5
Series Solutions of Linear
Differential Equations
EXERCISES 5.1
Solutions About Ordinary Points
n+1 n+1
2
x
/(n + 1) 2n
1. lim = lim
|x| = 2|x|
n
n
n→∞
n→∞
2 x /n
n+1
The series is absolutely convergent for 2|x| < 1 or |x| < 12 . The radius of convergence is R = 12 . At x = − 12 , the
∞
∞
series n=1 (−1)n /n converges by the alternating series test. At x = 12 , the series n=1 1/n is the harmonic
series which diverges. Thus, the given series converges on [− 12 , 12 ).
100n+1 (x + 7)n+1 /(n + 1)! = lim 100 |x + 7| = 0
2. lim n→∞ n + 1
n→∞
100n (x + 7)n /n!
The radius of convergence is R = ∞. The series is absolutely convergent on (−∞, ∞).
3. By the ratio test,
(x − 5)n+1 /10n+1 = lim 1 |x − 5| = 1 |x − 5|.
lim
n→∞ (x − 5)n /10n n→∞ 10
10
1
The series is absolutely convergent for 10
|x − 5| < 1, |x − 5| < 10, or on (−5, 15). The radius of convergence is
∞
∞
n
R = 10. At x = −5, the series n=1 (−1) (−10)n /10n = n=1 1 diverges by the nth term test. At x = 15, the
∞
∞
series n=1 (−1)n 10n /10n = n=1 (−1)n diverges by the nth term test. Thus, the series converges on (−5, 15).
(n + 1)!(x − 1)n+1 = lim (n + 1)|x − 1| = ∞, x = 1
4. lim n→∞
n→∞
n!(x − 1)n
0, x = 1
The radius of convergence is R = 0 and the series converges only for x = 1.
x3
x5
x7
x2
x4
x6
2x3
2x5
4x7
5. sin x cos x = x −
+
−
+ ···
1−
+
−
+ ··· = x −
+
−
+ ···
6
120 5040
2
24 720
3
15
315
x2
x3
x4
x2
x4
x3
x4
6. e−x cos x = 1 − x +
−
+
− ···
1−
+
− ··· = 1 − x +
−
+ ···
2
6
24
2
24
3
6
1
1
5x4
61x6
x2
=
+
+
+ ···
=1+
2
4
6
cos x
2
4!
6!
1 − x2 + x − x + · · ·
4!
6!
Since cos(π/2) = cos(−π/2) = 0, the series converges on (−π/2, π/2).
1−x
1 3
3
3
8.
= − x + x2 − x3 + · · ·
2+x
2 4
8
16
Since the function is undefined at x = −2, the series converges on (−2, 2).
7.
9. Let k = n + 2 so that n = k − 2 and
∞
n=1
ncn xn+2 =
∞
(k − 2)ck−2 xk .
k=3
252
5.1
Solutions About Ordinary Points
10. Let k = n − 3 so that n = k + 3 and
∞
∞
(2n − 1)cn xn−3 =
n=3
11.
∞
2ncn xn−1 +
n=1
∞
(2k + 5)ck+3 xk .
k=0
∞
6cn xn+1 = 2 · 1 · c1 x0 +
n=0
2ncn xn−1 +
n=2
∞
n=0
k=n−1
= 2c1 +
∞
= 2c1 +
k=n+1
∞
2(k + 1)ck+1 xk +
k=1
∞
6cn xn+1
6ck−1 xk
k=1
[2(k + 1)ck+1 + 6ck−1 ]xk
k=1
12.
∞
∞
n(n − 1)cn xn + 2
n=2
n=2
ncn xn
n=1
= 2 · 2 · 1c2 x + 2 · 3 · 2c3 x + 3 · 1 · c1 x +
0
∞
n(n − 1)cn xn−2 + 3
1
1
∞
n(n − 1)cn x +2
n
n=2
∞
n(n − 1)cn x
n−2
n=4
= 4c2 + (3c1 + 12c3 )x +
= 4c2 + (3c1 + 12c3 )x +
∞
k(k − 1)ck xk + 2
k=2
∞
k=2
∞
∞
n=2
k=n
= 4c2 + (3c1 + 12c3 )x +
+3
∞
k=n−2
(k + 2)(k + 1)ck+2 xk + 3
k=2
∞
ncn xn
k=n
kck xk
k=2
k(k − 1) + 3k ck + 2(k + 2)(k + 1)ck+2 xk
k(k + 2)ck + 2(k + 1)(k + 2)ck+2 xk
k=2
13. y =
∞
(−1)n+1 xn−1 ,
y =
n=1
∞
(−1)n+1 (n − 1)xn−2
n=2
(x + 1)y + y = (x + 1)
∞
(−1)n+1 (n − 1)xn−2 +
n=2
=
∞
∞
(−1)n+1 xn−1
n=1
(−1)n+1 (n − 1)xn−1 +
n=2
∞
(−1)n+1 (n − 1)xn−2 +
n=2
= −x0 + x0 +
∞
∞
(−1)n+1 (n − 1)xn−2 +
n=3
k=n−1
=
=
∞
k=1
∞
(−1)k+2 kxk +
∞
14. y =
k=n−2
(−1)k+3 (k + 1)xk +
k=1
∞
(−1)k+2 xk
k=1
(−1)k+2 k − (−1)k+2 k − (−1)k+2 + (−1)k+2 xk = 0
k=1
∞
(−1)n 2n 2n−1
x
,
22n (n!)2
n=1
y =
∞
(−1)n 2n(2n − 1)
n=1
22n (n!)2
(−1)n+1 xn−1
n=1
(−1)n+1 (n − 1)xn−1 +
n=2
∞
x2n−2
253
∞
n=2
(−1)n+1 xn−1
k=n−1
5.1
Solutions About Ordinary Points
xy + y + xy =
∞
(−1)n 2n(2n − 1)
22n (n!)2
n=1
x2n−1 +
∞
(−1)n 2n
∞
(−1)n 2n+1
2n−1
x
+
x
22n (n!)2
22n (n!)2
n=1
n=0
k=n
∞ k=n
k=n+1
(−1) 2k(2k − 1) (−1) 2k
(−1)
x2k−1
+ 2k
+ 2k−2
2k
2
2
2 (k!)
2 (k!)
2
[(k − 1)!]2
k=1
∞ (−1)k (2k)2
(−1)k
x2k−1
=
− 2k−2
22k (k!)2
2
[(k − 1)!]2
k=1
∞
(2k)2 − 22 k 2 2k−1
x
=
(−1)k
=0
22k (k!)2
=
k
k
k−1
k=1
15. The singular points of (x2 − 25)y + 2xy + y = 0 are −5 and 5. The distance from 0 to either of these points
is 5. The distance from 1 to the closest of these points is 4.
16. The singular points of (x2 − 2x + 10)y + xy − 4y = 0 are 1 + 3i and 1 − 3i. The distance from 0 to either of
√
these points is 10 . The distance from 1 to either of these points is 3.
∞
17. Substituting y = n=0 cn xn into the differential equation we have
y − xy =
∞
n(n − 1)cn xn−2 −
n=2
∞
n=0
k=n−2
= 2c2 +
∞
cn xn+1 =
∞
(k + 2)(k + 1)ck+2 xk −
k=0
∞
ck−1 xk
k=1
k=n+1
[(k + 2)(k + 1)ck+2 − ck−1 ]xk = 0.
k=1
Thus
c2 = 0
(k + 2)(k + 1)ck+2 − ck−1 = 0
and
ck+2 =
1
ck−1 ,
(k + 2)(k + 1)
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
1
6
c4 = c5 = 0
1
c6 =
180
c3 =
and so on. For c0 = 0 and c1 = 1 we obtain
c3 = 0
1
c4 =
12
c5 = c6 = 0
1
c7 =
504
and so on. Thus, two solutions are
1
1 6
y1 = 1 + x3 +
x + ···
6
180
and
254
y2 = x +
1 4
1 7
x +
x + ··· .
12
504
5.1
18. Substituting y =
Solutions About Ordinary Points
∞
n
n=0 cn x into the differential equation we have
2
y +x y =
∞
n(n − 1)cn x
n−2
n=2
+
∞
n+2
cn x
n=0
k=n−2
=
∞
k
(k + 2)(k + 1)ck+2 x +
k=0
∞
ck−2 xk
k=2
k=n+2
∞
= 2c2 + 6c3 x +
[(k + 2)(k + 1)ck+2 + ck−2 ]xk = 0.
k=2
Thus
c2 = c3 = 0
(k + 2)(k + 1)ck+2 + ck−2 = 0
and
ck+2 = −
1
ck−2 ,
(k + 2)(k + 1)
k = 2, 3, 4, . . . .
Choosing c0 = 1 and c1 = 0 we find
1
12
c5 = c6 = c7 = 0
1
c8 =
672
c4 = −
and so on. For c0 = 0 and c1 = 1 we obtain
c4 = 0
1
20
c6 = c7 = c8 = 0
1
c9 =
1440
c5 = −
and so on. Thus, two solutions are
y1 = 1 −
19. Substituting y =
1 4
1 8
x +
x − ···
12
672
and
y2 = x −
1 5
1 9
x +
x − ··· .
20
1440
∞
n
n=0 cn x into the differential equation we have
y − 2xy + y =
∞
n=2
n(n − 1)cn xn−2 − 2
∞
n=1
k=n−2
=
∞
ncn xn +
k=0
= 2c2 + c0 +
k=n
(k + 2)(k + 1)ck+2 xk − 2
k=n
kck xk +
k=1
∞
cn xn
n=0
∞
∞
∞
ck xk
k=0
[(k + 2)(k + 1)ck+2 − (2k − 1)ck ]xk = 0.
k=1
Thus
2c2 + c0 = 0
(k + 2)(k + 1)ck+2 − (2k − 1)ck = 0
255
5.1
Solutions About Ordinary Points
and
1
c2 = − c0
2
2k − 1
ck+2 =
ck ,
(k + 2)(k + 1)
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
1
2
c3 = c5 = c7 = · · · = 0
1
c4 = −
8
7
c6 = −
240
c2 = −
and so on. For c0 = 0 and c1 = 1 we obtain
c2 = c4 = c6 = · · · = 0
1
c3 =
6
1
c5 =
24
1
c7 =
112
and so on. Thus, two solutions are
1
1
7 6
y1 = 1 − x2 − x4 −
x − ···
2
8
240
20. Substituting y =
1
1
1 7
y2 = x + x3 + x5 +
x + ··· .
6
24
112
and
∞
n
n=0 cn x into the differential equation we have
y − xy + 2y =
∞
n(n − 1)cn xn−2 −
n=2
∞
n=1
k=n−2
=
∞
ncn xn + 2
k=n
k=0
∞
k=n
kck xk + 2
k=1
= 2c2 + 2c0 +
∞
∞
ck xk
k=0
[(k + 2)(k + 1)ck+2 − (k − 2)ck ]xk = 0.
k=1
Thus
2c2 + 2c0 = 0
(k + 2)(k + 1)ck+2 − (k − 2)ck = 0
and
c2 = −c0
ck+2 =
cn xn
n=0
(k + 2)(k + 1)ck+2 xk −
∞
k−2
ck ,
(k + 2)(k + 1)
256
k = 1, 2, 3, . . . .
5.1
Solutions About Ordinary Points
Choosing c0 = 1 and c1 = 0 we find
c2 = −1
c3 = c5 = c7 = · · · = 0
c4 = 0
c6 = c8 = c10 = · · · = 0.
For c0 = 0 and c1 = 1 we obtain
c2 = c4 = c6 = · · · = 0
1
c3 = −
6
1
c5 = −
120
and so on. Thus, two solutions are
y1 = 1 − x2
21. Substituting y =
and
1
1 5
y2 = x − x3 −
x − ··· .
6
120
∞
n
n=0 cn x into the differential equation we have
y + x2 y + xy =
∞
n(n − 1)cn xn−2 +
n=2
∞
n=1
k=n−2
=
∞
ncn xn+1 +
n=0
k=n+1
(k + 2)(k + 1)ck+2 xk +
k=0
∞
∞
cn xn+1
k=n+1
(k − 1)ck−1 xk +
k=2
= 2c2 + (6c3 + c0 )x +
∞
Thus
c2 = 0
6c3 + c0 = 0
(k + 2)(k + 1)ck+2 + kck−1 = 0
and
c2 = 0
1
c3 = − c0
6
k
ck−1 ,
(k + 2)(k + 1)
Choosing c0 = 1 and c1 = 0 we find
1
6
c4 = c5 = 0
1
c6 =
45
c3 = −
257
ck−1 xk
k=1
[(k + 2)(k + 1)ck+2 + kck−1 ]xk = 0.
k=2
ck+2 = −
∞
k = 2, 3, 4, . . . .
5.1
Solutions About Ordinary Points
and so on. For c0 = 0 and c1 = 1 we obtain
c3 = 0
1
6
c5 = c6 = 0
5
c7 =
252
c4 = −
and so on. Thus, two solutions are
1
1
y1 = 1 − x3 + x6 − · · ·
6
45
22. Substituting y =
1
5 7
x − ··· .
y2 = x − x4 +
6
252
and
∞
n
n=0 cn x into the differential equation we have
y + 2xy + 2y =
∞
n(n − 1)cn xn−2 + 2
n=2
∞
n=1
k=n−2
=
∞
ncn xn + 2
k=n
(k + 2)(k + 1)ck+2 xk + 2
k=0
k=n
kck xk + 2
k=1
= 2c2 + 2c0 +
∞
Thus
2c2 + 2c0 = 0
(k + 2)(k + 1)ck+2 + 2(k + 1)ck = 0
and
c2 = −c0
2
ck ,
k+2
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
c2 = −1
c3 = c5 = c7 = · · · = 0
c4 =
1
2
c6 = −
∞
ck xk
k=0
[(k + 2)(k + 1)ck+2 + 2(k + 1)ck ]xk = 0.
k=1
ck+2 = −
cn xn
n=0
∞
∞
1
6
and so on. For c0 = 0 and c1 = 1 we obtain
c2 = c4 = c6 = · · · = 0
2
c3 = −
3
4
c5 =
15
8
c7 = −
105
258
5.1
Solutions About Ordinary Points
and so on. Thus, two solutions are
1
1
2
4
8 7
y1 = 1 − x2 + x4 − x6 + · · · and y2 = x − x3 + x5 −
x + ··· .
2
6
3
15
105
∞
23. Substituting y = n=0 cn xn into the differential equation we have
(x − 1)y + y =
∞
n(n − 1)cn xn−1 −
n=2
∞
n(n − 1)cn xn−2 +
n=2
=
∞
k=n−2
(k + 1)kck+1 xk −
k=1
∞
= −2c2 + c1 +
k=n−1
(k + 2)(k + 1)ck+2 xk +
k=0
∞
ncn xn−1
n=1
k=n−1
∞
∞
(k + 1)ck+1 xk
k=0
[(k + 1)kck+1 − (k + 2)(k + 1)ck+2 + (k + 1)ck+1 ]xk = 0.
k=1
Thus
−2c2 + c1 = 0
(k + 1)2 ck+1 − (k + 2)(k + 1)ck+2 = 0
and
1
c1
2
k+1
ck+2 =
ck+1 ,
k+2
c2 =
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find c2 = c3 = c4 = · · · = 0. For c0 = 0 and c1 = 1 we obtain
1
1
1
c2 = ,
c3 = ,
c4 = ,
2
3
4
and so on. Thus, two solutions are
1
1
1
y1 = 1
and
y2 = x + x2 + x3 + x4 + · · · .
2
3
4
∞
n
24. Substituting y = n=0 cn x into the differential equation we have
(x + 2)y + xy − y =
∞
n=2
n(n − 1)cn xn−1 +
∞
2n(n − 1)cn xn−2 +
n=2
k=n−1
=
∞
(k + 1)kck+1 xk +
k=1
= 4c2 − c0 +
n=1
k=n−2
∞
ncn xn −
∞
∞
k=n
kck xk −
k=1
∞
ck xk
k=0
(k + 1)kck+1 + 2(k + 2)(k + 1)ck+2 + (k − 1)ck xk = 0.
k=1
Thus
4c2 − c0 = 0
(k + 1)kck+1 + 2(k + 2)(k + 1)ck+2 + (k − 1)ck = 0,
and
cn xn
n=0
k=n
2(k + 2)(k + 1)ck+2 xk +
k=0
∞
∞
1
c0
4
(k + 1)kck+1 + (k − 1)ck
,
ck+2 = −
2(k + 2)(k + 1)
k = 1, 2, 3, . . .
c2 =
259
k = 1, 2, 3, . . . .
5.1
Solutions About Ordinary Points
Choosing c0 = 1 and c1 = 0 we find
c1 = 0,
c2 =
1
,
4
c3 = −
1
,
24
c4 = 0,
c5 =
1
480
and so on. For c0 = 0 and c1 = 1 we obtain
c2 = 0
c3 = 0
c4 = c5 = c6 = · · · = 0.
Thus, two solutions are
1
1
1 5
y1 = c0 1 + x2 − x3 +
x + ···
4
24
480
25. Substituting y =
and
y2 = c1 x.
∞
n
n=0 cn x into the differential equation we have
y − (x + 1)y − y =
∞
n(n − 1)cn xn−2 −
n=2
n=1
k=n−2
=
∞
∞
ncn xn −
k=n
k=0
ncn xn−1 −
n=1
(k + 2)(k + 1)ck+2 xk −
∞
∞
k=1
= 2c2 − c1 − c0 +
∞
k=n−1
kck xk −
cn xn
n=0
∞
∞
k=n
(k + 1)ck+1 xk −
k=0
∞
ck xk
k=0
[(k + 2)(k + 1)ck+2 − (k + 1)ck+1 − (k + 1)ck ]xk = 0.
k=1
Thus
2c2 − c1 − c0 = 0
(k + 2)(k + 1)ck+2 − (k + 1)(ck+1 + ck ) = 0
and
c1 + c0
2
ck+1 + ck
ck+2 =
,
k+2
c2 =
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
1
,
2
c3 =
1
,
6
c4 =
1
,
6
1
,
2
c3 =
1
,
2
c4 =
1
,
4
1
1
1
y1 = 1 + x2 + x3 + x4 + · · ·
2
6
6
and
c2 =
and so on. For c0 = 0 and c1 = 1 we obtain
c2 =
and so on. Thus, two solutions are
260
1
1
1
y2 = x + x2 + x3 + x4 + · · · .
2
2
4
5.1
26. Substituting y =
Solutions About Ordinary Points
∞
n
n=0 cn x into the differential equation we have
x2 + 1 y − 6y =
∞
n(n − 1)cn xn +
n=2
∞
n(n − 1)cn xn−2 − 6
n=2
=
∞
k=n−2
k(k − 1)ck xk +
k=2
∞
cn xn
n=0
k=n
∞
k=n
(k + 2)(k + 1)ck+2 xk − 6
k=0
∞
ck xk
k=0
= 2c2 − 6c0 + (6c3 − 6c1 )x +
∞
k 2 − k − 6 ck + (k + 2)(k + 1)ck+2 xk = 0.
k=2
Thus
2c2 − 6c0 = 0
6c3 − 6c1 = 0
(k − 3)(k + 2)ck + (k + 2)(k + 1)ck+2 = 0
and
c2 = 3c0
c3 = c1
ck+2 = −
k−3
ck ,
k+1
k = 2, 3, 4, . . . .
Choosing c0 = 1 and c1 = 0 we find
c2 = 3
c3 = c5 = c7 = · · · = 0
c4 = 1
c6 = −
1
5
and so on. For c0 = 0 and c1 = 1 we obtain
c2 = c4 = c6 = · · · = 0
c3 = 1
c5 = c7 = c9 = · · · = 0.
Thus, two solutions are
27. Substituting y =
∞
1
y1 = 1 + 3x2 + x4 − x6 + · · ·
5
and
y2 = x + x3 .
n
n=0 cn x into the differential equation we have
x2 + 2 y + 3xy − y =
∞
n(n − 1)cn xn + 2
n=2
∞
n=2
k=n
=
∞
k=2
k(k − 1)ck xk + 2
∞
n(n − 1)cn xn−2 + 3
∞
n=1
k=n−2
(k + 2)(k + 1)ck+2 xk + 3
k=0
= (4c2 − c0 ) + (12c3 + 2c1 )x +
∞
kck xk −
∞
∞
cn xn
n=0
k=n
k=1
∞
ncn xn −
k=n
ck xk
k=0
2(k + 2)(k + 1)ck+2 + k 2 + 2k − 1 ck xk = 0.
k=2
261
5.1
Solutions About Ordinary Points
Thus
4c2 − c0 = 0
12c3 + 2c1 = 0
2(k + 2)(k + 1)ck+2 + k 2 + 2k − 1 ck = 0
and
1
c0
4
1
c3 = − c1
6
k 2 + 2k − 1
ck+2 = −
ck ,
2(k + 2)(k + 1)
c2 =
k = 2, 3, 4, . . . .
Choosing c0 = 1 and c1 = 0 we find
1
4
c3 = c5 = c7 = · · · = 0
7
c4 = −
96
c2 =
and so on. For c0 = 0 and c1 = 1 we obtain
c2 = c4 = c6 = · · · = 0
1
c3 = −
6
7
c5 =
120
and so on. Thus, two solutions are
1
7
y1 = 1 + x2 − x4 + · · ·
4
96
28. Substituting y =
1
7 5
y2 = x − x3 +
x − ··· .
6
120
and
∞
n
n=0 cn x into the differential equation we have
x2 − 1 y + xy − y =
∞
n=2
n(n − 1)cn xn −
∞
n(n − 1)cn xn−2 +
n=2
k=n
=
∞
k=2
∞
n=1
k=n−2
k(k − 1)ck xk −
∞
k=n
(k + 2)(k + 1)ck+2 xk +
k=0
= (−2c2 − c0 ) − 6c3 x +
ncn xn −
∞
k=1
cn xn
n=0
k=n
kck xk −
∞
ck xk
k=0
−(k + 2)(k + 1)ck+2 + k 2 − 1 ck xk = 0.
k=2
Thus
−2c2 − c0 = 0
−6c3 = 0
−(k + 2)(k + 1)ck+2 + (k − 1)(k + 1)ck = 0
262
∞
∞
5.1
and
Solutions About Ordinary Points
1
c2 = − c0
2
c3 = 0
ck+2 =
k−1
ck ,
k+2
k = 2, 3, 4, . . . .
Choosing c0 = 1 and c1 = 0 we find
1
2
c3 = c5 = c7 = · · · = 0
1
c4 = −
8
c2 = −
and so on. For c0 = 0 and c1 = 1 we obtain
c2 = c4 = c6 = · · · = 0
c3 = c5 = c7 = · · · = 0.
Thus, two solutions are
29. Substituting y =
1
1
y1 = 1 − x2 − x4 − · · ·
2
8
∞
and
y2 = x.
n
n=0 cn x into the differential equation we have
(x − 1)y − xy + y =
∞
n(n − 1)cn xn−1 −
n=2
∞
n(n − 1)cn xn−2 −
n=2
k=n−1
=
∞
(k + 1)kck+1 xk −
k=1
= −2c2 + c0 +
n=1
k=n−2
∞
(k + 2)(k + 1)ck+2 xk −
k=0
∞
∞
∞
kck xk +
k=1
∞
cn xn
n=0
k=n
k=n
∞
ck xk
k=0
[−(k + 2)(k + 1)ck+2 + (k + 1)kck+1 − (k − 1)ck ]xk = 0.
k=1
Thus
−2c2 + c0 = 0
−(k + 2)(k + 1)ck+2 + (k − 1)kck+1 − (k − 1)ck = 0
and
ncn xn +
1
c0
2
kck+1
(k − 1)ck
ck+2 =
−
,
k+2
(k + 2)(k + 1)
c2 =
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
1
1
,
c3 = ,
c4 = 0,
2
6
and so on. For c0 = 0 and c1 = 1 we obtain c2 = c3 = c4 = · · · = 0. Thus,
1 2 1 3
y = C1 1 + x + x + · · · + C2 x
2
6
and
1
y = C1 x + x2 + · · · + C2 .
2
c2 =
263
5.1
Solutions About Ordinary Points
The initial conditions imply C1 = −2 and C2 = 6, so
1
1
y = −2 1 + x2 + x3 + · · · + 6x = 8x − 2ex .
2
6
∞
n
n=0 cn x into the differential equation we have
30. Substituting y =
(x+1)y − (2 − x)y + y
=
∞
n=2
n(n − 1)cn xn−1 +
∞
n=2
k=n−1
=
∞
n(n − 1)cn xn−2 − 2
∞
n=1
k=n−2
(k + 1)kck+1 xk +
k=1
∞
(k + 2)(k + 1)ck+2 xk − 2
= 2c2 − 2c1 + c0 +
∞
∞
n=1
k=n−1
k=0
∞
ncn xn−1 +
ncn xn +
∞
n=0
k=n
(k + 1)ck+1 xk +
k=0
k=n
∞
kck xk +
k=1
[(k + 2)(k + 1)ck+2 − (k + 1)ck+1 + (k + 1)ck ]xk = 0.
k=1
Thus
2c2 − 2c1 + c0 = 0
(k + 2)(k + 1)ck+2 − (k + 1)ck+1 + (k + 1)ck = 0
and
1
c2 = c1 − c0
2
1
1
ck+1 −
ck ,
ck+2 =
k+2
k+2
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
1
c2 = − ,
2
1
c3 = − ,
6
c4 =
1
,
12
c3 = 0,
1
c4 = − ,
4
and so on. For c0 = 0 and c1 = 1 we obtain
c2 = 1,
and so on. Thus,
1
1
1
1
y = C1 1 − x2 − x3 + x4 + · · · + C2 x + x2 − x4 + · · ·
2
6
12
4
and
1
1
y = C1 −x − x2 + x3 + · · · + C2 1 + 2x − x3 + · · · .
2
3
The initial conditions imply C1 = 2 and C2 = −1, so
1 2 1 3
1 4
1 4
2
y = 2 1 − x − x + x + ··· − x + x − x + ···
2
6
12
4
1
5
= 2 − x − 2x2 − x3 + x4 + · · · .
3
12
264
cn xn
∞
k=0
ck xk
5.1
31. Substituting y =
Solutions About Ordinary Points
∞
n
n=0 cn x into the differential equation we have
y − 2xy + 8y =
∞
n(n − 1)cn xn−2 − 2
n=2
∞
n=1
k=n−2
=
∞
ncn xn + 8
∞
n=0
k=n
(k + 2)(k + 1)ck+2 xk − 2
k=0
∞
k=n
kck xk + 8
k=1
= 2c2 + 8c0 +
∞
cn xn
∞
ck xk
k=0
[(k + 2)(k + 1)ck+2 + (8 − 2k)ck ]xk = 0.
k=1
Thus
2c2 + 8c0 = 0
(k + 2)(k + 1)ck+2 + (8 − 2k)ck = 0
and
c2 = −4c0
ck+2 =
2(k − 4)
ck ,
(k + 2)(k + 1)
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
c2 = −4
c3 = c5 = c7 = · · · = 0
4
c4 =
3
c6 = c8 = c10 = · · · = 0.
For c0 = 0 and c1 = 1 we obtain
c2 = c4 = c6 = · · · = 0
c3 = −1
1
c5 =
10
and so on. Thus,
4
1
y = C1 1 − 4x2 + x4 + C2 x − x3 + x5 + · · ·
3
10
and
y = C1
16
−8x + x3
3
+ C2
1 4
1 − 3x + x + · · · .
2
2
The initial conditions imply C1 = 3 and C2 = 0, so
4
y = 3 1 − 4x2 + x4 = 3 − 12x2 + 4x4 .
3
265
5.1
Solutions About Ordinary Points
32. Substituting y =
∞
n
n=0 cn x into the differential equation we have
(x2 + 1)y + 2xy =
∞
n=2
n(n − 1)cn xn +
∞
n(n − 1)cn xn−2 +
n=2
k=n
=
∞
∞
n=1
k=n−2
k(k − 1)ck xk +
k=2
∞
2ncn xn
k=n
(k + 2)(k + 1)ck+2 xk +
k=0
∞
∞
2kck xk
k=1
k(k + 1)ck + (k + 2)(k + 1)ck+2 xk = 0.
= 2c2 + (6c3 + 2c1 )x +
k=2
Thus
2c2 = 0
6c3 + 2c1 = 0
k(k + 1)ck + (k + 2)(k + 1)ck+2 = 0
and
c2 = 0
1
c3 = − c1
3
k
ck+2 = −
ck ,
k+2
k = 2, 3, 4, . . . .
Choosing c0 = 1 and c1 = 0 we find c3 = c4 = c5 = · · · = 0. For c0 = 0 and c1 = 1 we obtain
1
3
c4 = c6 = c8 = · · · = 0
1
c5 = −
5
1
c7 =
7
c3 = −
and so on. Thus
y = C0 + C1
1 3 1 5 1 7
x − x + x − x + ···
3
5
7
and
y = c1 1 − x2 + x4 − x6 + · · · .
The initial conditions imply c0 = 0 and c1 = 1, so
33. Substituting y =
1
1
1
y = x − x3 + x5 − x7 + · · · .
3
5
7
∞
n
n=0 cn x into the differential equation we have
1
1 5
n(n − 1)cn xn−2 + x − x3 +
x − · · · c0 + c1 x + c2 x2 + · · ·
6
120
n=2
1
2
3
2
3
= 2c2 + 6c3 x + 12c4 x + 20c5 x + · · · + c0 x + c1 x + c2 − c0 x + · · ·
6
1
= 2c2 + (6c3 + c0 )x + (12c4 + c1 )x2 + 20c5 + c2 − c0 x3 + · · · = 0.
6
y + (sin x)y =
∞
266
5.1
Solutions About Ordinary Points
Thus
2c2 = 0
6c3 + c0 = 0
12c4 + c1 = 0
1
20c5 + c2 − c0 = 0
6
c2 = 0
and
1
c3 = − c0
6
1
c4 = − c1
12
1
1
c0 .
c5 = − c2 +
20
120
Choosing c0 = 1 and c1 = 0 we find
c2 = 0,
1
c3 = − ,
6
c4 = 0,
c3 = 0,
c4 = −
c5 =
1
120
and so on. For c0 = 0 and c1 = 1 we obtain
c2 = 0,
1
,
12
c5 = 0
and so on. Thus, two solutions are
34. Substituting y =
1
1 5
y1 = 1 − x3 +
x + ···
6
120
∞
and
y2 = x −
1 4
x + ··· .
12
n
n=0 cn x into the differential equation we have
y + ex y − y =
∞
n(n − 1)cn xn−2
n=2
∞
1
1
+ 1 + x + x2 + x3 + · · · c1 + 2c2 x + 3c3 x2 + 4c4 x3 + · · · −
cn xn
2
6
n=0
2
3
= 2c2 + 6c3 x + 12c4 x + 20c5 x + · · ·
1
2
+ c1 + (2c2 + c1 )x + 3c3 + 2c2 + c1 x + · · · − [c0 + c1 x + c2 x2 + · · ·]
2
1
= (2c2 + c1 − c0 ) + (6c3 + 2c2 )x + 12c4 + 3c3 + c2 + c1 x2 + · · · = 0.
2
Thus
2c2 + c1 − c0 = 0
6c3 + 2c2 = 0
1
12c4 + 3c3 + c2 + c1 = 0
2
267
5.1
Solutions About Ordinary Points
and
1
1
c0 − c1
2
2
1
c3 = − c2
3
1
1
1
c4 = − c3 + c2 − c1 .
4
12
24
c2 =
Choosing c0 = 1 and c1 = 0 we find
c2 =
1
,
2
1
c3 = − ,
6
c4 = 0
and so on. For c0 = 0 and c1 = 1 we obtain
1
c2 = − ,
2
c3 =
1
,
6
c4 = −
1
24
and so on. Thus, two solutions are
1
1
y1 = 1 + x2 − x3 + · · ·
2
6
1
1
1
y2 = x − x2 + x3 − x4 + · · · .
2
6
24
and
35. The singular points of (cos x)y + y + 5y = 0 are odd integer multiples of π/2. The distance from 0 to either
±π/2 is π/2. The singular point closest to 1 is π/2. The distance from 1 to the closest singular point is then
π/2 − 1.
36. Substituting y =
∞
n
n=0 cn x into the first differential equation leads to
y − xy =
∞
n(n − 1)cn xn−2 −
n=2
n=0
k=n−2
= 2c2 +
∞
∞
cn xn+1 =
∞
(k + 2)(k + 1)ck+2 xk −
k=0
k=n+1
[(k + 2)(k + 1)ck+2 − ck−1 ]xk = 1.
k=1
Thus
2c2 = 1
(k + 2)(k + 1)ck+2 − ck−1 = 0
and
c2 =
ck+2 =
1
2
ck−1
,
(k + 2)(k + 1)
k = 1, 2, 3, . . . .
Let c0 and c1 be arbitrary and iterate to find
1
2
1
c3 = c0
6
1
c4 =
c1
12
1
1
c5 =
c2 =
20
40
c2 =
268
∞
k=1
ck−1 xk
5.1
Solutions About Ordinary Points
and so on. The solution is
1
1
1
1
y = c0 + c1 x + x2 + c0 x3 + c1 x4 + c5 + · · ·
2
6
12
40
1 3
1 4
1
1
= c0 1 + x + · · · + c1 x + x + · · · + x2 + x5 + · · · .
6
12
2
40
Substituting y =
∞
n
n=0 cn x into the second differential equation leads to
y − 4xy − 4y =
∞
n(n − 1)cn xn−2 −
n=2
∞
n=1
k=n−2
=
∞
4ncn xn −
∞
n=0
k=n
(k + 2)(k + 1)ck+2 xk −
k=0
∞
= 2c2 − 4c0 +
k=n
4kck xk −
k=1
∞
4cn xn
∞
k=0
(k + 2)(k + 1)ck+2 − 4(k + 1)ck xk
k=1
= ex = 1 +
∞
1
k=1
k!
4ck xk
xk .
Thus
2c2 − 4c0 = 1
(k + 2)(k + 1)ck+2 − 4(k + 1)ck =
1
k!
and
1
+ 2c0
2
1
4
ck+2 =
+
ck ,
(k + 2)! k + 2
c2 =
k = 1, 2, 3, . . . .
Let c0 and c1 be arbitrary and iterate to find
1
+ 2c0
2
4
4
1
1
+ c1 =
+ c1
c3 =
3! 3
3! 3
c2 =
c4 =
1
1
13
4
1
+ c2 =
+ + 2c0 =
+ 2c0
4! 4
4! 2
4!
c5 =
4
4
16
1
1
17 16
+ c3 =
+
+ c1 =
+ c1
5! 5
5! 5 · 3! 15
5!
15
c6 =
1
1
261 4
4
4 · 13 8
+ c4 =
+
+ c0 =
+ c0
6! 6
6!
6 · 4!
6
6!
3
c7 =
4
4 · 17
64
64
1
1
409
+ c5 =
+
+
c1 =
+
c1
7! 7
7!
7 · 5!
105
7!
105
and so on. The solution is
269
5.1
Solutions About Ordinary Points
1
1
13
17 16
4
2
3
4
y = c0 + c1 x +
+ 2c0 x +
+ c1 x +
+ 2c0 x +
+ c1 x5
2
3! 3
4!
5!
15
64
261 4
409
+ c0 x6 +
+
c1 x7 + · · ·
+
6!
3
7!
105
4 3 16 5
4 6
64 7
2
4
= c0 1 + 2x + 2x + x + · · · + c1 x + x + x +
x + ···
3
3
15
105
1
1
13
17
261 6 409 7
+ x2 + x3 + x4 + x5 +
x +
x + ··· .
2
3!
4!
5!
6!
7!
37. We identify P (x) = 0 and Q(x) = sin x/x. The Taylor series representation for sin x/x is 1 − x2 /3! + x4 /5! − · · · ,
for |x| < ∞. Thus, Q(x) is analytic at x = 0 and x = 0 is an ordinary point of the differential equation.
38. If x > 0 and y > 0, then y = −xy < 0 and the graph of a solution curve is concave down. Thus, whatever
portion of a solution curve lies in the first quadrant is concave down. When x > 0 and y < 0, y = −xy > 0,
so whatever portion of a solution curve lies in the fourth quadrant is concave up.
∞
39. (a) Substituting y = n=0 cn xn into the differential equation we have
y + xy + y =
∞
n(n − 1)cn xn−2 +
n=2
∞
n=1
k=n−2
=
∞
ncn xn +
k=n
k=0
∞
k=n
kck xk +
k=1
= (2c2 + c0 ) +
∞
cn xn
n=0
(k + 2)(k + 1)ck+2 xk +
∞
∞
ck xk
k=0
(k + 2)(k + 1)ck+2 + (k + 1)ck xk = 0.
k=1
Thus
2c2 + c0 = 0
(k + 2)(k + 1)ck+2 + (k + 1)ck = 0
and
1
c2 = − c0
2
1
ck+2 = −
ck ,
k+2
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
1
2
c3 = c5 = c7 = · · · = 0
1
1 1
c4 = − −
= 2
4
2
2 ·2
1 1 1
c6 = −
=− 3
6 22 · 2
2 · 3!
c2 = −
and so on. For c0 = 0 and c1 = 1 we obtain
270
5.1
Solutions About Ordinary Points
c2 = c4 = c6 = · · · = 0
1
2
c3 = − = −
3
3!
1 1
1
4·2
c5 = − −
=
=
5
3
5·3
5!
6·4·2
14 · 2
c7 = −
=−
7 5!
7!
and so on. Thus, two solutions are
y1 =
∞
(−1)k
k=0
2k · k!
x2k
and
y2 =
∞
(−1)k 2k k!
k=0
(2k + 1)!
x2k+1 .
(b) For y1 , S3 = S2 and S5 = S4 , so we plot S2 , S4 , S6 , S8 , and S10 .
y
4
2
-4 -2
-2
N=2
y
y
4
4
2
x
2 4
-4 -2
-2
-4
2
2 4
x
-4 -2
-2
N=4
-4
N=6
2 4
x
y
y
4
4
2
2
-4 -2
-2
-4
2 4
x
N=8
-4 -2
-2
-4
-4
y
y
y
4
4
N=10
2 4
x
2 4
x
For y2 , S3 = S4 and S5 = S6 , so we plot S2 , S4 , S6 , S8 , and S10 .
y
y
4
4
2
-4 -2
-2
2
x
2 4
N=2
-4 -2
-2
-4
2 4
2
x
-4 -2
-2
-4
(c)
-4
N=4
y2
4
4
2
2
2
4
2 4
N=6
-4 -2
-2
-4
y1
-2
2
x
x
-4
-2
-4
2
-2
-2
-4
-4
4
4
N=8
2 4
2
x
-4 -2
-2
N=10
-4
x
The graphs of y1 and y2 obtained from a numerical solver are shown. We see that the partial sum representations indicate the even and odd natures of the solution, but don’t really give a very accurate representation
of the true solution. Increasing N to about 20 gives a much more accurate representation on [−4, 4].
∞ k
∞
∞
−x2 /2
2
k
k 2k
k
(d) From ex =
=
k=0 x /k! we see that e
k=0 (−x /2) /k! =
k=0 (−1) x /2 k! . From (5) of
Section 3.2 we have
271
5.1
Solutions About Ordinary Points
−x2 /2
2
2
e−x /2
e
−x2 /2
−x2 /2
ex /2 dx
dx
=
e
dx
=
e
y12
e−x2
(e−x2 /2 )2
∞ ∞
∞
∞
(−1)k
(−1)k 2k 1
1
2k
2k
2k
=
x
x dx =
x
x dx
2k k!
2k k!
2k k!
2k k!
k=0
k=0
k=0
k=0
∞
∞
(−1)k
1
2k
2k+1
=
x
x
2k k!
(2k + 1)2k k!
y2 = y1
e−
x dx
dx = e−x /2
k=0
2
k=0
1
1
1
1 3
1
1
= 1 − x2 + 2 x4 − 3
x6 + · · · x +
x +
x5 +
x7 + · · ·
2
3
2
2 ·2
2 · 3!
3·2
5·2 ·2
7 · 2 · 3!
∞
k
k
(−1) 2 k!
2
4·2 5 6·4·2 7
= x − x3 +
x −
x + ··· =
x2k+1 .
3!
5!
7!
(2k + 1)!
k=0
40. (a) We have
y + (cos x)y = 2c2 + 6c3 x + 12c4 x2 + 20c5 x3 + 30c6 x4 + 42c7 x5 + · · ·
x2
x4
x6
+ 1−
+
−
+ · · · (c0 + c1 x + c2 x2 + c3 x3 + c4 x4 + c5 x5 + · · · )
2!
4!
6!
1 1 = (2c2 + c0 ) + (6c3 + c1 )x + 12c4 + c2 − c0 x2 + 20c5 + c3 − c1 x3
2
2
1
1 4 1
1 5
+ 30c6 + c4 + c0 − c2 x + 42c7 + c5 + c1 − c3 x + · · · .
24
2
24
2
Then
30c6 + c4 +
1
1
c0 − c2 = 0
24
2
and
42c7 + c5 +
1
1
c1 − c3 = 0,
24
2
which gives c6 = −c0 /80 and c7 = −19c1 /5040. Thus
1
1
1
y1 (x) = 1 − x2 + x4 − x6 + · · ·
2
12
80
and
1
1
19 7
y2 (x) = x − x3 + x5 −
x + ··· .
6
30
5040
(b) From part (a) the general solution of the differential equation is y = c1 y1 + c2 y2 . Then y(0) = c1 + c2 · 0 = c1
and y (0) = c1 · 0 + c2 = c2 , so the solution of the initial-value problem is
1
1
1
1
1
19 7
y = y1 + y2 = 1 + x − x2 − x3 + x4 + x5 − x6 −
x + ··· .
2
6
12
30
80
5040
272
5.2
(c)
y
y
y
4
4
4
2
2
2
-6 -4 -2
2
4
6
x
-6 -4 -2
2
4
6
x
-6 -4 -2
-2
-2
-2
-4
-4
-4
y
y
y
4
4
4
2
2
2
-6 -4 -2
(d)
Solutions About Singular Points
2
4
6
x
-6 -4 -2
2
4
6
x
-6 -4 -2
-2
-2
-2
-4
-4
-4
2
4
6
x
2
4
6
x
y
6
4
2
-6 -4 -2
-2
2
4
6
x
-4
-6
EXERCISES 5.2
Solutions About Singular Points
1. Irregular singular point: x = 0
2. Regular singular points: x = 0, −3
3. Irregular singular point: x = 3; regular singular point: x = −3
4. Irregular singular point: x = 1; regular singular point: x = 0
5. Regular singular points: x = 0, ±2i
6. Irregular singular point: x = 5; regular singular point: x = 0
273
5.2
Solutions About Singular Points
7. Regular singular points: x = −3, 2
8. Regular singular points: x = 0, ±i
9. Irregular singular point: x = 0; regular singular points: x = 2, ±5
10. Irregular singular point: x = −1; regular singular points: x = 0, 3
11. Writing the differential equation in the form
y +
5
x
y +
y=0
x−1
x+1
we see that x0 = 1 and x0 = −1 are regular singular points. For x0 = 1 the differential equation can be put in
the form
(x − 1)2 y + 5(x − 1)y +
x(x − 1)2
y = 0.
x+1
In this case p(x) = 5 and q(x) = x(x − 1)2 /(x + 1). For x0 = −1 the differential equation can be put in the form
(x + 1)2 y + 5(x + 1)
x+1 y + x(x + 1)y = 0.
x−1
In this case p(x) = (x + 1)/(x − 1) and q(x) = x(x + 1).
12. Writing the differential equation in the form
y +
x+3 y + 7xy = 0
x
we see that x0 = 0 is a regular singular point. Multiplying by x2 , the differential equation can be put in the
form
x2 y + x(x + 3)y + 7x3 y = 0.
We identify p(x) = x + 3 and q(x) = 7x3 .
13. We identify P (x) = 5/3x + 1 and Q(x) = −1/3x2 , so that p(x) = xP (x) = 53 + x and q(x) = x2 Q(x) = − 13 .
Then a0 = 53 , b0 = − 13 , and the indicial equation is
5
2
1
1
1
1
r(r − 1) + r − = r2 + r − = (3r2 + 2r − 1) = (3r − 1)(r + 1) = 0.
3
3
3
3
3
3
The indicial roots are 13 and −1. Since these do not differ by an integer we expect to find two series solutions
using the method of Frobenius.
14. We identify P (x) = 1/x and Q(x) = 10/x, so that p(x) = xP (x) = 1 and q(x) = x2 Q(x) = 10x. Then a0 = 1,
b0 = 0, and the indicial equation is
r(r − 1) + r = r2 = 0.
The indicial roots are 0 and 0. Since these are equal, we expect the method of Frobenius to yield a single series
solution.
∞
15. Substituting y = n=0 cn xn+r into the differential equation and collecting terms, we obtain
2xy − y + 2y = 2r2 − 3r c0 xr−1 +
∞
[2(k + r − 1)(k + r)ck − (k + r)ck + 2ck−1 ]xk+r−1 = 0,
k=1
which implies
2r2 − 3r = r(2r − 3) = 0
and
(k + r)(2k + 2r − 3)ck + 2ck−1 = 0.
274
5.2
Solutions About Singular Points
The indicial roots are r = 0 and r = 3/2. For r = 0 the recurrence relation is
ck = −
2ck−1
,
k(2k − 3)
k = 1, 2, 3, . . . ,
and
c2 = −2c0 ,
c1 = 2c0 ,
c3 =
4
c0 ,
9
and so on. For r = 3/2 the recurrence relation is
ck = −
2ck−1
,
(2k + 3)k
k = 1, 2, 3, . . . ,
and
2
2
4
c1 = − c0 ,
c2 =
c3 = −
c0 ,
c0 ,
5
35
945
and so on. The general solution on (0, ∞) is
2
4 3
4 3
2 2
2
3/2
y = C1 1 + 2x − 2x + x + · · · + C2 x
1− x+ x −
x + ··· .
9
5
35
945
16. Substituting y =
∞
n+r
into the differential equation and collecting terms, we obtain
n=0 cn x
2xy + 5y + xy = 2r2 + 3r c0 xr−1 + 2r2 + 7r + 5 c1 xr
+
∞
[2(k + r)(k + r − 1)ck + 5(k + r)ck + ck−2 ]xk+r−1
k=2
= 0,
which implies
2r2 + 3r = r(2r + 3) = 0,
2r2 + 7r + 5 c1 = 0,
and
(k + r)(2k + 2r + 3)ck + ck−2 = 0.
The indicial roots are r = −3/2 and r = 0, so c1 = 0 . For r = −3/2 the recurrence relation is
ck = −
ck−2
,
(2k − 3)k
k = 2, 3, 4, . . . ,
and
1
c2 = − c0 ,
2
and so on. For r = 0 the recurrence relation is
ck = −
c3 = 0,
ck−2
,
k(2k + 3)
c4 =
1
c0 ,
40
k = 2, 3, 4, . . . ,
and
1
1
c0 ,
c0 ,
c3 = 0,
c4 =
14
616
and so on. The general solution on (0, ∞) is
1
1 4
1
1
y = C1 x−3/2 1 − x2 + x4 + · · · + C2 1 − x2 +
x + ··· .
2
40
14
616
c2 = −
275
5.2
Solutions About Singular Points
17. Substituting y =
∞
n+r
into the differential equation and collecting terms, we obtain
n=0 cn x
1
4xy + y + y =
2
∞ 7
1
r−1
4(k + r)(k + r − 1)ck + (k + r)ck + ck−1 xk+r−1
4r − r c0 x
+
2
2
2
k=1
= 0,
which implies
7
7
4r − r = r 4r −
=0
2
2
2
and
1
(k + r)(8k + 8r − 7)ck + ck−1 = 0.
2
The indicial roots are r = 0 and r = 7/8. For r = 0 the recurrence relation is
ck = −
2ck−1
,
k(8k − 7)
and
c1 = −2c0 ,
c2 =
k = 1, 2, 3, . . . ,
2
c0 ,
9
c3 = −
4
c0 ,
459
and so on. For r = 7/8 the recurrence relation is
and
c1 = −
ck = −
2ck−1
,
(8k + 7)k
2
c0 ,
15
c2 =
k = 1, 2, 3, . . . ,
2
c0 ,
345
c3 = −
4
c0 ,
32,085
and so on. The general solution on (0, ∞) is
2
2
4 3
4
2 2
y = C1 1 − 2x + x2 −
x + · · · + C2 x7/8 1 − x +
x −
x3 + · · · .
9
459
15
345
32,085
∞
18. Substituting y = n=0 cn xn+r into the differential equation and collecting terms, we obtain
2x2 y − xy + x2 + 1 y = 2r2 − 3r + 1 c0 xr + 2r2 + r c1 xr+1
+
∞
[2(k + r)(k + r − 1)ck − (k + r)ck + ck + ck−2 ]xk+r
k=2
= 0,
which implies
2r2 − 3r + 1 = (2r − 1)(r − 1) = 0,
2r2 + r c1 = 0,
and
[(k + r)(2k + 2r − 3) + 1]ck + ck−2 = 0.
The indicial roots are r = 1/2 and r = 1, so c1 = 0. For r = 1/2 the recurrence relation is
ck = −
and
ck−2
,
k(2k − 1)
1
c2 = − c0 ,
6
k = 2, 3, 4, . . . ,
c3 = 0,
276
c4 =
1
c0 ,
168
5.2
Solutions About Singular Points
and so on. For r = 1 the recurrence relation is
ck = −
ck−2
,
k(2k + 1)
k = 2, 3, 4, . . . ,
and
1
1
c0 ,
c0 ,
c3 = 0,
c4 =
10
360
and so on. The general solution on (0, ∞) is
1 2
1 4
1 2
1 4
1/2
y = C1 x
1− x +
x + · · · + C2 x 1 − x +
x + ··· .
6
168
10
360
∞
19. Substituting y = n=0 cn xn+r into the differential equation and collecting terms, we obtain
c2 = −
3xy + (2 − x)y − y = 3r2 − r c0 xr−1
+
∞
[3(k + r − 1)(k + r)ck + 2(k + r)ck − (k + r)ck−1 ]xk+r−1
k=1
= 0,
which implies
3r2 − r = r(3r − 1) = 0
and
(k + r)(3k + 3r − 1)ck − (k + r)ck−1 = 0.
The indicial roots are r = 0 and r = 1/3. For r = 0 the recurrence relation is
ck−1
ck =
, k = 1, 2, 3, . . . ,
3k − 1
and
1
1
1
c2 =
c3 =
c0 ,
c0 ,
c0 ,
2
10
80
and so on. For r = 1/3 the recurrence relation is
ck−1
ck =
, k = 1, 2, 3, . . . ,
3k
and
1
1
1
c1 = c0 ,
c2 =
c3 =
c0 ,
c0 ,
3
18
162
and so on. The general solution on (0, ∞) is
1
1
1
1 3
1
1
y = C1 1 + x + x2 + x3 + · · · + C2 x1/3 1 + x + x2 +
x + ··· .
2
10
80
3
18
162
∞
20. Substituting y = n=0 cn xn+r into the differential equation and collecting terms, we obtain
∞ 2
2
2
2 2
r
(k + r)(k + r − 1)ck + ck − ck−1 xk+r
x y − x−
y = r −r+
c0 x +
9
9
9
c1 =
k=1
= 0,
which implies
2
r −r+ =
9
2
and
2
r−
3
1
r−
3
=0
2
(k + r)(k + r − 1) +
ck − ck−1 = 0.
9
277
5.2
Solutions About Singular Points
The indicial roots are r = 2/3 and r = 1/3. For r = 2/3 the recurrence relation is
ck =
3ck−1
,
3k 2 + k
3
c0 ,
4
c2 =
k = 1, 2, 3, . . . ,
and
c1 =
9
c0 ,
56
c3 =
9
c0 ,
560
and so on. For r = 1/3 the recurrence relation is
ck =
3ck−1
,
3k 2 − k
3
c0 ,
2
c2 =
k = 1, 2, 3, . . . ,
and
c1 =
9
c0 ,
20
c3 =
9
c0 ,
160
and so on. The general solution on (0, ∞) is
3
3
9 3
9 3
9
9
y = C1 x2/3 1 + x + x2 +
x + · · · + C2 x1/3 1 + x + x2 +
x + ··· .
4
56
560
2
20
160
21. Substituting y =
∞
n+r
into the differential equation and collecting terms, we obtain
n=0 cn x
2xy − (3 + 2x)y + y = 2r2 − 5r c0 xr−1 +
∞
[2(k + r)(k + r − 1)ck
k=1
− 3(k + r)ck − 2(k + r − 1)ck−1 + ck−1 ]xk+r−1
= 0,
which implies
2r2 − 5r = r(2r − 5) = 0
and
(k + r)(2k + 2r − 5)ck − (2k + 2r − 3)ck−1 = 0.
The indicial roots are r = 0 and r = 5/2. For r = 0 the recurrence relation is
ck =
(2k − 3)ck−1
,
k(2k − 5)
k = 1, 2, 3, . . . ,
and
c1 =
1
c0 ,
3
1
c2 = − c0 ,
6
1
c3 = − c0 ,
6
and so on. For r = 5/2 the recurrence relation is
ck =
2(k + 1)ck−1
,
k(2k + 5)
k = 1, 2, 3, . . . ,
and
c1 =
4
c0 ,
7
c2 =
4
c0 ,
21
c3 =
32
c0 ,
693
and so on. The general solution on (0, ∞) is
1
32 3
1
4
1
4
y = C1 1 + x − x2 − x3 + · · · + C2 x5/2 1 + x + x2 +
x + ··· .
3
6
6
7
21
693
278
5.2
22. Substituting y =
Solutions About Singular Points
∞
n+r
into the differential equation and collecting terms, we obtain
n=0 cn x
4
4
5
x2 y + xy + x2 −
y = r2 −
c0 xr + r2 + 2r +
c1 xr+1
9
9
9
∞ 4
(k + r)(k + r − 1)ck + (k + r)ck − ck + ck−2 xk+r
+
9
k=2
= 0,
which implies
r2 −
and
4
2
2
= r+
r−
= 0,
9
3
3
5
r2 + 2r +
c1 = 0,
9
4
(k + r)2 −
ck + ck−2 = 0.
9
The indicial roots are r = −2/3 and r = 2/3, so c1 = 0. For r = −2/3 the recurrence relation is
ck = −
9ck−2
,
3k(3k − 4)
k = 2, 3, 4, . . . ,
and
3
c2 = − c0 ,
4
and so on. For r = 2/3 the recurrence relation is
ck = −
and
c2 = −
c3 = 0,
9ck−2
,
3k(3k + 4)
3
c0 ,
20
c3 = 0,
c4 =
9
c0 ,
128
k = 2, 3, 4, . . . ,
c4 =
9
c0 ,
1,280
and so on. The general solution on (0, ∞) is
3 2
3 2
9 4
9
−2/3
2/3
4
y = C1 x
1− x +
1− x +
x + · · · + C2 x
x + ··· .
4
128
20
1,280
∞
23. Substituting y = n=0 cn xn+r into the differential equation and collecting terms, we obtain
9x2 y + 9x2 y + 2y = 9r2 − 9r + 2 c0 xr +
∞
[9(k + r)(k + r − 1)ck + 2ck + 9(k + r − 1)ck−1 ]xk+r = 0,
k=1
which implies
9r2 − 9r + 2 = (3r − 1)(3r − 2) = 0
and
[9(k + r)(k + r − 1) + 2]ck + 9(k + r − 1)ck−1 = 0.
The indicial roots are r = 1/3 and r = 2/3. For r = 1/3 the recurrence relation is
ck = −
and
(3k − 2)ck−1
,
k(3k − 1)
1
c1 = − c0 ,
2
c2 =
1
c0 ,
5
279
k = 1, 2, 3, . . . ,
c3 = −
7
c0 ,
120
5.2
Solutions About Singular Points
and so on. For r = 2/3 the recurrence relation is
ck = −
(3k − 1)ck−1
,
k(3k + 1)
k = 1, 2, 3, . . . ,
and
1
5
1
c1 = − c0 ,
c2 =
c3 = − c0 ,
c0 ,
2
28
21
and so on. The general solution on (0, ∞) is
1
1
7 3
1 3
1 2
5 2
1/3
2/3
y = C1 x
1− x+ x −
1 − x + x − x + ··· .
x + · · · + C2 x
2
5
120
2
28
21
∞
24. Substituting y = n=0 cn xn+r into the differential equation and collecting terms, we obtain
2x2 y + 3xy + (2x − 1)y = 2r2 + r − 1 c0 xr +
∞
[2(k + r)(k + r − 1)ck + 3(k + r)ck − ck + 2ck−1 ]xk+r = 0,
k=1
which implies
2r2 + r − 1 = (2r − 1)(r + 1) = 0
and
[(k + r)(2k + 2r + 1) − 1]ck + 2ck−1 = 0.
The indicial roots are r = −1 and r = 1/2. For r = −1 the recurrence relation is
ck = −
2ck−1
,
k(2k − 3)
k = 1, 2, 3, . . . ,
and
c2 = −2c0 ,
c1 = 2c0 ,
c3 =
4
c0 ,
9
and so on. For r = 1/2 the recurrence relation is
ck = −
2ck−1
,
k(2k + 3)
k = 1, 2, 3, . . . ,
and
2
2
4
c1 = − c0 ,
c2 =
c3 = −
c0 ,
c0 ,
5
35
945
and so on. The general solution on (0, ∞) is
2
2
4
4 3
y = C1 x−1 1 + 2x − 2x2 + x3 + · · · + C2 x1/2 1 − x + x2 −
x + ··· .
9
5
35
945
∞
25. Substituting y = n=0 cn xn+r into the differential equation and collecting terms, we obtain
xy + 2y − xy = r + r c0 x
2
r−1
2
r
+ r + 3r + 2 c1 x +
∞
[(k + r)(k + r − 1)ck + 2(k + r)ck − ck−2 ]xk+r−1 = 0,
k=2
which implies
r2 + r = r(r + 1) = 0,
r2 + 3r + 2 c1 = 0,
and
(k + r)(k + r + 1)ck − ck−2 = 0.
280
5.2
Solutions About Singular Points
The indicial roots are r1 = 0 and r2 = −1, so c1 = 0. For r1 = 0 the recurrence relation is
ck−2
ck =
, k = 2, 3, 4, . . . ,
k(k + 1)
and
1
c0
3!
c3 = c5 = c7 = · · · = 0
1
c4 = c0
5!
1
c2n =
c0 .
(2n + 1)!
c2 =
For r2 = −1 the recurrence relation is
ck =
and
ck−2
,
k(k − 1)
k = 2, 3, 4, . . . ,
1
c0
2!
c3 = c5 = c7 = · · · = 0
1
c4 = c0
4!
1
c0 .
c2n =
(2n)!
c2 =
The general solution on (0, ∞) is
y = C1
∞
1
1
x2n + C2 x−1
x2n
(2n
+
1)!
(2n)!
n=0
n=0
∞
∞
∞
1
1
1
2n+1
2n
=
+ C2
C1
x
x
x
(2n + 1)!
(2n)!
n=0
n=0
=
26. Substituting y =
∞
n+r
into the differential equation and collecting terms, we obtain
n=0 cn x
1
x y + xy + x −
4
2 1
[C1 sinh x + C2 cosh x].
x
2
y=
1
r −
4
2
3
c0 x + r + 2r +
4
r
2
c1 xr+1
∞ 1
(k + r)(k + r − 1)ck + (k + r)ck − ck + ck−2 xk+r
+
4
k=2
= 0,
which implies
r2 −
and
1
1
1
= r−
r+
= 0,
4
2
2
3
r2 + 2r +
c1 = 0,
4
1
(k + r)2 −
ck + ck−2 = 0.
4
281
5.2
Solutions About Singular Points
The indicial roots are r1 = 1/2 and r2 = −1/2, so c1 = 0. For r1 = 1/2 the recurrence relation is
ck = −
ck−2
,
k(k + 1)
k = 2, 3, 4, . . . ,
and
1
c0
3!
c3 = c5 = c7 = · · · = 0
1
c4 = c0
5!
(−1)n
c2n =
c0 .
(2n + 1)!
c2 = −
For r2 = −1/2 the recurrence relation is
ck = −
ck−2
,
k(k − 1)
k = 2, 3, 4, . . . ,
and
1
c0
2!
c3 = c5 = c7 = · · · = 0
1
c4 = c0
4!
(−1)n
c0 .
c2n =
(2n)!
c2 = −
The general solution on (0, ∞) is
y = C1 x1/2
∞
∞
(−1)n 2n
(−1)n 2n
x + C2 x−1/2
x
(2n + 1)!
(2n)!
n=0
n=0
= C1 x−1/2
∞
∞
(−1)n 2n+1
(−1)n 2n
+ C2 x−1/2
x
x
(2n + 1)!
(2n)!
n=0
n=0
= x−1/2 [C1 sin x + C2 cos x].
27. Substituting y =
∞
n+r
into the differential equation and collecting terms, we obtain
n=0 cn x
xy − xy + y = r2 − r c0 xr−1 +
∞
[(k + r + 1)(k + r)ck+1 − (k + r)ck + ck ]xk+r = 0
k=0
which implies
r2 − r = r(r − 1) = 0
and
(k + r + 1)(k + r)ck+1 − (k + r − 1)ck = 0.
The indicial roots are r1 = 1 and r2 = 0. For r1 = 1 the recurrence relation is
ck+1 =
kck
,
(k + 2)(k + 1)
282
k = 0, 1, 2, . . . ,
5.2
Solutions About Singular Points
and one solution is y1 = c0 x. A second solution is
− −1 dx
x
1 2
e
e
1
1 3
y2 = x
1 + x + x + x + · · · dx
dx = x
dx = x
x2
x2
x2
2
3!
1
1
1 2
1
1 2
1 1
1
1 3
=x
+ + + x + x + · · · dx = x − + ln x + x + x + x + · · ·
x2
x 2 3!
4!
x
2
12
72
1
1
1
= x ln x − 1 + x2 + x3 + x4 + · · · .
2
12
72
The general solution on (0, ∞) is
y = C1 x + C2 y2 (x).
∞
28. Substituting y = n=0 cn xn+r into the differential equation and collecting terms, we obtain
3
y + y − 2y = r2 + 2r c0 xr−2 + r2 + 4r + 3 c1 xr−1
x
+
∞
[(k + r)(k + r − 1)ck + 3(k + r)ck − 2ck−2 ]xk+r−2
k=2
= 0,
which implies
r2 + 2r = r(r + 2) = 0
r2 + 4r + 3 c1 = 0
(k + r)(k + r + 2)ck − 2ck−2 = 0.
The indicial roots are r1 = 0 and r2 = −2, so c1 = 0. For r1 = 0 the recurrence relation is
2ck−2
ck =
, k = 2, 3, 4, . . . ,
k(k + 2)
and
1
c2 = c0
4
c3 = c5 = c7 = · · · = 0
1
c4 =
c0
48
1
c0 .
c6 =
1,152
The result is
1 2
1 4
1
y1 = c0 1 + x + x +
x6 + · · · .
4
48
1,152
A second solution is
− (3/x)dx
e
dx
y2 = y 1
dx
=
y
1
2
1 2
1 4
3
y12
x 1 + 4 x + 48
x + ···
dx
1
1 2
7 4
19 6
x
x
x
1
−
=
y
= y1
+
+
+
·
·
·
dx
1
5 4
7
x3
2
48
576
x3 1 + 12 x2 + 48
x + 576
x6 + · · ·
1
1
19 4
1
1
7
19 3
7 2
−
+
·
·
·
dx
=
y
−
−
+
·
·
·
= y1
−
+
x
−
x
ln
x
+
x
x
1
x3
2x 48
576
2x2
2
96
2,304
1
1
7
19 4
= − y1 ln x + y − 2 + x2 −
x + ··· .
2
2x
96
2,304
283
5.2
Solutions About Singular Points
The general solution on (0, ∞) is
y = C1 y1 (x) + C2 y2 (x).
29. Substituting y =
∞
n+r
into the differential equation and collecting terms, we obtain
n=0 cn x
xy + (1 − x)y − y = r c0 x
2
r−1
∞
+
[(k + r)(k + r − 1)ck + (k + r)ck − (k + r)ck−1 ]xk+r−1 = 0,
k=1
which implies r2 = 0 and
(k + r)2 ck − (k + r)ck−1 = 0.
The indicial roots are r1 = r2 = 0 and the recurrence relation is
ck =
One solution is
y1 = c0
A second solution is
ck−1
,
k
k = 1, 2, 3, . . . .
1 2
1 3
1 + x + x + x + · · · = c0 ex .
2
3!
ex /x
1 −x
x
e dx
dx
=
e
2x
2x
e
e
x
1
1
1 3
1 2
1
1 2
x
x
=e
1 − x + x − x + · · · dx = e
− 1 + x − x + · · · dx
x
2
3!
x
2
3!
∞
1 2
1 3
(−1)n+1 n
= ex ln x − x +
x −
x + · · · = ex ln x − ex
x .
2·2
3 · 3!
n · n!
n=1
y2 = y 1
e−
(1/x−1)dx
dx = ex
The general solution on (0, ∞) is
y = C1 ex + C2 ex
ln x −
∞
(−1)n+1
n=1
30. Substituting y =
n · n!
xn
.
∞
n+r
into the differential equation and collecting terms, we obtain
n=0 cn x
xy + y + y = r2 c0 xr−1 +
∞
[(k + r)(k + r − 1)ck + (k + r)ck + ck−1 ]xk+r−1 = 0
k=1
which implies r2 = 0 and
(k + r)2 ck + ck−1 = 0.
The indicial roots are r1 = r2 = 0 and the recurrence relation is
ck = −
One solution is
ck−1
,
k2
k = 1, 2, 3, . . . .
∞
1
1 3
1 4
(−1)n n
y1 = c0 1 − x + 2 x2 −
x
+
x
−
·
·
·
=
c
x .
0
2
2
2
(3!)
(4!)
(n!)2
n=0
284
5.2
A second solution is
y2 = y 1
e−
(1/x)dx
y12
dx = y1
Solutions About Singular Points
dx
x
2
1 3
1 − x + 14 x2 − 36
x + ···
dx
35 4
x 1 − 2x + 32 x2 − 59 x3 + 288
x − ···
1
5 2 23 3 677 4
= y1
1 + 2x + x + x +
x + · · · dx
x
2
9
288
1
677 3
5
23
= y1
+ 2 + x + x2 +
x + · · · dx
x
2
9
288
5
23
677 4
= y1 ln x + 2x + x2 + x3 +
x + ···
4
27
1,152
5
23
677 4
= y1 ln x + y1 2x + x2 + x3 +
x + ··· .
4
27
1,152
= y1
The general solution on (0, ∞) is
y = C1 y1 (x) + C2 y2 (x).
31. Substituting y =
∞
n+r
into the differential equation and collecting terms, we obtain
n=0 cn x
xy + (x − 6)y − 3y = (r2 − 7r)c0 xr−1 +
∞
(k + r)(k + r − 1)ck + (k + r − 1)ck−1
k=1
− 6(k + r)ck − 3ck−1 xk+r−1 = 0,
which implies
r2 − 7r = r(r − 7) = 0
and
(k + r)(k + r − 7)ck + (k + r − 4)ck−1 = 0.
The indicial roots are r1 = 7 and r2 = 0. For r1 = 7 the recurrence relation is
(k + 7)kck + (k + 3)ck−1 = 0,
k = 1, 2, 3, . . . ,
or
ck = −
k+3
ck−1 ,
k(k + 7)
k = 1, 2, 3, . . . .
Taking c0 = 0 we obtain
1
c1 = − c0
2
5
c0
c2 =
18
1
c3 = − c0 ,
6
and so on. Thus, the indicial root r1 = 7 yields a single solution. Now, for r2 = 0 the recurrence relation is
k(k − 7)ck + (k − 4)ck−1 = 0,
285
k = 1, 2, 3, . . . .
5.2
Solutions About Singular Points
Then
−6c1 − 3c0 = 0
−10c2 − 2c1 = 0
−12c3 − c2 = 0
−12c4 + 0c3 = 0 =⇒ c4 = 0
−10c5 + c4 = 0 =⇒ c5 = 0
−6c6 + 2c5 = 0 =⇒ c6 = 0
0c7 + 3c6 = 0 =⇒ c7 is arbitrary
and
ck = −
Taking c0 = 0 and c7 = 0 we obtain
k−4
ck−1 ,
k(k − 7)
k = 8, 9, 10, . . . .
1
c1 = − c0
2
1
c2 =
c0
10
1
c3 = −
c0
120
c4 = c5 = c6 = · · · = 0.
Taking c0 = 0 and c7 = 0 we obtain
c1 = c2 = c3 = c4 = c5 = c6 = 0
1
c8 = − c7
2
5
c9 =
c7
36
1
c10 = − c7 ,
36
and so on. In this case we obtain the two solutions
1
1
1 3
1
5
1
y1 = 1 − x + x2 −
x
and
y2 = x7 − x8 + x9 − x10 + · · · .
2
10
120
2
36
36
∞
32. Substituting y = n=0 cn xn+r into the differential equation and collecting terms, we obtain
x(x − 1)y + 3y − 2y
= 4r − r2 c0 xr−1 +
∞
[(k + r − 1)(k + r − 12)ck−1 − (k + r)(k + r − 1)ck + 3(k + r)ck − 2ck−1 ]xk+r−1
k=1
= 0,
which implies
4r − r2 = r(4 − r) = 0
and
−(k + r)(k + r − 4)ck + [(k + r − 1)(k + r − 2) − 2]ck−1 = 0.
The indicial roots are r1 = 4 and r2 = 0. For r1 = 4 the recurrence relation is
−(k + 4)kck + [(k + 3)(k + 2) − 2]ck−1 = 0
286
5.2
Solutions About Singular Points
or
k+1
ck−1 ,
k
ck =
k = 1, 2, 3, . . . .
Taking c0 = 0 we obtain
c1 = 2c0
c2 = 3c0
c3 = 4c0 ,
and so on. Thus, the indicial root r1 = 4 yields a single solution. For r2 = 0 the recurrence relation is
−k(k − 4)ck + k(k − 3)ck−1 = 0,
k = 1, 2, 3, . . . ,
−(k − 4)ck + (k − 3)ck−1 = 0,
k = 1, 2, 3, . . . .
or
Then
3c1 − 2c0 = 0
2c2 − c1 = 0
c3 + 0c2 = 0 ⇒ c3 = 0
0c4 + c3 = 0 ⇒ c4 is arbitrary
and
ck =
(k − 3)ck−1
,
k−4
k = 5, 6, 7, . . . .
Taking c0 = 0 and c4 = 0 we obtain
2
c0
3
1
c2 = c0
3
c3 = c4 = c5 = · · · = 0.
c1 =
Taking c0 = 0 and c4 = 0 we obtain
c1 = c2 = c3 = 0
c5 = 2c4
c6 = 3c4
c7 = 4c4 ,
and so on. In this case we obtain the two solutions
2
1
y1 = 1 + x + x2
3
3
y2 = x4 + 2x5 + 3x6 + 4x7 + · · · .
and
33. (a) From t = 1/x we have dt/dx = −1/x2 = −t2 . Then
dy
dy
dy dt
=
= −t2
dx
dt dx
dt
and
d2 y
d
=
2
dx
dx
dy
dx
=
d
dx
−t2
Now
x4
d2 y
1
+ λy = 4
dx2
t
dy
dt
t4
= −t2
dy
d2 y dt
−
2
dt dx
dt
d2 y
dy
+ 2t3
dt2
dt
287
+ λy =
2t
dt
dx
= t4
d2 y
dy
+ 2t3
.
2
dt
dt
d2 y 2 dy
+
+ λy = 0
dt2
t dt
5.2
Solutions About Singular Points
becomes
t
(b) Substituting y =
t
∞
n=0 cn t
n+r
d2 y
dy
+2
+ λty = 0.
2
dt
dt
into the differential equation and collecting terms, we obtain
d2 y
dy
+2
+ λty = (r2 + r)c0 tr−1 + (r2 + 3r + 2)c1 tr
2
dt
dt
+
∞
[(k + r)(k + r − 1)ck + 2(k + r)ck + λck−2 ]tk+r−1
k=2
= 0,
which implies
r2 + r = r(r + 1) = 0,
r2 + 3r + 2 c1 = 0,
and
(k + r)(k + r + 1)ck + λck−2 = 0.
The indicial roots are r1 = 0 and r2 = −1, so c1 = 0. For r1 = 0 the recurrence relation is
ck = −
λck−2
,
k(k + 1)
k = 2, 3, 4, . . . ,
and
λ
c0
3!
c3 = c5 = c7 = · · · = 0
c2 = −
c4 =
λ2
c0
5!
..
.
c2n = (−1)n
λn
c0 .
(2n + 1)!
For r2 = −1 the recurrence relation is
ck = −
λck−2
,
k(k − 1)
k = 2, 3, 4, . . . ,
and
λ
c0
2!
c3 = c5 = c7 = · · · = 0
c2 = −
c4 =
λ2
c0
4!
..
.
c2n = (−1)n
λn
c0 .
(2n)!
288
5.2
Solutions About Singular Points
The general solution on (0, ∞) is
y(t) = c1
∞
∞
(−1)n √
(−1)n √
( λ t)2n + c2 t−1
( λ t)2n
(2n
+
1)!
(2n)!
n=0
n=0
=
∞
∞
(−1)n √
(−1)n √
1
C1
( λ t)2n+1 + C2
( λ t)2n
t
(2n + 1)!
(2n)!
n=0
n=0
=
√
√
1
[C1 sin λ t + C2 cos λ t ].
t
(c) Using t = 1/x, the solution of the original equation is
√
√
λ
λ
y(x) = C1 x sin
+ C2 x cos
.
x
x
34. (a) From the boundary conditions y(a) = 0, y(b) = 0 we find
√
√
λ
λ
C1 sin
+ C2 cos
=0
a
a
√
√
λ
λ
C1 sin
+ C2 cos
= 0.
b
b
Since this is a homogeneous system of linear equations, it will have nontrivial solutions for C1 and C2 if
√
λ
sin
a
√
λ
sin
b
√ λ
√
√
√
√
cos
λ
λ
λ
λ
a cos
−
cos
sin
=
sin
√ a
b
a
b
λ
cos
b
√
√ √ b−a
λ
λ
= sin
λ
−
= sin
= 0.
a
b
ab
This will be the case if
√
b−a
= nπ
λ
ab
or
√
λ=
nπab
nπab
=
, n = 1, 2, . . . ,
b−a
L
or, if
λn =
n2 π 2 a2 b2
Pn b 4
=
.
2
L
EI
√
√
The critical loads are then Pn = n2 π 2 (a/b)2 EI0 /L2 . Using C2 = −C1 sin( λ/a)/ cos( λ/a) we have
√ √
√
λ
λ
sin( λ/a)
√
y = C1 x sin
cos
−
x
x
cos( λ/a)
√
√
√
√ λ
λ
λ
λ
= C3 x sin
cos
− cos
sin
x
a
x
a
√
1
1
= C3 x sin λ
−
,
x a
and
yn (x) = C3 x sin
nπab
L
1
1
−
x a
= C3 x sin
nπab a
nπab a
− 1 = C4 x sin
1−
.
La x
L
x
289
5.2
Solutions About Singular Points
(b) When n = 1, b = 11, and a = 1, we have,
for C4 = 1,
1
y1 (x) = x sin 1.1π 1 −
x
y
2
.
1
1
3
5
7
9
11
x
35. Express the differential equation in standard form:
y + P (x)y + Q(x)y + R(x)y = 0.
Suppose x0 is a singular point of the differential equation. Then we say that x0 is a regular singular point if
(x − x0 )P (x), (x − x0 )2 Q(x), and (x − x0 )3 R(x) are analytic at x = x0 .
36. Substituting y =
∞
n+r
into the first differential equation and collecting terms, we obtain
n=0 cn x
x3 y + y = c0 xr +
∞
[ck + (k + r − 1)(k + r − 2)ck−1 ]xk+r = 0.
k=1
It follows that c0 = 0 and
ck = −(k + r − 1)(k + r − 2)ck−1 .
The only solution we obtain is y(x) = 0.
Substituting y =
∞
n+r
into the second differential equation and collecting terms, we obtain
n=0 cn x
x2 y + (3x − 1)y + y = −rc0 +
∞
[(k + r + 1)2 ck − (k + r + 1)ck+1 ]xk+r = 0,
k=0
which implies
−rc0 = 0
(k + r + 1) ck − (k + r + 1)ck+1 = 0.
2
If c0 = 0, then the solution of the differential equation is y = 0. Thus, we take r = 0, from which we obtain
ck+1 = (k + 1)ck , k = 0, 1, 2, . . . .
Letting c0 = 1 we get c1 = 2, c2 = 3!, c3 = 4!, and so on. The solution of the differential equation is then
∞
y = n=0 (n + 1)!xn , which converges only at x = 0.
37. We write the differential equation in the form x2 y + (b/a)xy + (c/a)y = 0 and identify a0 = b/a and b0 = c/a
as in (12) in the text. Then the indicial equation is
r(r − 1) +
b
c
r+ =0
a
a
or
ar2 + (b − a)r + c = 0,
which is also the auxiliary equation of ax2 y + bxy + cy = 0.
290
5.3
Special Functions
EXERCISES 5.3
Special Functions
1. Since ν 2 = 1/9 the general solution is y = c1 J1/3 (x) + c2 J−1/3 (x).
2. Since ν 2 = 1 the general solution is y = c1 J1 (x) + c2 Y1 (x).
3. Since ν 2 = 25/4 the general solution is y = c1 J5/2 (x) + c2 J−5/2 (x).
4. Since ν 2 = 1/16 the general solution is y = c1 J1/4 (x) + c2 J−1/4 (x).
5. Since ν 2 = 0 the general solution is y = c1 J0 (x) + c2 Y0 (x).
6. Since ν 2 = 4 the general solution is y = c1 J2 (x) + c2 Y2 (x).
7. We identify α = 3 and ν = 2. Then the general solution is y = c1 J2 (3x) + c2 Y2 (3x).
8. We identify α = 6 and ν = 12 . Then the general solution is y = c1 J1/2 (6x) + c2 J−1/2 (6x).
9. We identify α = 5 and ν = 23 . Then the general solution is y = c1 J2/3 (5x) + c2 J−2/3 (5x).
√
√
√
10. We identify α = 2 and ν = 8. Then the general solution is y = c1 J8 ( 2x) + c2 Y8 ( 2x).
11. If y = x−1/2 v(x) then
1
y = x−1/2 v (x) − x−3/2 v(x),
2
3
y = x−1/2 v (x) − x−3/2 v (x) + x−5/2 v(x),
4
and
1
x2 y + 2xy + α2 x2 y = x3/2 v (x) + x1/2 v (x) + α2 x3/2 − x−1/2 v(x) = 0.
4
Multiplying by x1/2 we obtain
1
x v (x) + xv (x) + α x −
4
2 2 2
v(x) = 0,
whose solution is v = c1 J1/2 (αx) + c2 J−1/2 (αx). Then y = c1 x−1/2 J1/2 (αx) + c2 x−1/2 J−1/2 (αx).
√
12. If y = x v(x) then
1
y = x1/2 v (x) + x−1/2 v(x)
2
1
1/2 y = x v (x) + x−1/2 v (x) − x−3/2 v(x)
4
and
1
1
1
x2 y + α2 x2 − ν 2 +
y = x5/2 v (x) + x3/2 v (x) − x1/2 v(x) + α2 x2 − ν 2 +
x1/2 v(x)
4
4
4
= x5/2 v (x) + x3/2 v (x) + (α2 x5/2 − ν 2 x1/2 )v(x) = 0.
Multiplying by x−1/2 we obtain
x2 v (x) + xv (x) + (α2 x2 − ν 2 )v(x) = 0,
√
√
whose solution is v(x) = c1 Jν (αx) + c2 Yν (αx). Then y = c1 x Jν (αx) + c2 x Yν (αx).
291
5.3
Special Functions
13. Write the differential equation in the form y + (2/x)y + (4/x)y = 0. This is the form of (18) in the text with
a = − 12 , c = 12 , b = 4, and p = 1, so, by (19) in the text, the general solution is
y = x−1/2 [c1 J1 (4x1/2 ) + c2 Y1 (4x1/2 )].
14. Write the differential equation in the form y + (3/x)y + y = 0. This is the form of (18) in the text with a = −1,
c = 1, b = 1, and p = 1, so, by (19) in the text, the general solution is
y = x−1 [c1 J1 (x) + c2 Y1 (x)].
15. Write the differential equation in the form y − (1/x)y + y = 0. This is the form of (18) in the text with a = 1,
c = 1, b = 1, and p = 1, so, by (19) in the text, the general solution is
y = x[c1 J1 (x) + c2 Y1 (x)].
16. Write the differential equation in the form y − (5/x)y + y = 0. This is the form of (18) in the text with a = 3,
c = 1, b = 1, and p = 2, so, by (19) in the text, the general solution is
y = x3 [c1 J3 (x) + c2 Y3 (x)].
17. Write the differential equation in the form y + (1 − 2/x2 )y = 0. This is the form of (18) in the text with a = 12 ,
c = 1, b = 1, and p = 32 , so, by (19) in the text, the general solution is
y = x1/2 [c1 J3/2 (x) + c2 Y3/2 (x)] = x1/2 [C1 J3/2 (x) + C2 J−3/2 (x)].
18. Write the differential equation in the form y + (4 + 1/4x2 )y = 0. This is the form of (18) in the text with
a = 12 , c = 1, b = 2, and p = 0, so, by (19) in the text, the general solution is
y = x1/2 [c1 J0 (2x) + c2 Y0 (2x)].
19. Write the differential equation in the form y + (3/x)y + x2 y = 0. This is the form of (18) in the text with
a = −1, c = 2, b = 12 , and p = 12 , so, by (19) in the text, the general solution is
1 2
1 2
y = x−1 c1 J1/2
x + c2 Y1/2
x
2
2
or
−1
y=x
1 2
1 2
x + C2 J−1/2
x
C1 J1/2
.
2
2
20. Write the differential equation in the form y + (1/x)y + ( 19 x4 − 4/x2 )y = 0. This is the form of (18) in the
text with a = 0, c = 3, b = 19 , and p = 23 , so, by (19) in the text, the general solution is
y = c1 J2/3
or
y = C1 J2/3
1 3
x
9
1 3
x
9
+ c2 Y2/3
1 3
x
9
+ C2 J−2/3
292
1 3
x .
9
5.3
Special Functions
21. Using the fact that i2 = −1, along with the definition of Jν (x) in (7) in the text, we have
Iν (x) = i−ν Jν (ix) = i−ν
∞
(−1)n
n!Γ(1 + ν + n)
n=0
=
x 2n+ν
(−1)n
i2n+ν−ν
n!Γ(1 + ν + n)
2
n=0
=
x 2n+ν
(−1)n
(i2 )n
n!Γ(1 + ν + n)
2
n=0
=
x 2n+ν
(−1)2n
n!Γ(1 + ν + n) 2
n=0
=
x 2n+ν
1
,
n!Γ(1 + ν + n) 2
n=0
ix
2
2n+ν
∞
∞
∞
∞
which is a real function.
22. (a) The differential equation has the form of (18) in the text with
1
2
2c − 2 = 2 =⇒ c = 2
1
1
b2 c2 = −β 2 c2 = −1 =⇒ β =
and b = i
2
2
1
2
2 2
a − p c = 0 =⇒ p = .
4
1 − 2a = 0 =⇒ a =
Then, by (19) in the text,
1 2
1 2
y = x1/2 c1 J1/4
.
ix + c2 J−1/4
ix
2
2
In terms of real functions the general solution can be written
1 2
1 2
1/2
y=x
x + C2 K1/4
x
C1 I1/4
.
2
2
(b) Write the differential equation in the form y + (1/x)y − 7x2 y = 0. This is the form of (18) in the text
with
1 − 2a = 1 =⇒ a = 0
2c − 2 = 2 =⇒ c = 2
1√
b2 c2 = −β 2 c2 = −7 =⇒ β =
7
2
a2 − p2 c2 = 0 =⇒ p = 0.
Then, by (19) in the text,
y = c1 J0
1√
7 ix2
2
+ c2 Y0
and b =
1√
7 ix2 .
2
In terms of real functions the general solution can be written
1√ 2
1√ 2
y = C1 I0
7x + C2 K0
7x .
2
2
293
1√
7i
2
5.3
Special Functions
23. The differential equation has the form of (18) in the text with
1
2
2c − 2 = 0 =⇒ c = 1
1 − 2a = 0 =⇒ a =
b2 c2 = 1 =⇒ b = 1
1
a2 − p2 c2 = 0 =⇒ p = .
2
Then, by (19) in the text,
y = x1/2 [c1 J1/2 (x) + c2 J−1/2 (x)] = x1/2 c1
2
sin x + c2
πx
2
cos x = C1 sin x + C2 cos x.
πx
24. Write the differential equation in the form y + (4/x)y + (1 + 2/x2 )y = 0. This is the form of (18) in the text
with
1 − 2a = 4 =⇒ a = −
3
2
2c − 2 = 0 =⇒ c = 1
b2 c2 = 1 =⇒ b = 1
1
a2 − p2 c2 = 2 =⇒ p = .
2
Then, by (19), (23), and (24) in the text,
−3/2
y=x
−3/2
[c1 J1/2 (x) + c2 J−1/2 (x)] = x
c1
2
sin x + c2
πx
2
1
1
cos x = C1 2 sin x + C2 2 cos x.
πx
x
x
1 2
25. Write the differential equation in the form y + (2/x)y + ( 16
x − 3/4x2 )y = 0. This is the form of (18) in the
text with
1 − 2a = 2 =⇒ a = −
1
2
2c − 2 = 2 =⇒ c = 2
1
1
b2 c2 =
=⇒ b =
16
8
3
1
2
2 2
a − p c = − =⇒ p = .
4
2
Then, by (19) in the text,
1 2
1 2
y=x
x + c2 J−1/2
x
c1 J1/2
8
8
16
16
1 2
1 2
−1/2
=x
c1
x + c2
x
sin
cos
πx2
8
πx2
8
1 2
1 2
−3/2
−3/2
= C1 x
x + C2 x
x .
sin
cos
8
8
−1/2
26. Write the differential equation in the form y − (1/x)y + (4 + 3/4x2 )y = 0. This is the form of (18) in the text
with
1 − 2a = −1 =⇒ a = 1
2c − 2 = 0 =⇒ c = 1
b2 c2 = 4 =⇒ b = 2
3
1
a2 − p2 c2 =
=⇒ p = .
4
2
294
5.3
Special Functions
Then, by (19) in the text,
y = x[c1 J1/2 (2x) + c2 J−1/2 (2x)]
2
2
= x c1
sin 2x + c2
cos 2x
π2x
π2x
= C1 x1/2 sin 2x + C2 x1/2 cos 2x.
27. (a) The recurrence relation follows from
−νJν (x) + xJν−1 (x) = −
=−
=
∞
x 2n+ν
(−1)n ν
(−1)n x 2n+ν−1
+x
n!Γ(1 + ν + n) 2
n!Γ(ν + n) 2
n=0
n=0
∞
∞
x 2n+ν x x 2n+ν−1
(−1)n ν
(−1)n (ν + n)
+
·2
n!Γ(1 + ν + n) 2
n!Γ(1 + ν + n)
2
2
n=0
n=0
∞
∞
(−1)n (2n + ν) x 2n+ν
n!Γ(1 + ν + n)
n=0
2
= xJν (x).
(b) The formula in part (a) is a linear first-order differential equation in Jν (x). An integrating factor for this
equation is xν , so
d ν
[x Jν (x)] = xν Jν−1 (x).
dx
28. Subtracting the formula in part (a) of Problem 27 from the formula in Example 5 we obtain
0 = 2νJν (x) − xJν+1 (x) − xJν−1 (x)
29. Letting ν = 1 in (21) in the text we have
d
xJ0 (x) =
[xJ1 (x)]
dx
30.
or
x
so
2νJν (x) = xJν+1 (x) + xJν−1 (x).
r=x
rJ0 (r) dr = rJ1 (r) = xJ1 (x).
0
r=0
From (20) we obtain J0 (x) = −J1 (x), and from (21) we obtain J0 (x) = J−1 (x). Thus J0 (x) = J−1 (x) = −J1 (x).
31. Since Γ( 12 ) =
we obtain
√
π and
1
(2n − 1)! √
Γ 1− +n =
π
2
(n − 1)!22n−1
n = 1, 2, 3, . . . ,
∞
x 2n−1/2
(−1)n
1 x −1/2 (−1)n (n − 1)!22n−1 x2n−1/2
√
=
+
n!(2n − 1)!22n−1/2 π
n!Γ(1 − 12 + n) 2
Γ( 12 ) 2
n=0
n=1
∞
∞
1
2 (−1)n 21/2 x−1/2 2n
2
2 (−1)n 2n
2
√ x =
=√
+
+
x =
cos x.
πx
πx n=1 (2n)!
πx
π x n=1 2n(2n − 1)! π
J−1/2 (x) =
∞
32. (a) By Problem 28, with ν = 1/2, we obtain J1/2 (x) = xJ3/2 (x) + xJ−1/2 (x) so that
2 sin x
J3/2 (x) =
− cos x ;
πx
x
with ν = −1/2 we obtain −J−1/2 (x) = xJ1/2 (x) + xJ−3/2 (x) so that
2 cos x
J−3/2 (x) = −
+ sin x ;
πx
x
and with ν = 3/2 we obtain 3J3/2 (x) = xJ5/2 (x) + xJ1/2 (x) so that
2 3 sin x 3 cos x
J5/2 (x) =
−
−
sin
x
.
πx
x2
x
295
5.3
Special Functions
(b)
y
1
0.5
y
1
0.5
ν = 1/2
5
10
15
ν = −1/2
20 x
-0.5
-1
5
5
20 x
10
15
20 x
5
10
15
20 x
-0.5
-1
2
s=
α
k −αt/2
,
e
m
dx
dx
k −αt/2
dx ds
dx 2 k α −αt/2
=
=
=
−
e
−
e
dt
ds dt
dt α m
2
ds
m
d dx
k −αt/2
dx
dx α k −αt/2
+
=
e
−
e
dt
ds 2 m
dt ds
m
dx α k −αt/2
d2 x ds
k −αt/2
=
+ 2
e
−
e
ds 2 m
ds dt
m
dx α k −αt/2
d2 x k −αt
=
+ 2
.
e
e
ds 2 m
ds
m
d2 x
d
=
2
dt
dt
Then
d2 x
d2 x mα
m 2 + ke−αt x = ke−αt 2 +
dt
ds
2
Multiplying by 22 /α2 m we have
22 k −αt d2 x
2
e
+
2
2
α m
ds
α
or, since s = (2/α) k/m e−αt/2 ,
s2
34. Differentiating y = x1/2 w
k −αt/2 dx
e
+ ke−αt x = 0.
m
ds
k −αt/2 dx
22 k −αt
e
+ 2
e x=0
m
ds
α m
d2 x
dx
+s
+ s2 x = 0.
ds2
ds
2
3/2
3 αx
with respect to 23 αx3/2 we obtain
2 3/2
1
2 3/2
αx1/2 + x−1/2 w
y = x1/2 w
αx
αx
3
2
3
2 3/2
2 3/2
1/2
y = αxw
αx + αw
αx
αx
3
3
1
2 3/2
1 −3/2
2 3/2
+ αw
− x
.
w
αx
αx
2
3
4
3
Then, after combining terms and simplifying, we have
3
1
y + α2 xy = α αx3/2 w + w + αx3/2 −
w = 0.
2
4αx3/2
296
ν = 3/2
5
ν = 5/2
33. Letting
and
15
-0.5
-1
y
1
0.5
ν = −3/2
-0.5
-1
and
10
-0.5
-1
y
1
0.5
we have
y
1
0.5
10
15
20 x
5.3
Letting t = 23 αx3/2 or αx3/2 = 32 t this differential equation becomes
3 α 2 1
t w (t) + tw (t) + t2 −
w(t) = 0,
2 t
9
Special Functions
t > 0.
35. (a) By Problem 34, a solution of Airy’s equation is y = x1/2 w( 23 αx3/2 ), where
w(t) = c1 J1/3 (t) + c2 J−1/3 (t)
is a solution of Bessel’s equation of order 13 . Thus, the general solution of Airy’s equation for x > 0 is
2 3/2
2 3/2
2 3/2
= c1 x1/2 J1/3
+ c2 x1/2 J−1/3
.
y = x1/2 w
αx
αx
αx
3
3
3
(b) Airy’s equation, y + α2 xy = 0, has the form of (18) in the text with
1
2
3
2c − 2 = 1 =⇒ c =
2
2
2 2
2
b c = α =⇒ b = α
3
1
2
2 2
a − p c = 0 =⇒ p = .
3
1 − 2a = 0 =⇒ a =
Then, by (19) in the text,
1/2
y=x
2 3/2
2 3/2
c1 J1/3
+ c2 J−1/3
.
αx
αx
3
3
36. The general solution of the differential equation is
y(x) = c1 J0 (αx) + c2 Y0 (αx).
In order to satisfy the conditions that limx→0+ y(x) and limx→0+ y (x) are finite we are forced to define c2 = 0.
Thus, y(x) = c1 J0 (αx). The second boundary condition, y(2) = 0, implies c1 = 0 or J0 (2α) = 0. In order to
have a nontrivial solution we require that J0 (2α) = 0. From Table 5.1, the first three positive zeros of J0 are
found to be
2α1 = 2.4048, 2α2 = 5.5201, 2α3 = 8.6537
and so α1 = 1.2024, α2 = 2.7601, α3 = 4.3269. The eigenfunctions corresponding to the eigenvalues λ1 = α12 ,
λ2 = α22 , λ3 = α32 are J0 (1.2024x), J0 (2.7601x), and J0 (4.3269x).
37. (a) The differential equation y + (λ/x)y = 0 has the form of (18) in the text with
1
2
1
2c − 2 = −1 =⇒ c =
2
√
2 2
b c = λ =⇒ b = 2 λ
1 − 2a = 0 =⇒ a =
a2 − p2 c2 = 0 =⇒ p = 1.
Then, by (19) in the text,
√
√
y = x1/2 [c1 J1 (2 λx ) + c2 Y1 (2 λx )].
(b) We first note that y = J1 (t) is a solution of Bessel’s equation, t2 y + ty + (t2 − 1)y = 0, with ν = 1. That
is,
t2 J1 (t) + tJ1 (t) + (t2 − 1)J1 (t) = 0,
297
5.3
Special Functions
√
or, letting t = 2 x ,
Now, if y =
√
√
xJ1 (2 x ), we have
y =
and
√
√
√
√
4xJ1 (2 x ) + 2 xJ1 (2 x ) + (4x − 1)J1 (2 x ) = 0.
√
√
√
√
√
1
1
1
x J1 (2 x ) √ + √ J1 (2 x ) = J1 (2 x ) + x−1/2 J1 (2 x )
2
x 2 x
√
√
1 √
1
y = x−1/2 J1 (2 x ) +
J1 (2 x ) − x−3/2 J1 (2 x ).
2x
4
Then
√
√
√
√
√
1
1
x J1 2 x + J1 (2 x ) − x−1/2 J1 (2 x ) + x J(2 x )
2
4
√
√
√
√
√
1
= √ [4xJ1 (2 x ) + 2 x J1 (2 x ) − J1 (2 x ) + 4xJ(2 x )]
4 x
xy + y =
√
= 0,
√
and y =
√
x J1 (2 x ) is a solution of Airy’s differential equation.
38. We see from the graphs below that the graphs of the modified Bessel functions are not oscillatory, while those
of the Bessel functions, shown in Figures 5.3 and 5.4 in the text, are oscillatory.
I0
I1
I2
20
20
20
15
15
15
10
10
10
5
5
5
1
2
3
4
5 x
1
2
3
5 x
4
K0
5
K1
5
K2
5
4
4
4
3
3
3
2
2
2
1
1
1
1
2
3
4
5 x
1
2
3
4
5 x
39. (a) We identify m = 4, k = 1, and α = 0.1. Then
x(t) = c1 J0 (10e−0.05t ) + c2 Y0 (10e−0.05t )
and
x (t) = −0.5c1 J0 (10e−0.05t ) − 0.5c2 Y0 (10e−0.05t ).
Now x(0) = 1 and x (0) = −1/2 imply
c1 J0 (10) + c2 Y0 (10) = 1
c1 J0 (10) + c2 Y0 (10) = 1.
298
1
2
3
4
5 x
1
2
3
4
5 x
5.3
Special Functions
Using Cramer’s rule we obtain
c1 =
Y0 (10) − Y0 (10)
J0 (10)Y0 (10) − J0 (10)Y0 (10)
c2 =
J0 (10) − J0 (10)
.
J0 (10)Y0 (10) − J0 (10)Y0 (10)
and
Using Y0 = −Y1 and J0 = −J1 and Table 5.2 we find c1 = −4.7860 and c2 = −3.1803. Thus
x(t) = −4.7860J0 (10e−0.05t ) − 3.1803Y0 (10e−0.05t ).
x
(b)
10
5
t
−5
50
100
150
200
40. (a) Identifying α = 12 , the general solution of x + 14 tx = 0 is
1 3/2
1 3/2
1/2
1/2
x(t) = c1 x J1/3
+ c2 x J−1/3
.
x
x
3
3
Solving the system x(0.1) = 1, x (0.1) = − 12 we find c1 = −0.809264 and c2 = 0.782397.
x
(b)
1
t
−1
50
150
100
200
41. (a) Letting t = L − x, the boundary-value problem becomes
d2 θ
+ α2 tθ = 0,
dt2
θ (0) = 0,
θ(L) = 0,
where α2 = δg/EI. This is Airy’s differential equation, so by Problem 35 its solution is
2 3/2
2 3/2
1/2
1/2
y = c1 t J1/3
αt
αt
+ c2 t J−1/3
= c1 θ1 (t) + c2 θ2 (t).
3
3
(b) Looking at the series forms of θ1 and θ2 we see that θ1 (0) = 0, while θ2 (0) = 0. Thus, the boundary
condition θ (0) = 0 implies c1 = 0, and so
√
2 3/2
.
θ(t) = c2 t J−1/3
αt
3
From θ(L) = 0 we have
√
c2 L J−1/3
2 3/2
αL
3
= 0,
so either c2 = 0, in which case θ(t) = 0, or J−1/3 ( 23 αL3/2 ) = 0. The column will just start to bend when L
is the length corresponding to the smallest positive zero of J−1/3 .
299
5.3
Special Functions
(c) Using Mathematica, the first positive root of J−1/3 (x) is x1 ≈ 1.86635. Thus 23 αL3/2 = 1.86635 implies
1/3
2/3 9EI
3(1.86635)
=
L=
(1.86635)2
2α
4δg
1/3
9(2.6 × 107 )π(0.05)4 /4
2
=
(1.86635)
≈ 76.9 in.
4(0.28)π(0.05)2
42. (a) Writing the differential equation in the form xy + (P L/M )y = 0, we identify λ = P L/M .
From
Problem 37 the solution of this differential equation is
√
√
y = c1 x J1 2 P Lx/M + c2 x Y1 2 P Lx/M .
Now J1 (0) = 0, so y(0) = 0 implies c2 = 0 and
√
y = c1 x J1 2 P Lx/M .
√
(b) From y(L) = 0 we have y = J1 (2L P M ) = 0. The first positive zero of J1 is 3.8317 so, solving
2L P1 /M = 3.8317, we find P1 = 3.6705M/L2 . Therefore,
√
√
3.8317 √
3.6705x
√
y1 (x) = c1 x J1 2
x .
= c1 x J1
L
L
(c) For c1 = 1 and L = 1 the graph of y1 =
√
√
x J1 (3.8317 x )
y
0.3
is shown.
0.2
0.1
x
0.2
0.4
0.6
0.8
1
43. (a) Since l = v, we integrate to obtain l(t) = vt + c. Now l(0) = l0 implies c = l0 , so l(t) = vt + l0 . Using
sin θ ≈ θ in l d2 θ/dt2 + 2l dθ/dt + g sin θ = 0 gives
(l0 + vt)
d2 θ
dθ
+ 2v
+ gθ = 0.
dt2
dt
(b) Dividing by v, the differential equation in part (a) becomes
l0 + vt d2 θ
dθ g
+2
+ θ = 0.
2
v
dt
dt
v
Letting x = (l0 + vt)/v = t + l0 /v we have dx/dt = 1, so
dθ
dθ dx
dθ
=
=
dt
dx dt
dx
and
d2 θ
d(dθ/dx) dx
d2 θ
d(dθ/dt)
=
=
=
.
dt2
dt
dx
dt
dx2
Thus, the differential equation becomes
x
d2 θ
dθ
g
+2
+ θ=0
2
dx
dx v
or
300
2 dθ
d2 θ
g
+
+
θ = 0.
2
dx
x dx vx
5.3
Special Functions
(c) The differential equation in part (b) has the form of (18) in the text with
1 − 2a = 2 =⇒ a = −
1
2
1
2
g
g
b2 c2 =
=⇒ b = 2
v
v
2c − 2 = −1 =⇒ c =
a2 − p2 c2 = 0 =⇒ p = 1.
Then, by (19) in the text,
g 1/2
g 1/2
θ(x) = x
c1 J1 2
+ c2 Y1 2
x
x
v
v
2
2
v
θ(t) =
c1 J1
g(l0 + vt) + c2 Y1
g(l0 + vt)
.
l0 + vt
v
v
−1/2
or
(d) To simplify calculations, let
2
g 1/2
u=
g(l0 + vt) = 2
x ,
v
v
√
and at t = 0 let u0 = 2 gl0 /v. The general solution for θ(t) can then be written
θ = C1 u−1 J1 (u) + C2 u−1 Y1 (u).
(1)
Before applying the initial conditions, note that
dθ
dθ du
=
dt
du dt
so when dθ/dt = 0 at t = 0 we have dθ/du = 0 at u = u0 . Also,
dθ
d −1
d −1
= C1
[u J1 (u)] + C2
[u Y1 (u)]
du
du
du
which, in view of (20) in the text, is the same as
dθ
= −C1 u−1 J2 (u) − C2 u−1 Y2 (u).
du
Now at t = 0, or u = u0 , (1) and (2) give the system
−1
C1 u−1
0 J1 (u0 ) + C2 u0 Y1 (u0 ) = θ0
−1
C1 u−1
0 J2 (u0 ) + C2 u0 Y2 (u0 ) = 0
whose solution is easily obtained using Cramer’s rule:
C1 =
u0 θ0 Y2 (u0 )
,
J1 (u0 )Y2 (u0 ) − J2 (u0 )Y1 (u0 )
C2 =
−u0 θ0 J2 (u0 )
.
J1 (u0 )Y2 (u0 ) − J2 (u0 )Y1 (u0 )
In view of the given identity these results simplify to
π
C1 = − u20 θ0 Y2 (u0 )
2
The solution is then
θ=
and
C2 =
π 2
u θ0 J2 (u0 ).
2 0
π 2
J1 (u)
Y1 (u)
u0 θ0 −Y2 (u0 )
+ J2 (u0 )
.
2
u
u
301
(2)
5.3
Special Functions
√
Returning to u = (2/v) g(l0 + vt) and u0 = (2/v) gl0 , we have


2
2
√
J1
Y
g(l
+
vt)
g(l
+
vt)
0
1
0

π gl0 θ0 
2
v
v
−Y2 2 gl0
.
√
√
θ(t) =
gl
+
J
2
0


v
v
v
l0 + vt
l0 + vt
1
1
(e) When l0 = 1 ft, θ0 = 10
radian, and v = 60
ft/s, the above function is
√
√
J1 (480 2(1 + t/60))
Y1 (480 2(1 + t/60))
θ(t) = −1.69045
− 2.79381
.
1 + t/60
1 + t/60
The plots of θ(t) on [0, 10], [0, 30], and [0, 60] are
Θ t
0.1
Θ t
0.1
Θ t
0.1
0.05
0.05
0.05
2
4
6
8
10
t
5
10
15
20
25
30
t
10
-0.05
-0.05
-0.05
-0.1
-0.1
-0.1
(f ) The graphs indicate that θ(t) decreases as l increases. The
20
30
150
200
40
50
60
t
Θ t
0.1
graph of θ(t) on [0, 300] is shown.
0.05
50
100
-0.05
-0.1
44. (a) From (26) in the text, we have
6·7 2 4·6·7·9 4
2 · 4 · 6 · 7 · 9 · 11 6
P6 (x) = c0 1 −
x +
x =
x ,
2!
4!
6!
where
c0 = (−1)3
Thus,
5
P6 (x) = −
16
1·3·5
5
=− .
2·4·6
16
231 6
1 − 21x + 63x −
x
5
2
4
=
1
(231x6 − 315x4 + 105x2 − 5).
16
Also, from (26) in the text we have
6 · 9 3 4 · 6 · 9 · 11 5 2 · 4 · 6 · 9 · 11 · 13 7
P7 (x) = c1 x −
x +
x −
x
3!
5!
7!
where
c1 = (−1)3
Thus
35
P7 (x) = −
16
1·3·5·7
35
=− .
2·4·6
16
99
429 7
x − 9x + x5 −
x
5
35
3
=
1
(429x7 − 693x5 + 315x3 − 35x).
16
(b) P6 (x) satisfies 1 − x2 y − 2xy + 42y = 0 and P7 (x) satisfies 1 − x2 y − 2xy + 56y = 0.
302
250
300
t
5.3
Special Functions
45. The recurrence relation can be written
Pk+1 (x) =
2k + 1
k
xPk (x) −
Pk−1 (x),
k+1
k+1
k = 2, 3, 4, . . . .
3 2 1
x −
2
2
5
3
3 2 1
2
5
k = 2: P3 (x) = x
x −
− x = x3 − x
3
2
2
3
2
2
7
5 3 3
3 3 2 1
35 4 30 2 3
k = 3: P4 (x) = x
x − x −
x −
=
x − x +
4
2
2
4 2
2
8
8
8
35 4 30 2 3
4 5 3 3
63 5 35 3 15
9
k = 4: P5 (x) = x
x − x +
−
x − x =
x − x + x
5
8
8
8
5 2
2
8
4
8
11
5
63 5 35 3 15
5 35 4 30 2 3
231 6 315 4 105 2
k = 5: P6 (x) =
x
x − x + x −
x − x +
=
x −
x +
x −
6
8
4
8
6 8
8
8
16
16
16
16
13
5
231 6 315 4 105 2
6 63 5 35 3 15
k = 6: P7 (x) =
x
x −
x +
x −
−
x − x + x
7
16
16
16
16
7 8
4
8
429 7 693 5 315 3 35
=
x −
x +
x − x
16
16
16
16
46. If x = cos θ then
dy
dy
= − sin θ
,
dθ
dx
k = 1: P2 (x) =
d2 y
d2 y
dy
= sin2 θ 2 − cos θ
,
2
dθ
dx
dx
and
d2 y
dy
sin θ 2 + cos θ
+ n(n + 1)(sin θ)y = sin θ
dθ
dθ
d2 y
dy
− 2 cos θ
1 − cos θ
+ n(n + 1)y = 0.
dx2
dx
2
That is,
1 − x2
dy
d2 y
− 2x
+ n(n + 1)y = 0.
2
dx
dx
47. The only solutions bounded on [−1, 1] are y = cPn (x), c a constant and n = 0, 1, 2, . . . . By (iv) of the
properties of the Legendre polynomials, y(0) = 0 or Pn (0) = 0 implies n must be odd. Thus the first three
positive eigenvalues correspond to n = 1, 3, and 5 or λ1 = 1 · 2, λ2 = 3 · 4 = 12, and λ3 = 5 · 6 = 30. We can
take the eigenfunctions to be y1 = P1 (x), y2 = P3 (x), and y3 = P5 (x).
48. Using a CAS we find
1 d
P1 (x) =
(x2 − 1)1 = x
2 dx
1 d2
1
P2 (x) = 2
(x2 − 1)2 = (3x2 − 1)
2 2! dx2
2
3
1 d
1
P3 (x) = 3
(x2 − 1)3 = (5x3 − 3x)
2 3! dx3
2
1 d4
1
P4 (x) = 4
(x2 − 1)4 = (35x4 − 30x2 + 3)
2 4! dx4
8
1 d5
1
P5 (x) = 5
(x2 − 1)5 = (63x5 − 70x3 + 15x)
2 5! dx5
8
6
1 d
1
P6 (x) = 6
(x2 − 1)6 =
(231x6 − 315x4 + 105x2 − 5)
2 6! dx6
16
1 d7
1
P7 (x) = 7
(x2 − 1)7 =
(429x7 − 693x5 + 315x3 − 35x)
7
2 7! dx
16
303
5.3
Special Functions
P1
1
P2
1
P3
1
P4
1
0.5
0.5
0.5
0.5
49.
-1 -0.5
0.5
1 x
-1 -0.5
0.5
1 x
-1 -0.5
0.5
1 x
-1 -0.5
-0.5
-0.5
-0.5
-0.5
-1
-1
-1
-1
P5
1
P6
1
P7
1
0.5
0.5
0.5
-1 -0.5
0.5
1 x
-1 -0.5
0.5
1 x
-1 -0.5
-0.5
-0.5
-0.5
-1
-1
-1
0.5
0.5
1 x
1 x
50. Zeros of Legendre polynomials for n ≥ 1 are
P1 (x) : 0
P2 (x) : ±0.57735
P3 (x) : 0, ±0.77460
P4 (x) : ±0.33998, ±0.86115
P5 (x) : 0, ±0.53847, ±0.90618
P6 (x) : ±0.23862, ±0.66121, ±0.93247
P7 (x) : 0, ±0.40585, ±0.74153 , ±0.94911
P10 (x) : ±0.14887, ±0.43340, ±0.67941, ±0.86506, ±0.097391
The zeros of any Legendre polynomial are in the interval (−1, 1) and are symmetric with respect to 0.
CHAPTER 5 REVIEW EXERCISES
1. False; J1 (x) and J−1 (x) are not linearly independent when ν is a positive integer. (In this case ν = 1). The
general solution of x2 y + xy + (x2 − 1)y = 0 is y = c1 J1 (x) + c2 Y1 (x).
2. False; y = x is a solution that is analytic at x = 0.
3. x = −1 is the nearest singular point to the ordinary point x = 0. Theorem 5.1 guarantees the existence of two
∞
power series solutions y = n=1 cn xn of the differential equation that converge at least for −1 < x < 1. Since
− 12 ≤ x ≤ 12 is properly contained in −1 < x < 1, both power series must converge for all points contained in
− 12 ≤ x ≤ 12 .
304
CHAPTER 5 REVIEW EXERCISES
4. The easiest way to solve the system
2c2 + 2c1 + c0 = 0
6c3 + 4c2 + c1 = 0
1
12c4 + 6c3 − c1 + c2 = 0
3
2
20c5 + 8c4 − c2 + c3 = 0
3
is to choose, in turn, c0 = 0, c1 = 0 and c0 = 0, c1 = 0. Assuming that c0 = 0, c1 = 0, we have
1
c2 = − c0
2
2
1
c3 = − c2 = c0
3
3
1
1
1
c4 = − c3 − c2 = − c0
2
12
8
2
1
1
1
c5 = − c4 + c2 − c3 =
c0 ;
5
30
20
60
whereas the assumption that c0 = 0, c1 = 0 implies
c2 = −c1
2
1
1
c3 = − c2 − c1 = c1
3
6
2
1
1
1
5
c4 = − c3 + c1 − c2 = − c1
2
36
12
36
2
1
1
1
c5 = − c4 + c2 − c3 = −
c1 .
5
30
20
360
five terms of two power series solutions are then
1
1
1
1
y1 (x) = c0 1 − x2 + x3 − x4 + x5 + · · ·
2
3
8
60
and
1
5
1 5
y2 (x) = c1 x − x2 + x3 − x4 −
x + ··· .
2
36
360
5. The interval of convergence is centered at 4. Since the series converges at −2, it converges at least on the
interval [−2, 10). Since it diverges at 13, it converges at most on the interval [−5, 13). Thus, at −7 it does not
converge, at 0 and 7 it does converge, and at 10 and 11 it might converge.
6. We have
x5
x3
+
− ···
x−
sin x
2x5
x3
6
120
f (x) =
=
+
+ ··· .
=x+
2
4
cos x
3
15
x
x
1−
+
− ···
2
24
7. The differential equation (x3 − x2 )y + y + y = 0 has a regular singular point at x = 1 and an irregular singular
point at x = 0.
8. The differential equation (x − 1)(x + 3)y + y = 0 has regular singular points at x = 1 and x = −3.
∞
9. Substituting y = n=0 cn xn+r into the differential equation we obtain
2xy + y + y = 2r2 − r c0 xr−1 +
∞
[2(k + r)(k + r − 1)ck + (k + r)ck + ck−1 ]xk+r−1 = 0
k=1
305
CHAPTER 5 REVIEW EXERCISES
which implies
2r2 − r = r(2r − 1) = 0
and
(k + r)(2k + 2r − 1)ck + ck−1 = 0.
The indicial roots are r = 0 and r = 1/2. For r = 0 the recurrence relation is
ck−1
ck = −
, k = 1, 2, 3, . . . ,
k(2k − 1)
so
1
1
c1 = −c0 ,
c2 = c0 ,
c3 = − c0 .
6
90
For r = 1/2 the recurrence relation is
ck−1
ck = −
, k = 1, 2, 3, . . . ,
k(2k + 1)
so
1
1
1
c2 =
c3 = −
c1 = − c0 ,
c0 ,
c0 .
3
30
630
Two linearly independent solutions are
1
1
y1 = 1 − x + x2 − x3 + · · ·
6
90
and
1
1 3
1
y2 = x1/2 1 − x + x2 −
x + ··· .
3
30
630
∞
10. Substituting y = n=0 cn xn into the differential equation we have
y − xy − y =
∞
n(n − 1)cn xn−2 −
n=2
∞
n=1
k=n−2
=
∞
ncn xn −
∞
n=0
k=n
(k + 2)(k + 1)ck+2 xk −
k=0
∞
k=n
kck xk −
k=1
= 2c2 − c0 +
∞
2c2 − c0 = 0
(k + 2)(k + 1)ck+2 − (k + 1)ck = 0
and
∞
ck xk
k=0
[(k + 2)(k + 1)ck+2 − (k + 1)ck ]xk = 0.
k=1
Thus
cn xn
1
c0
2
1
ck+2 =
ck ,
k+2
c2 =
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
1
2
c3 = c5 = c7 = · · · = 0
1
c4 =
8
1
c6 =
48
c2 =
306
CHAPTER 5 REVIEW EXERCISES
and so on. For c0 = 0 and c1 = 1 we obtain
c2 = c4 = c6 = · · · = 0
1
c3 =
3
1
c5 =
15
1
c7 =
105
and so on. Thus, two solutions are
1
1
1
y1 = 1 + x2 + x4 + x6 + · · ·
2
8
48
and
11. Substituting y =
1
1
1 7
y2 = x + x3 + x5 +
x + ···.
3
15
105
∞
n
n=0 cn x into the differential equation we obtain
(x − 1)y + 3y = (−2c2 + 3c0 ) +
∞
[(k + 1)kck+1 − (k + 2)(k + 1)ck+2 + 3ck ]xk = 0
k=1
which implies c2 = 3c0 /2 and
ck+2 =
(k + 1)kck+1 + 3ck
,
(k + 2)(k + 1)
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
3
,
2
c3 =
1
,
2
c4 =
5
8
c2 = 0,
c3 =
1
,
2
c4 =
1
4
c2 =
and so on. For c0 = 0 and c1 = 1 we obtain
and so on. Thus, two solutions are
3
1
5
y1 = 1 + x2 + x3 + x4 + · · ·
2
2
8
and
12. Substituting y =
1
1
y2 = x + x3 + x4 + · · · .
2
4
∞
n
n=0 cn x into the differential equation we obtain
y − x2 y + xy = 2c2 + (6c3 + c0 )x +
∞
[(k + 3)(k + 2)ck+3 − (k − 1)ck ]xk+1 = 0
k=1
which implies c2 = 0, c3 = −c0 /6, and
ck+3 =
k−1
ck ,
(k + 3)(k + 2)
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
1
6
c4 = c7 = c10 = · · · = 0
c3 = −
c5 = c8 = c11 = · · · = 0
1
c6 = −
90
307
CHAPTER 5 REVIEW EXERCISES
and so on. For c0 = 0 and c1 = 1 we obtain
c3 = c6 = c9 = · · · = 0
c4 = c7 = c10 = · · · = 0
c5 = c8 = c11 = · · · = 0
and so on. Thus, two solutions are
1
1
y1 = 1 − x3 − x6 − · · ·
6
90
13. Substituting y =
and y2 = x.
∞
n+r
into the differential equation, we obtain
n=0 cn x
xy − (x + 2)y + 2y = (r2 − 3r)c0 xr−1 +
∞
[(k + r)(k + r − 3)ck − (k + r − 3)ck−1 ]xk+r−1 = 0,
k=1
which implies
r2 − 3r = r(r − 3) = 0
and
(k + r)(k + r − 3)ck − (k + r − 3)ck−1 = 0.
The indicial roots are r1 = 3 and r2 = 0. For r2 = 0 the recurrence relation is
k(k − 3)ck − (k − 3)ck−1 = 0,
Then
k = 1, 2, 3, . . . .
c1 − c0 = 0
2c2 − c1 = 0
0c3 − 0c2 = 0 =⇒ c3 is arbitrary
and
ck =
1
ck−1 ,
k
k = 4, 5, 6, . . . .
Taking c0 = 0 and c3 = 0 we obtain
c1 = c0
1
c2 = c0
2
c3 = c4 = c5 = · · · = 0.
Taking c0 = 0 and c3 = 0 we obtain
c0 = c1 = c2 = 0
1
6
c4 = c3 = c3
4
4!
1
6
c3 = c3
c5 =
5·4
5!
1
6
c6 =
c3 = c3 ,
6·5·4
6!
and so on. In this case we obtain the two solutions
and
1
y1 = 1 + x + x2
2
y2 = x3 +
6 4
6
6
1
x + x5 + x6 + · · · = 6ex − 6 1 + x + x2 .
4!
5!
6!
2
308
CHAPTER 5 REVIEW EXERCISES
14. Substituting y =
∞
n
n=0 cn x into the differential equation we have
∞
1 2
1 4
1 6
2
3
4
x + · · · (2c2 + 6c3 x + 12c4 x + 20c5 x + 30c6 x + · · ·) +
(cos x)y + y = 1 − x + x −
cn xn
2
24
720
n=0
1
= 2c2 + 6c3 x + (12c4 − c2 )x2 + (20c5 − 3c3 )x3 + 30c6 − 6c4 + c2 x4 + · · ·
12
+ [c0 + c1 x + c2 x2 + c3 x3 + c4 x4 + · · · ]
1
= (c0 + 2c2 ) + (c1 + 6c3 )x + 12c4 x2 + (20c5 − 2c3 )x3 + 30c6 − 5c4 + c2 x4 + · · ·
12
= 0.
Thus
c0 + 2c2 = 0
c1 + 6c3 = 0
12c4 = 0
20c5 − 2c3 = 0
1
30c6 − 5c4 + c2 = 0
12
and
1
c2 = − c0
2
1
c3 = − c1
6
c4 = 0
1
c5 =
c3
10
1
1
c6 = c4 −
c2 .
6
360
Choosing c0 = 1 and c1 = 0 we find
1
1
c2 = − , c3 = 0, c4 = 0, c5 = 0, c6 =
2
720
and so on. For c0 = 0 and c1 = 1 we find
1
1
c2 = 0, c3 = − , c4 = 0, c5 = − , c6 = 0
6
60
and so on. Thus, two solutions are
1
1 6
y1 = 1 − x2 +
x + ···
2
720
15. Substituting y =
1
1
y2 = x − x3 − x5 + · · · .
6
60
and
∞
n
n=0 cn x into the differential equation we have
y + xy + 2y =
∞
n=2
n(n − 1)cn xn−2 +
n=1
k=n−2
=
∞
∞
ncn xn + 2
= 2c2 + 2c0 +
k=n
k=0
∞
k=1
∞
cn xn
n=0
(k + 2)(k + 1)ck+2 xk +
∞
k=n
kck xk + 2
∞
ck xk
k=0
[(k + 2)(k + 1)ck+2 + (k + 2)ck ]xk = 0.
k=1
309
CHAPTER 5 REVIEW EXERCISES
Thus
2c2 + 2c0 = 0
(k + 2)(k + 1)ck+2 + (k + 2)ck = 0
and
c2 = −c0
ck+2 = −
1
ck ,
k+1
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
c2 = −1
c3 = c5 = c7 = · · · = 0
1
c4 =
3
1
c6 = −
15
and so on. For c0 = 0 and c1 = 1 we obtain
c2 = c4 = c6 = · · · = 0
1
c3 = −
2
1
c5 =
8
1
c7 = −
48
and so on. Thus, the general solution is
1
1
1
1
1
y = C0 1 − x2 + x4 − x6 + · · · + C1 x − x3 + x5 − x7 + · · ·
3
15
2
8
48
and
4
3
2
5
7
y = C0 −2x + x3 − x5 + · · · + C1 1 − x2 + x4 − x6 + · · · .
3
5
2
8
48
Setting y(0) = 3 and y (0) = −2 we find c0 = 3 and c1 = −2. Therefore, the solution of the initial-value problem
is
16. Substituting y =
∞
1
1
1
y = 3 − 2x − 3x2 + x3 + x4 − x5 − x6 + x7 + · · · .
4
5
24
n
n=0 cn x into the differential equation we have
(x + 2)y + 3y =
∞
n=2
n(n − 1)cn xn−1 + 2
∞
n(n − 1)cn xn−2 + 3
n=2
k=n−1
=
∞
k=n−2
(k + 1)kck+1 xk + 2
k=1
= 4c2 + 3c0 +
∞
cn xn
n=0
k=n
(k + 2)(k + 1)ck+2 xk + 3
k=0
∞
∞
∞
ck xk
k=0
[(k + 1)kck+1 + 2(k + 2)(k + 1)ck+2 + 3ck ]xk = 0.
k=1
Thus
4c2 + 3c0 = 0
(k + 1)kck+1 + 2(k + 2)(k + 1)ck+2 + 3ck = 0
310
CHAPTER 5 REVIEW EXERCISES
and
3
c2 = − c0
4
ck+2 = −
k
3
ck+1 −
ck ,
2(k + 2)
2(k + 2)(k + 1)
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
c2 = −
3
4
1
8
1
c4 =
16
c3 =
c5 = −
9
320
and so on. For c0 = 0 and c1 = 1 we obtain
c2 = 0
c3 = −
1
4
1
16
c5 = 0
c4 =
and so on. Thus, the general solution is
3
1
1
1
9 5
1
y = C0 1 − x2 + x3 + x4 −
x + · · · + C1 x − x3 + x4 + · · ·
4
8
16
320
4
16
and
3
3 2 1 3
3 2 1 3
9 4
y = C0 − x + x + x − x + · · · + C1 1 − x + x + · · · .
2
8
4
64
4
4
Setting y(0) = 0 and y (0) = 1 we find c0 = 0 and c1 = 1. Therefore, the solution of the initial-value problem is
1
1
y = x − x3 + x4 + · · · .
4
16
17. The singular point of (1 − 2 sin x)y + xy = 0 closest to x = 0 is π/6. Hence a lower bound is π/6.
18. While we can find two solutions of the form
y1 = c0 [1 + · · · ] and y2 = c1 [x + · · · ],
the initial conditions at x = 1 give solutions for c0 and c1 in terms of infinite series. Letting t = x − 1 the
initial-value problem becomes
d2 y
dy
+ (t + 1)
+ y = 0,
y(0) = −6, y (0) = 3.
dt2
dt
∞
Substituting y = n=0 cn tn into the differential equation, we have
∞
∞
∞
∞
d2 y
dy
n−2
n
n−1
+
y
=
+
(t
+
1)
n(n
−
1)c
t
+
nc
t
+
nc
t
+
cn tn
n
n
n
dt2
dt
n=2
n=1
n=1
n=0
k=n−2
=
∞
k=n
(k + 2)(k + 1)ck+2 tk +
k=0
= 2c2 + c1 + c0 +
∞
k=1
∞
k=n−1
kck tk +
∞
k=0
k=n
(k + 1)ck+1 tk +
∞
ck tk
k=0
[(k + 2)(k + 1)ck+2 + (k + 1)ck+1 + (k + 1)ck ]tk = 0.
k=1
311
CHAPTER 5 REVIEW EXERCISES
2c2 + c1 + c0 = 0
Thus
(k + 2)(k + 1)ck+2 + (k + 1)ck+1 + (k + 1)ck = 0
and
c1 + c0
2
ck+1 + ck
ck+2 = −
,
k+2
c2 = −
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
1
c2 = − ,
2
1
,
6
c4 =
1
c3 = − ,
6
c4 =
c3 =
1
,
12
and so on. For c0 = 0 and c1 = 1 we find
1
c2 = − ,
2
1
,
6
and so on. Thus, the general solution is
1 2 1 3
1 2 1 3 1 4
1 4
y = c0 1 − t + t + t + · · · + c1 t − t − t + t + · · · .
2
6
12
2
6
6
The initial conditions then imply c0 = −6 and c1 = 3. Thus the solution of the initial-value problem is
1
1
1
2
3
4
y = −6 1 − (x − 1) + (x − 1) + (x − 1) + · · ·
2
6
12
1
1
1
2
3
4
+ 3 (x − 1) − (x − 1) − (x − 1) + (x − 1) + · · · .
2
6
6
19. Writing the differential equation in the form
y +
1 − cos x y + xy = 0,
x
and noting that
1 − cos x
x x3
x5
= −
+
− ···
x
2
24 720
is analytic at x = 0, we conclude that x = 0 is an ordinary point of the differential equation.
20. Writing the differential equation in the form
y +
x
y=0
ex − 1 − x
and noting that
x
2 2
x
x2
=
−
+
+
− ···
ex − 1 − x
x 3 18 270
we see that x = 0 is a singular point of the differential equation. Since
2x2
x3
x4
x
2
x
=
2x
−
+
+
− ··· ,
ex − 1 − x
3
18 270
we conclude that x = 0 is a regular singular point.
312
CHAPTER 5 REVIEW EXERCISES
21. Substituting y =
∞
n
n=0 cn x into the differential equation we have
y + x2 y + 2xy =
∞
n=2
n(n − 1)cn xn−2 +
∞
n=1
k=n−2
=
∞
ncn xn+1 + 2
∞
n=0
k=n+1
(k + 2)(k + 1)ck+2 xk +
k=0
∞
cn xn+1
k=n+1
(k − 1)ck−1 xk + 2
k=2
= 2c2 + (6c3 + 2c0 )x +
∞
∞
ck−1 xk
k=1
[(k + 2)(k + 1)ck+2 + (k + 1)ck−1 ]xk = 5 − 2x + 10x3 .
k=2
Thus, equating coefficients of like powers of x gives
2c2 = 5
6c3 + 2c0 = −2
12c4 + 3c1 = 0
20c5 + 4c2 = 10
(k + 2)(k + 1)ck+2 + (k + 1)ck−1 = 0,
k = 4, 5, 6, . . . ,
and
c2 =
5
2
1
1
c3 = − c0 −
3
3
1
c4 = − c1
4
1 1
1 1 5
c5 = − c2 = −
=0
2 5
2 5 2
1
ck+2 = −
ck−1 .
k+2
Using the recurrence relation, we find
1
1
1
1
c6 = − c3 =
(c0 + 1) = 2
c0 + 2
6
3·6
3 · 2!
3 · 2!
1
1
c7 = − c4 =
c1
7
4·7
c8 = c11 = c14 = · · · = 0
1
1
1
c9 = − c6 = − 3
c0 − 3
9
3 · 3!
3 · 3!
1
1
c10 = − c7 = −
c1
10
4 · 7 · 10
1
1
1
c0 + 4
c12 = − c9 = 4
12
3 · 4!
3 · 4!
1
1
c13 = − c0 =
c1
13
4 · 7 · 10 · 13
313
CHAPTER 5 REVIEW EXERCISES
and so on. Thus
22. (a) From y = −
1 3
1
1
1
6
9
12
y = c0 1 − x + 2
x − 3
x + 4
x − ···
3
3 · 2!
3 · 3!
3 · 4!
1
1 7
1
1
+ c1 x − x4 +
x −
x10 +
x13 − · · ·
4
4·7
4 · 7 · 10
4 · 7 · 10 · 13
5
1
1
1
1
+ x2 − x3 + 2
x6 − 3
x9 + 4
x12 − · · · .
2
3
3 · 2!
3 · 3!
3 · 4!
1 du
we obtain
u dx
dy
1
1 d2 u
+ 2
=−
dx
u dx2
u
du
dx
2
.
Then dy/dx = x2 + y 2 becomes
−
1 d2 u
1
+ 2
2
u dx
u
du
dx
2
= x2 +
1
u2
du
dx
2
,
d2 u
+ x2 u = 0.
dx2
so
(b) The differential equation u + x2 u = 0 has the form (18) in the text with
1
2
2c − 2 = 2 =⇒ c = 2
1
b2 c2 = 1 =⇒ b =
2
1
a2 − p2 c2 = 0 =⇒ p = .
4
1 − 2a = 0 =⇒ a =
Then, by (19) in the text,
1/2
u=x
(c) We have
1 2
1 2
c1 J1/4
.
x + c2 J−1/4
x
2
2
1 du
dw dt
1
d 1/2
1
1
x1/2
= − 1/2
x w(t) = − 1/2
+ x−1/2 w
u dx
dt dx 2
x w(t) dx
x w
1
dw
dw
1
1
1
dw
= − 1/2
x3/2
+ 1/2 w = −
2x2
+w =−
4t
+w .
dt
2xw
dt
2xw
dt
x w
2x
y=−
Now
4t
dw
d
+ w = 4t [c1 J1/4 (t) + c2 J−1/4 (t)] + c1 J1/4 (t) + c2 J−1/4 (t)
dt
dt
1
1
= 4t c1 J−3/4 (t) − J1/4 (t) + c2 − J−1/4 (t) − J3/4 (t) + c1 J1/4 (t) + c2 J−1/4 (t)
4t
4t
1 2
1 2
2
2
= 4c1 tJ−3/4 (t) − 4c2 tJ3/4 (t) = 2c1 x J−3/4
x − 2c2 x J3/4
x ,
2
2
so
y=−
2c1 x2 J−3/4 ( 12 x2 ) − 2c2 x2 J3/4 ( 12 x2 )
−c1 J−3/4 ( 12 x2 ) + c2 J3/4 ( 12 x2 )
=
x
.
2x[c1 J1/4 ( 12 x2 ) + c2 J−1/4 ( 12 x2 )]
c1 J1/4 ( 12 x2 ) + c2 J−1/4 ( 12 x2 )
314
CHAPTER 5 REVIEW EXERCISES
Letting c = c1 /c2 we have
y=x
J3/4 ( 12 x2 ) − cJ−3/4 ( 12 x2 )
.
cJ1/4 ( 12 x2 ) + J−1/4 ( 12 x2 )
23. (a) Equations (10) and (24) of Section 5.3 in the text imply
cos π2 J1/2 (x) − J−1/2 (x)
2
Y1/2 (x) =
cos x.
= −J−1/2 (x) = −
π
sin 2
πx
(b) From (15) of Section 5.3 in the text
I1/2 (x) = i−1/2 J1/2 (ix)
and
I−1/2 (x) = i1/2 J−1/2 (ix)
so
I1/2 (x) =
and
I−1/2 (x) =
∞
2 1
x2n+1 =
πx n=0 (2n + 1)!
∞
2 1
x2n =
πx n=0 (2n)!
2
sinh x
πx
2
cosh x.
πx
(c) Equation (16) of Section 5.3 in the text and part (b) imply
π I−1/2 (x) − I1/2 (x)
π
K1/2 (x) =
=
2
sin π2
2
=
2
cosh x −
πx
2
sinh x
πx
π ex + e−x
π −x
ex − e−x
−
=
e .
2x
2
2
2x
24. (a) Using formula (5) of Section 3.2 in the text, we find that a second solution of (1 − x2 )y − 2xy = 0 is
2
2
e 2x dx/(1−x )
y2 (x) = 1 ·
dx = e− ln(1−x ) dx
12
dx
1+x
1
=
ln
,
=
1 − x2
2
1−x
where partial fractions was used to obtain the last integral.
(b) Using formula (5) of Section 3.2 in the text, we find that a second solution of
(1 − x2 )y − 2xy + 2y = 0 is
2
− ln(1−x2 )
e 2x dx/(1−x )
e
dx = x
dx
y2 (x) = x ·
2
x
x2
dx
1
1+x
1
x
1+x
=x
dx
=
x
ln
−
=
ln
− 1,
x2 (1 − x2 )
2
1−x
x
2
1−x
where partial fractions was used to obtain the last integral.
315
CHAPTER 5 REVIEW EXERCISES
(c)
y2
2
y2
2
1
1
1x
-1
y2 (x) =
1x
-1
-1
-1
-2
-2
1
ln
2
1+x
1−x
y2 =
x
ln
2
1+x
1−x
−1
25. (a) By the binomial theorem we have
−1/2
1 + t2 − 2xt
1 2
(−1/2)(−3/2) 2
(−1/2)(−3/2)(−5/2) 2
t − 2xt +
(t − 2xt)2 +
(t − 2xt)3 + · · ·
2
2!
3!
1
3
5
= 1 − (t2 − 2xt) + (t2 − 2xt)2 − (t2 − 2xt)3 + · · ·
2
8
16
∞
1
1
= 1 + xt + (3x2 − 1)t2 + (5x3 − 3x)t3 + · · · =
Pn (x)tn .
2
2
n=0
=1−
(b) Letting x = 1 in (1 − 2xt + t2 )−1/2 , we have
(1 − 2t + t2 )−1/2 = (1 − t)−1 =
From part (a) we have
∞
1
= 1 + t + t2 + t 3 + . . .
1−t
Pn (1)tn = (1 − 2t + t2 )−1/2 =
n=0
∞
(|t| < 1) =
∞
tn .
n=0
tn .
n=0
Equating the coefficients of corresponding terms in the two series, we see that Pn (1) = 1. Similarly, letting
x = −1 we have
1
= 1 − t + t2 − 3t3 + . . .
1+t
∞
∞
=
(−1)n tn =
Pn (−1)tn ,
(1 + 2t + t2 )−1/2 = (1 + t)−1 =
n=0
n=0
n
so that Pn (−1) = (−1) .
316
(|t| < 1)