
=0
0
! = 0
∞
=0
! = 0.
2. Let () =
∞
= 0
. Then 0 = ⇒ 0 − = 0 ⇒
∞
= 1
−1 −
∞
= 0
= 0 or
∞
= 1
−1 −
∞
= 0
+1 = 0. Replacing with + 1 in the first sum and with − 1 in the second
gives
∞
= 0
( + 1)+1 −
∞
= 1
−1 = 0 or 1 +
∞
= 1
( + 1)+1 −
∞
= 1
−1 = 0. Thus,
1 +
∞
= 1
[( + 1)+1 − −1] = 0. Equating coefficients gives 1 = 0 and ( + 1) +1 − −1 = 0. Thus, the
recursion relation is +1 = −1
+ 1, = 12, . But 1 = 0, so 3 = 0 and 5 = 0 and in general 2+1 = 0. Also,
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© Cengage Learning. All Rights Reserved.734 ¤ CHAPTER 17 SECOND-ORDER DIFFERENTIAL EQUATIONS
2 =
0
2 , 4 =
2
4 =
0
4 · 2 =
0
22 · 2!, 6 =
4
6
=
0
6 · 4 · 2 =
0
23 · 3! and in general 2 = 2·0!. Thus, the solution
is () =
∞
= 0
=
∞
= 0
22 =
∞
= 0
0
2 · ! 2 = 0
∞
= 0
22
! = 022.
3. Assuming () =
∞
= 0
, we have 0() =
∞
= 1
−1 =
∞
= 0
( + 1)+1 and
−2 = −
∞
= 0
+2 = −
∞
= 2
−2. Hence, the equation 0 = 2 becomes
∞
= 0
( + 1)+1 −
∞
= 2
−2 = 0
or 1 + 22 +
∞
= 2
[( + 1)+1 − −2] = 0. Equating coefficients gives 1 = 2 = 0 and +1 = −2
+ 1
for = 23, . But 1 = 0, so 4 = 0 and 7 = 0 and in general 3+1 = 0. Similarly 2 = 0 so 3+2 = 0. Finally
3 =
0
3 , 6 =
3
6
=
0
6 · 3
=
0
32 · 2!, 9 =
6
9
=
0
9 · 6 · 3
=
0
33 · 3!, , and 3 = 3·0!. Thus, the solution
is () =
∞
= 0
=
∞
= 0
33 =
∞
= 0
0
3 · !3 = 0
∞
= 0
3
3! = 0
∞
= 0
33
! = 033.
4. Let () =
∞
=0
⇒ 0 () =
∞
=1
−1 =
∞
=0
( + 1)+1. Then the differential equation becomes
( − 3)
∞
=0
( + 1)+1 + 2
∞
=0
= 0 ⇒
∞
=0
( + 1)+1+1 − 3
∞
=0
( + 1)+1 + 2
∞
=0
= 0 ⇒
∞
=1
−
∞
=0
3( + 1)+1 +
∞
=0
2 = 0 ⇒
∞
=0
[( + 2) − 3( + 1)+1] = 0
since ∞=1 = ∞=0 . Equating coefficients gives ( + 2) − 3( + 1)+1 = 0, thus the recursion relation is
+1 =
( + 2)
3( + 1) , = 012 . Then 1 = 230 , 2 = 3(2) 31 = 3320 , 3 = 3(3) 42 = 4330 , 4 = 3(4) 53 = 5340 , and
in general, = ( + 1)0
3
. Thus the solution is () =
∞
=0
= 0
∞
=0
+ 1
3
.
Note that 0 ∞=0 3+ 1 = (3 9−0)2 for || 3.
5. Let () =
∞
=0
⇒ 0 () =
∞
=1
−1 and 00 () =
∞
=0
( + 2)( + 1)+2. The differential equation
becomes
∞
=0
( + 2)( + 1)+2 +
∞
=1
−1 +
∞
=0
= 0 or
∞
=0
[( + 2)( + 1)+2 + + ] = 0
since ∞=1 = ∞=0 . Equating coefficients gives ( + 2)( + 1)+2 + ( + 1) = 0, thus the recursion
relation is +2 = −( + 1)
( + 2)( + 1) = −
+ 2, = 012 . Then the even
coefficients are given by 2 = −0
2 , 4 = −
2
4 =
0
2 · 4, 6 = −
4
6
= −
0
2 · 4 · 6, and in general,
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© Cengage Learning. All Rights Reserved.SECTION 17.4 SERIES SOLUTIONS ¤ 735
2 = (−1) 0
2 · 4 · · · · · 2 =
(−1)0
2 ! . The odd coefficients are 3 = −31 , 5 = −53 = 3·15, 7 = −75 = −3 ·51· 7,
and in general, 2+1 = (−1) 1
3 · 5 · 7 · · · · · (2 + 1) =
(−2) !1
(2 + 1)! . The solution is
() = 0
∞
=0
(−1)
2 ! 2 + 1
∞
=0
(−2) !
(2 + 1)! 2+1.
6. Let () =
∞
= 0
. Then 00() =
∞
= 2
( − 1)−2 =
∞
= 0
( + 2)( + 1)+2. Hence, the equation 00 =
becomes
∞
= 0
( + 2)( + 1)+2 −
∞
= 0
= 0 or
∞
= 0
[( + 2)( + 1)+2 − ] = 0. So the recursion relation
is +2 =
( + 2)( + 1), = 0 1 . Given 0 and 1, 2 = 2·01, 4 = 4·23 = 4! 0 , 6 = 6·45 = 6! 0 , ,
2 =
0
(2)! and 3 = 3·12, 5 = 5·34 = 5 · 4·13 · 2 = 5! 1 , 7 = 7·56 = 7! 1 , , 2+1 = (2+ 1)! 1 . Thus, the solution
is () =
∞
= 0
=
∞
= 0
22 +
∞
= 0
2+12+1 = 0
∞
= 0
2
(2)! + 1
∞
= 0
2+1
(2 + 1)!. The solution can be written
as () = 0 cosh + 1 sinh or () = 0 +2− + 1 −2− = 0 +2 1 + 0 −2 1 −.
7. Let () =
∞
=0
⇒ 0 () =
∞
=1
−1 =
∞
=0
( + 1)+1 and 00 () =
∞
=0
( + 2)( + 1)+2. Then
(−1)00() =
∞
=0
(+2)(+1)+2+1−
∞
=0
(+2)(+1)+2 =
∞
=1
(+1)+1−
∞
=0
(+2)(+1)+2.
Since
∞
=1
( + 1)+1 =
∞
=0
( + 1)+1, the differential equation becomes
∞
=0
( + 1)+1 −
∞
=0
( + 2)( + 1)+2 +
∞
=0
( + 1)+1 = 0 ⇒
∞
=0
[( + 1)+1 − ( + 2)( + 1)+2 + ( + 1)+1] = 0 or
∞
=0
[( + 1)2+1 − ( + 2)( + 1)+2] = 0.
Equating coefficients gives ( + 1)2+1 − ( + 2)( + 1)+2 = 0 for = 012, . Then the recursion relation is
+2 =
( + 1)2
( + 2)( + 1)+1 =
+ 1
+ 2
+1, so given 0 and 1, we have 2 = 1 2 1, 3 = 2 3 2 = 1 3 1, 4 = 3 4 3 = 1 4 1, and
in general = 1
, = 123, . Thus the solution is () = 0 + 1
∞
=1
. Note that the solution can be expressed as
0 − 1 ln(1 − ) for || 1.
8. Assuming () =
∞
= 0
, 00() =
∞
= 2
( − 1)−2 =
∞
= 0
( + 2)( + 1)+2 and
−() = −
∞
= 0
+1 = −
∞
= 1
−1. The equation 00 = becomes
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© Cengage Learning. All Rights Reserved.736 ¤ CHAPTER 17 SECOND-ORDER DIFFERENTIAL EQUATIONS
∞
= 0
( + 2)( + 1)+2 −
∞
= 1
−1 = 0 or 22 +
∞
= 1
[( + 2)( + 1)+2 − −1] = 0. Equating coefficients
gives 2 = 0 and +2 = −1
( + 2)( + 1) for = 1 2, . Since 2 = 0, 3+2 = 0 for = 0 1 2 . Given 0,
3 =
0
3 · 2, 6 =
3
6 · 5
=
0
6 · 5 · 3 · 2, , 3 =
0
3(3 − 1)(3 − 3)(3 − 4) · · · · · 6 · 5 · 3 · 2. Given 1, 4 = 4·13,
7 =
4
7 · 6 =
1
7 · 6 · 4 · 3, , 3+1 =
1
(3 + 1)3(3 − 2)(3 − 3) 7 · 6 · 4 · 3. The solution can be written
as () = 0
∞
= 0
(3 − 2)(3 − 5) · · · · · 7 · 4 · 1
(3)! 3 + 1
∞
= 0
(3 − 1)(3 − 4) · · · · · 8 · 5 · 2
(3 + 1)! 3+1.
9. Let () =
∞
= 0
. Then −0() = −
∞
= 1
−1 = −
∞
= 1
= −
∞
= 0
,
00() =
∞
= 0
( + 2)( + 1)+2, and the equation 00 − 0 − = 0 becomes
∞
= 0
[( + 2)( + 1)+2 − − ] = 0. Thus, the recursion relation is
+2 =
+
( + 2)( + 1) =
( + 1)
( + 2)( + 1) =
+ 2
for = 012, . One of the given conditions is (0) = 1. But
(0) =
∞
=0
(0) = 0 + 0 + 0 + · · · = 0, so 0 = 1. Hence, 2 = 0
2
=
1 2
, 4 =
2
4 =
1
2 · 4, 6 =
4
6
=
1
2 · 4 · 6, ,
2 =
1
2!. The other given condition is 0(0) = 0. But 0(0) =
∞
=1
(0)−1 = 1 + 0 + 0 + · · · = 1, so 1 = 0.
By the recursion relation, 3 = 1
3
= 0, 5 = 0, , 2+1 = 0 for = 0, 1, 2, . Thus, the solution to the initial-value
problem is () =
∞
= 0
=
∞
= 0
22 =
∞
= 0
2
2! =
∞
= 0
(22)
! = 22.
10. Assuming that () =
∞
= 0
, we have 2 =
∞
= 0
+2 and
00() =
∞
= 2
( − 1)−2 =
∞
=−2
( + 4)( + 3)+4+2 = 22 + 63 +
∞
= 0
( + 4)( + 3)+4+2.
Thus, the equation 00 + 2 = 0 becomes 22 + 63 +
∞
= 0
[( + 4)( + 3)+4 + ]+2 = 0. So 2 = 3 = 0 and
the recursion relation is +4 = −
( + 4)( + 3), = 01 2, . But 1 = 0(0) = 0 = 2 = 3 and by the recursion
relation, 4+1 = 4+2 = 4+3 = 0 for = 012, . Also, 0 = (0) = 1, so 4 = − 0
4 · 3 = −
1
4 · 3,
8 = −
4
8 · 7 =
(−1)2
8 · 7 · 4 · 3, , 4 =
(−1)
4(4 − 1)(4 − 4)(4 − 5) · · · · · 4 · 3. Thus, the solution to the initial-value
problem is () =
∞
= 0
= 0 +
∞
= 0
44 = 1 +
∞
= 1
(−1) 4
4(4 − 1)(4 − 4)(4 − 5) · · · · · 4 · 3.
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© Cengage Learning. All Rights Reserved.SECTION 17.4 SERIES SOLUTIONS ¤ 737
11. Assuming that () =
∞
= 0
, we have =
∞
= 0
=
∞
= 0
+1, 20 = 2
∞
= 1
−1 =
∞
= 0
+1,
00() =
∞
= 2
( − 1)−2 =
∞
=−1
( + 3)( + 2)+3+1 [replace with + 3]
= 22 +
∞
= 0
( + 3)( + 2)+3+1,
and the equation 00 + 20 + = 0 becomes 22 +
∞
= 0
[( + 3)( + 2)+3 + + ]+1 = 0. So 2 = 0 and the
recursion relation is +3 = − −
( + 3)( + 2) = −
( + 1)
( + 3)( + 2), = 012, . But 0 = (0) = 0 = 2 and by the
recursion relation, 3 = 3+2 = 0 for = 0, 1, 2, . Also, 1 = 0(0) = 1, so 4 = − 21
4 · 3 = −
2
4 · 3,
7 = −
54
7 · 6 = (−1)2 7 · 2 6 · · 5 4 · 3 = (−1)2 227! 52 , , 3+1 = (−1) 2252 · · · · · (3 + 1)! (3 − 1)2 . Thus, the solution is
() =
∞
= 0
= +
∞
= 1 (−1) 2252 · · · · · (3(3+ 1)! − 1)23+1 .
12. (a) Let () =
∞
= 0
. Then 200() =
∞
= 2
( − 1) =
∞
= 0
( + 2)( + 1)+2+2,
0() =
∞
= 1
=
∞
=−1
( + 2)+2+2 = 1 +
∞
= 0
( + 2)+2+2, and the equation
200 + 0 + 2 = 0 becomes 1 +
∞
= 0
{[( + 2)( + 1) + ( + 2)]+2 + }+2 = 0. So 1 = 0 and the
recursion relation is +2 = −
( + 2)2 , = 012, . But 1 = 0(0) = 0 so 2+1 = 0 for = 0 1 2 .
Also, 0 = (0) = 1, so 2 = − 1
22 , 4 = −
2
42 = (−1)2 42122 = (−1)2 24 (2!) 1 2 , 6 = −642 = (−1)3 26 (3!) 1 2 , ,
2 = (−1) 1
22 (!)2 . The solution is () =
∞
= 0
=
∞
= 0
(−1) 2
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.
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© Cengage Learning. All Rights Reserved.738 ¤ CHAPTER 17 SECOND-ORDER DIFFERENTIAL EQUATIONS
17 Review
1. True. See Theorem 17.1.3.
2. False. The differential equation is not homogeneous.
3. True. cosh and sinh are linearly independent solutions of this linear homogeneous equation.
4. False. = is a solution of the complementary equation, so we have to take () = .
1. The auxiliary equation is 42 − 1 = 0 ⇒ (2 + 1)(2 − 1) = 0 ⇒ = ± 1 2. Then the general solution
is = 12 + 2−2.
2. The auxiliary equation is 2 − 2 + 10 = 0 ⇒ = 1 ± 3, so = (1 cos 3 + 2 sin 3).
3. The auxiliary equation is 2 + 3 = 0 ⇒ = ±√3. Then the general solution is = 1 cos√3 + 2 sin√3.
4. The auxiliary equation is 2 + 8 + 16 = 0 ⇒ ( + 4)2 = 0 ⇒ = −4, so the general solution is
= 1−4 + 2−4.
5. 2 − 4 + 5 = 0 ⇒ = 2 ± , so () = 2(1 cos + 2 sin). Try () = 2 ⇒ 0 = 22
and 00 = 42. Substitution into the differential equation gives 42 − 82 + 52 = 2 ⇒ = 1 and
the general solution is () = 2(1 cos + 2 sin) + 2.
6. 2 + − 2 = 0 ⇒ = 1, = −2 and () = 1 + 2−2. Try () = 2 + + ⇒ 0 = 2 +
and 00 = 2. Substitution gives 2 + 2 + − 22 − 2 − 2 = 2 ⇒ = = − 1 2, = − 3 4 so the
general solution is () = 1 + 2−2 − 1 22 − 1 2 − 3 4.
7. 2 − 2 + 1 = 0 ⇒ = 1 and () = 1 + 2. Try () = ( + )cos + ( + )sin ⇒
0 = ( − − )sin + ( + + )cos and 00 = (2 − − )cos + (−2 − − )sin. Substitution
gives (−2 + 2 − 2 − 2)cos + (2 − 2 + 2 − 2)sin = cos ⇒ = 0, = = = − 1 2.
The general solution is () = 1 + 2 − 1 2 cos − 1 2( + 1) sin.
8. 2 + 4 = 0 ⇒ = ±2 and () = 1 cos 2 + 2 sin 2. Try () = cos 2 + sin 2 so that no term
of is a solution of the complementary equation. Then 0 = ( + 2)cos 2 + ( − 2)sin2 and
00 = (4 − 4)cos 2 + (−4 − 4)sin 2. Substitution gives 4 cos 2 − 4sin 2 = sin 2 ⇒
= − 1
4 and = 0. The general solution is () = 1 cos 2 + 2 sin 2 − 1 4cos 2.
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© Cengage Learning. All Rights Reserved.CHAPTER 17 REVIEW ¤ 739
9. 2 − − 6 = 0 ⇒ = −2, = 3 and () = 1−2 + 23. For 00 − 0 − 6 = 1, try 1() = . Then
0 1() = 001 () = 0 and substitution into the differential equation gives = − 1 6 . For 00 − 0 − 6 = −2 try
2() = −2 [since = −2 satisfies the complementary equation]. Then 0 2 = ( − 2)−2 and
002 = (4 − 4)−2, and substitution gives −5−2 = −2 ⇒ = − 1 5 . The general solution then is
() = 1−2 + 23 + 1() + 2() = 1−2 + 23 − 1 6 − 1 5 −2.
10. Using variation of parameters, () = 1 cos + 2 sin, 0 1() = −cscsin = −1 ⇒ 1() = −, and
0 2() = csccos
= cot ⇒ 2() = ln|sin| ⇒ = −cos + sinln|sin|. The solution is
() = (1 − )cos + (2 + ln|sin|)sin.
11. The auxiliary equation is 2 + 6 = 0 and the general solution is () = 1 + 2−6 = 1 + 2−6(−1). But
3 = (1) = 1 + 2 and 12 = 0(1) = −62. Thus 2 = −2, 1 = 5 and the solution is () = 5 − 2−6(−1).
12. The auxiliary equation is 2 − 6 + 25 = 0 and the general solution is () = 3(1 cos 4 + 2 sin 4). But
2 = (0) = 1 and 1 = 0(0) = 31 + 42. Thus the solution is () = 32cos 4 − 5 4 sin 4.
13. The auxiliary equation is 2 − 5 + 4 = 0 and the general solution is () = 1 + 24. But 0 = (0) = 1 + 2
and 1 = 0(0) = 1 + 42, so the solution is () = 1 3(4 − ).
14. () = 1 cos(3) + 2 sin(3). For 900 + = 3, try 1() = + . Then 1() = 3. For 900 + = −,
try 2() = −. Then 9− + − = − or 2() = 10 1 −. Thus the general solution is
() = 1 cos(3) + 2 sin(3) + 3 + 10 1 −. But 1 = (0) = 1 + 10 1 and 2 = 0(0) = 1 3 2 + 3 − 10 1 , so
1 = 9
10 and 2 = − 27 10 . Hence the solution is () = 10 1 [9 cos(3) − 27 sin(3)] + 3 + 10 1 −.
15. 2 + 4 + 29 = 0 ⇒ = −2 ± 5 and the general solution is = −2(1 cos 5 + 2 sin 5). But 1 = (0) = 1 and
−1 = () = −1−2 ⇒ 1 = 2, so there is no solution.
16. 2 + 4 + 29 = 0 ⇒ = −2 ± 5 and the general solution is = −2(1 cos 5 + 2 sin 5). But 1 = (0) = 1 and
−−2 = () = −1−2 ⇒ 1 = 1, so 2 can vary and the solution of the boundary-value problem is
= −2(cos 5 + sin 5), where is any constant.
17. Let () =
∞
=0
. Then 00 () =
∞
=0
( − 1)−2 =
∞
=0
( + 2)( + 1)+2 and the differential equation
becomes
∞
=0
[( + 2)( + 1)+2 + ( + 1)] = 0. Thus the recursion relation is +2 = −( + 2)
for = 012, . But 0 = (0) = 0, so 2 = 0 for = 012, . Also 1 = 0(0) = 1, so 3 = −1
3, 5 =
(−1)2
3 · 5 ,
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© Cengage Learning. All Rights Reserved.740 ¤ CHAPTER 17 SECOND-ORDER DIFFERENTIAL EQUATIONS
7 =
(−1)3
3 · 5 · 7 =
(−1)3233!
7! , , 2+1 =
(−1) 2 !
(2 + 1)! for = 0 1 2 . Thus the solution to the initial-value problem
is () =
∞
=0
=
∞
=0
(−1) 2 !
(2 + 1)! 2+1.
18. Let () =
∞
=0
. Then 00 () =
∞
=0
( − 1)−2 =
∞
=0
( + 2)( + 1)+2 and the differential equation
becomes
∞
=0
[( + 2)( + 1)+2 − ( + 2)] = 0. Thus the recursion relation is +2 =
+ 1
for
= 0, 1, 2, . Given 0 and 1, we have 2 = 0
1 , 4 =
2
3
=
0
1 · 3, 6 =
4
5
=
0
1 · 3 · 5, ,
2 =
0
1 · 3 · 5 · · · · · (2 − 1) = 0
2−1( − 1)!
(2 − 1)! . Similarly 3 = 21 , 5 = 43 = 2·14,
7 =
5
6
=
1
2 · 4 · 6, , 2+1 =
1
2 · 4 · 6 · · · · · 2 =
1
2 !. Thus the general solution is
() =
∞
=0
= 0 + 0
∞
=1
2−1( − 1)! 2
(2 − 1)! +
∞
=0
2+1
2 ! . But
∞
=0
2+1
2 ! =
∞
=0
1 2 2
! = 22,
so () = 122 + 0 + 0 ∞
=1
2−1( − 1)!2
(2 − 1)! .
19. Here the initial-value problem is 200 + 400 + 400 = 12, (0) = 001, 0(0) = 0. Then
() = −10(1 cos 10 + 2 sin 10) and we try () = . Thus the general solution is
() = −10(1 cos 10 + 2 sin 10) + 100 3 . But 001 = 0(0) = 1 + 003 and 0 = 00(0) = −101 + 102,
so 1 = −002 = 2. Hence the charge is given by () = −002−10(cos 10 + sin 10) + 003.
20. By Hooke’s Law the spring constant is = 64 and the initial-value problem is 200 + 160 + 64 = 0, (0) = 0,
0(0) = 24. Thus the general solution is () = −4(1 cos 4 + 2 sin 4). But 0 = (0) = 1 and
24 = 0(0) = −41 + 42 ⇒ 1 = 0, 2 = 06. Thus the position of the mass is given by () = 06−4 sin 4.
21. (a) Since we are assuming that the earth is a solid sphere of uniform density, we can calculate the density as follows:
=
mass of earth
volume of earth =
4 3
3
. If is the volume of the portion of the earth which lies within a distance of the
center, then = 4 3 3 and = = 33 . Thus = −2 = − 3 .
(b) The particle is acted upon by a varying gravitational force during its motion. By Newton’s Second Law of Motion,
2
2 = = − 3 , so 00() = −2 () where 2 = 3 . At the surface, − = = − 2 , so
=
2 . Therefore 2 = .
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© Cengage Learning. All Rights Reserved.CHAPTER 17 REVIEW ¤ 741
(c) The differential equation 00 + 2 = 0 has auxiliary equation 2 + 2 = 0. (This is the of Section 17.1,
not the measuring distance from the earth’s center.) The roots of the auxiliary equation are ±, so by (11) in
Section 17.1, the general solution of our differential equation for is () = 1 cos + 2 sin. It follows that
0() = −1 sin + 2 cos. Now (0) = and 0(0) = 0, so 1 = and 2 = 0. Thus () = cos and
0() = −sin. This is simple harmonic motion (see Section 17.3) with amplitude , frequency , and phase angle 0.
The period is = 2. ≈ 3960 mi = 3960 · 5280 ft and = 32 fts2, so = ≈ 124 × 10−3 s−1 and
= 2 ≈ 5079 s ≈ 85 min.
(d) () = 0 ⇔ cos = 0 ⇔ = 2 + for some integer ⇒ 0() = −sin 2 + = ±. Thus the
particle passes through the center of the earth with speed ≈ 4899 mis ≈ 17,600 mih.
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