CHAPTER 5
Series Solutions of Linear
Differential Equations
Chapter Contents
5.1 Solutions about Ordinary Points
5.2 Solution about Singular Points
5.3 Special Functions
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Ch5_2
5.1 Solutions about Ordinary Point
Review of Power Series
Recall from that a power series in x – a has the form

n
2
c
(
x

a
)

c

c
(
x

a
)

c
(
x

a
)

n
0
1
2
n 0
Such a series is said to be a power series centered at
a.
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Ch5_3
Convergence
lim N  S N ( x)  lim N   n0 cn ( x  a) n exists.
N
Interval of Convergence
The set of all real numbers for which the series
converges.
Radius of Convergence
If R is the radius of convergence, the power series
converges for |x – a| < R and
diverges for |x – a| > R.
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Ch5_4
Absolute Convergence
Within its interval of convergence, a power series
converges absolutely. That is, the following
converges.

n
n0 | cn ( x  a)
|
Ratio Test
Suppose cn  0 for all n, and
cn1 ( x  a ) n1
cn1
lim
 | x  a | lim
L
n
n
n c
cn ( x  a )
n
If L < 1, this series converges absolutely, if L > 1, this
series diverges, if L = 1, the test is inclusive.
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Ch5_5
A Power Defines a Function

Suppose y  n0 cn x n
then
y '   n0 cn nx

n1
and
y "   n0 cn n(n  1) x n2

(1)
Identity Property
If all cn = 0, then the series = 0.
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Ch5_6
Analytic at a Point
A function f is analytic at a point a, if it can be
represented by a power series in x – a with a positive
radius of convergence. For example:
x x2
x3 x5
e  1    , sin x  x    
1! 2!
3! 5!
x2 x4 x6
cos x  1     
2! 4! 6!
x
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(2)
Ch5_7
Arithmetic of Power Series
Power series can be combined through the operations
of addition, multiplication and division.
e x sin x



x 2 x3 x 4
x3 x5
x7
 1  x       x  

 
2 6 24
6 120 5040



1 1
 1 1
 1 1
 1
 (1) x  (1) x 2      x 3      x 4  
   x5  
 6 2
 6 6
 120 12 24 
x3 x5
 x  x   
3 30
2
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Ch5_8
Example 1 Adding Two Power Series
Write n2 n(n  1)cn x

n2
 n0 cn x n1 as one power series.

Solution:
Since

 n(n  1)cn x
n2
n2

  cn x
n 0
n 1

series starts with
x for n = 3
↓
 2．1c2 x   n(n  1)cn x
0
n 3
n2
series starts with
x for n = 0
↓

  cn x n1
n 0
we let k = n – 2 for the first series and k = n + 1 for the
second series,
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Ch5_9
Example 1 (2)
then we can get the right-hand side as
same


2c2   (k  2)(k  1)ck 2 x   ck 1 x k
k
k 1
k 1
(3)
same
We now obtain

 n(n  1)cn x
n2
n2

  cn x n1
n 0

 2c2   [(k  2)(k  1)ck 2  ck 1 ]x
k
(4)
k 1
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Ch5_10
A Solution
Suppose the linear DE
a2 ( x) y  a1 ( x) y  a0 ( x) y  0
is put into
y  P( x) y  Q( x) y  0
(5)
(6)
Definition 5.1.1 Ordinary and Singular Points
A point x0 is said to be an ordinary point of (5) if both
P(x) and Q(x) in (6) are analytic at x0. A point that is
not an ordinary point is said to be a singular point.
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Ch5_11
Polynomial Coefficients
Since P(x) and Q(x) in (6) is a rational function,
P(x) = a1(x)/a2(x), Q(x) = a0(x)/a2(x)
It follows that x = x0 is an ordinary point of (5) if
a2(x0)  0.
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Ch5_12
Theorem 5.1.2 Existence of Power Series Solutions
If x = x0 is an ordinary point of (5), we can always find
two linearly independent solutions in the form of power
series centered at x0, that is,
y  n0 cn ( x  x0 ) n

A series solution converges at least of some interval
defined by |x – x0| < R, where R is the distance from x0
to the closest singular point.
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Ch5_13
Example 2 Power Series Solutions
Solve y"xy  0
Solution:
We know there are no finite singular points.


n
n2
y

c
x
y
"

n
(
n

1
)
c
x
Now,
n0 n and
 n2
n
then the DE gives

y  xy   cn n(n  1) x
n2
n2

  cn n(n  1) x
n2
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
 x  cn x n
n 0
n2

  cn x
n 1
(7)
n 0
Ch5_14
Example 2 (2)
From the result given in (4),

y  xy  2c2   [(k  1)(k  2)ck 2  ck 1 ]x k  0
k 1
(8)
Since (8) is identically zero, it is necessary all the
coefficients are zero, 2c2 = 0, and
(k  1)(k  2)ck 2  ck 1  0 ,
k 1, 2 , 3 ,
(9)
Now (9) is a recurrence relation, since
(k + 1)(k + 2)  0, then from (9)
ck 2
ck 1

,
(k  1)(k  2)
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k 1, 2 , 3 ,
(10)
Ch5_15
Example 2 (3)
Thus we obtain
k  1,
c0
c3  
2．3
k  2,
k  4,
c1
3．4
c2
c5  
 0 ← c2 is zero
4．5
c3
1
c6  

c0
5．6 2．3．5．6
k  5,
c7  
k  3,
c4  
c4
1

c1
6．7 3．4．6．7
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Ch5_16
Example 2 (4)
k  6,
k  7,
c5
c8  
 0 ← c5 is zero
7．8
c6
1
c9  

c0
8．9
2．3．5．6．8．9
k  8,
c7
1
c10  

c1
9．10
3．4．6．7．9．10
k  9,
c8
c11  
0
10．11
← c8 is zero
and so on.
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Ch5_17
Example 2 (5)
Then the power series solutions are
y = c0y1 + c1y2
c0 3 c1 4
c0
y  c0  c1 x  0 
x 
x 0
x6
2． 3
3． 4
2． 3． 5． 6
c1

x 7  0  ....
3． 4． 6． 7
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Ch5_18
Example 2 (6)
1 3
1
1
6
y1 ( x)  1 
x 
x 
x9  
2．3
2．3．5．6
2．3．5．6．8．9
(1) k
 1 
x 3k
k 1 2．3 (3n  1)(3n)

1 4
1
1
7
y2 ( x)  1 
x 
x 
x10  
3．4
3．4．6．7
3．4．6．7．9．10
(1) k
 x
x 3k 1
k 1 3．4 (3n )(3n  1)

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Ch5_19
Example 3 Power Series Solution
Solve ( x 2  1) y" xy' y  0
Solution:
Since x2 + 1 = 0, then x = i, −i are singular points. A
power series solution centered at 0 will converge at least
for |x| < 1. Using the power series form of y, y’ and y”,
then



n2
n 1
n 0
( x 2  1) n(n  1)cn x n2  x  ncn x n1   cn x n


  n(n  1)cn x   n(n  1)cn x
n
n2
n2
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n2


  ncn x   cn x n
n
n 1
n 0
Ch5_20
Example 3 (2)

 2c2 x 0  c0 x 0  6c3 x  c1 x  c1 x   n(n  1)cn x n
n2



k n



  n(n  1)cn x
  ncn x   cn x n
n4
n2
n2




 

n2
n
k n2
k n
k n

 2c2  c0  6c3 x   [k (k  1)ck  (k  2)(k  1)ck 2  kck  ck ]x k
k 2

 2c2  c0  6c3 x   [(k  1)(k  1)ck  (k  2)(k  1)ck 2 ]x k  0
k 2
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Ch5_21
Example 3 (3)
From the above, we get 2c2－c0 = 0, 6c3 = 0 , and
(k  1)(k  1)ck  (k  2)(k  1)ck 2  0
Thus c2 = c0/2, c3 = 0, ck+2 = (1 – k)ck/(k + 2)
Then
1
1
1
c4   c2  
c0   2 c0
4
2．4
2 2!
2
c5   c3  0 ← c3 is zero
5
3
3
1．3
c6   c4  
c0  3 c0
6
2．4．6
2 3!
4
c7   c5  0 ← c5 is zero
7
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Ch5_22
Example 3 (4)
5
3．5
1．3．5
c8   c6  
c0   4 c0
8
2．4．6．8
2 4!
6
c9   c7  0
9
← c7 is zero
7
3．5．7
1．3．5．7
c10   c8 
c0 
c0
5
10
2．4．6．8．10
2 ．5!
and so on.
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Ch5_23
Example 3 (5)
Therefore,
y  c0  c1 x  c2 x 2  c3 x 3  c4 x 4  c5 x 5
 c6 x 6  c7 x 7  c8 x 8  c9 x 9  c10 x10  
1 4 1．3 6 1．3．5 8 1．3．5．7 10
 1 2

 c0 1  x  2 x  3 x  4 x 
x


 c1 x
5

2 2!
2 3!
2 4!
2 5!
 2

 c0 y1 ( x)  c1 y2 ( x)
1 2 
n 1 1．3．5 ( 2n  3) 2 n
y1 ( x)  1  x   (1)
x ,
n
2
2 n!
n2
| x | 1
y2 ( x )  x
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Ch5_24
Example 4 Three-Term Recurrence Relation
If we seek a power series solution y(x) for
y  (1  x) y  0
we obtain c2 = c0/2 and the recurrence relation is
ck 2
ck  ck 1

,
(k  1)(k  2)
k 1, 2 , 3 ,
Examination of the formula shows c3, c4, c5, … are
expresses in terms of both c1 and c2. However it is more
complicated. To simplify it, we can first choose c0  0,
c1 = 0.
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Ch5_25
Example 4 (2)
Then we have
1
c2  c0
2
c1  c0
c0
1
c3 

 c0
2．3 2．3 6
c0
c2  c1
1
c4 

 c0
3．4
2．3．4 24
c3  c2
c0  1 1  1
c5 

   c0

4．5
4．5  6 2  30
and so on. Next, we choose c0 = 0, c1  0, then
1
c2  c0  0
2
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Ch5_26
Example 4 (3)
c1  c0
c1
1
c3 

 c1
2．3 2．3 6
c4 
c2  c1
c
1
 1  c1
3．4 3．4 12
c3  c2
c1
1
c5 


c1
4．5
4．5．6 120
and so on. Thus we have y = c0y1 + c1y2, where
1 2 1 3 1 4 1 5
y1 ( x)  1  x  x  x  x  
2
6
24
30
1 3 1 4 1 5
y2 ( x )  x  x  x 
x 
6
12
120
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Ch5_27
Example 5 ODE with Nonpolynomial
Coefficients
Solve y"(cos x) y  0
Solution:
We see x = 0 is an ordinary point of the equation. Using

n
y

c
x
the Maclaurin series for cos x, and using
 n 0 n ,
we find
y  (cos x) y
2
4
6



x
x
x
  n(n  1)cn x n2  1       cn x n
2! 4! 6!
n2

 n 0
1  2 
1  3

 2c2  c0  (6c3  c1 ) x  12c4  c2  c0  x   20c5  c3  c1  x  
2 
2 


0

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Ch5_28
Example 5 (2)
It follows that
1
2c2  c0  0 , 6c3  c1  0 , 12c4  c2  c0  0 ,
2
1
20c5  c3  c1  0
2
and so on. This gives c2 = –1/2c0, c3 = –1/6c1, c4 =
1/12c0, c5 = 1/30c1,…. By grouping terms we get the
general solution y = c0y1 + c1y2, where the convergence
is |x| < , and
1 2 1 4
y1 ( x)  1  x  x  
2
12
1
1
y2 ( x )  1  x 3  x 5  
6
30
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Ch5_29
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Ch5_30
5.2 Solutions about Singular Points
A Definition
A singular point x0 of a linear DE
a2 ( x) y  a1 ( x) y  a0 ( x) y  0
(1)
is further classified as either regular or irregular. This
classification depends on
y  P( x) y  Q( x) y  0
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(2)
Ch5_31
Definition 5.2.1 Regular/Irregular Singular Points
A singular point x0 is said to be a regular singular
point of (1), if p(x) = (x – x0)P(x), q(x) = (x – x0)2Q(x)
are both analytic at x0. A singular point that is not
regular is said to be irregular singular point.
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Ch5_32
Polynomial Coefficients
If x – x0 appears at most to the first power in the
denominator of P(x) and at most to the second power
in the denominator of Q(x), then x – x0 is a regular
singular point.
If (2) is multiplied by (x – x0)2,
( x  x0 ) 2 y  ( x  x0 ) p( x) y  q( x) y  0
(3)
where p, q are analytic at x = x0
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Ch5_33
Example 1 Classification of Singular Points
It should be clear x = 2, x = – 2 are singular points of
(x2 – 4)2y” + 3(x – 2)y’ + 5y = 0
According to (2), we have
3
P( x) 
( x  2)( x  2) 2
5
Q( x) 
( x  2) 2 ( x  2) 2
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Ch5_34
Example 1 (2)
For x = 2, the power of (x – 2) in the denominator of P
is 1, and the power of (x – 2) in the denominator of Q is
2. Thus x = 2 is a regular singular point.
For x = −2, the power of (x + 2) in the denominator of P
and Q are both 2. Thus x = − 2 is a irregular singular
point.
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Ch5_35
Theorem 5.2.1 Frobenius’ Theorem
If x = x0 is a regular singular point of (1), then there
exists one solution of the form
y  ( x  x0 )

r

n r
c
(
x

x
)

c
(
x

x
)
n
n
0
0
n
n 0
n 0
(4)
where the number r is a constant to be determined.
The series will converge at least on some interval
0 < x – x0 < R.
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Ch5_36
Example 2 Two Series Solutions
Because x = 0 is a regular singular point of
(5)
3xy  y  y  0
we try to find a solution y  n0 cn x nr . Now,


y   (n  r )cn x nr 1
n 0

y   (n  r )(n  r  1)cn x nr 2
n 0
Copyright © Jones and Bartlett；滄海書局
Ch5_37
Example 2 (2)
3 xy  y  y

 3 (n  r )(n  r  1)cn x
n  r 1
n 0

  (n  r )(3n  3r  2)cn x
n 0

  ( n  r ) cn x
n 0
n  r 1
n  r 1

  cn x n  r
n 0

  cn x n  r
n 0




r
1
n 1
n
 x r (3r  2)c0 x   (n  r )(3n  3r  2)cn x   cn x 
n 1
n 1

 
 



k  n 1
k n


1
k
 x r (3r  2)c0 x   [(k  r  1)(3k  3r  1)ck 1  ck ]x   0


k 0
r
Copyright © Jones and Bartlett；滄海書局
Ch5_38
Example 2 (3)
which implies
r(3r – 2)c0 = 0
(k + r + 1)(3k + 3r + 1)ck+1 – ck = 0, k = 0, 1, 2, …
Since nothing is gained by taking c0 = 0, then
r(3r – 2) = 0
(6)
and
ck
ck 1 
,
(k  r  1)(3k  3r  1)
k  0, 1, 2, 
(7)
From (6), r = 0, 2/3, when substituted into (7),
Copyright © Jones and Bartlett；滄海書局
Ch5_39
Example 2 (4)
r1 = 2/3,
r2 = 0,
ck 1 
ck
,
(3k  5)(k  1)
ck
ck 1 
,
(k  1)(3k  1)
Copyright © Jones and Bartlett；滄海書局
k = 0, 1, 2, …
(8)
k = 0, 1, 2, …
(9)
Ch5_40
Example 2 (5)
From (8)
c0
5．1
c0
c1


8．2 2! 5．8
c0
c2


11．3 3! 5．8．11
c3
c0


14．4 4! 5．8．11．14

c0

n! 5．8．11  (3n  2)
From(9)
c0
1．1
c0
c1


2．4 2!1．4
c0
c2


3．7 3!1．4．7
c3
c0


4．10 4!1．4．7．10

(1) n c0

n!1．4．7  (3n  2)
c1 
c1 
c2
c2
c3
c4
cn
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c3
c4
cn
Ch5_41
Example 2 (6)
These two series both contain the same multiple c0.
Omitting this term, we have



1
y1 ( x)  x 2 / 3 1  
xn 
 n1 n! 5．8．11 (3n  2) 

1
0
n
y2 ( x)  x 1  
x 
 n1 n!1．4．7  (3n  2) 
(10)
(11)
By the ratio test, both (10) and (11) converges for all
finite value of x, that is, |x| < . Also, from the forms of
(10) and (11), they are linearly independent. Thus the
solution is
y(x) = C1y1(x) + C2y2(x), 0 < x < 
Copyright © Jones and Bartlett；滄海書局
Ch5_42
Indicial Equation
Equation (6) is called the indicial equation, where
the values of r are called the indicial roots, or
exponents.
If x = 0 is a regular singular point of (1), then p =
xP(x) and q = x2Q(x) are analytic at x = 0.
Copyright © Jones and Bartlett；滄海書局
Ch5_43
Thus the power series expansions
p(x) = xP(x) = a0 + a1x + a2x2 + …
q(x) = x2Q(x) = b0 + b1x + b2x2 + …
(12)
are valid on intervals that have a positive radius of
convergence.
By multiplying (2) by x2, we have
(13)
x 2 y  x[ xP( x)] y  [ x 2Q( x)] y  0
After some substitutions, we find the indicial equation,
r(r – 1) + a0r + b0 = 0
(14)
Copyright © Jones and Bartlett；滄海書局
Ch5_44
Example 3 Two Series Solutions
Solve 2 xy"(1  x) y' y  0
Solution:

Let y  n0 cn x nr , then
2 xy  (1  x) y  y

 2 (n  r )(n  r  1)cn x
n  r 1
n 0


n 0
n 0

  (n  r )cn x nr 1
n 0
  ( n  r ) cn x n  r   cn x n  r

  (n  r )(2n  2r  1)cn x
n 0
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n  r 1

  (n  r  1)cn x nr
n 0
Ch5_45
Example 3 (2)




r
1
n 1
n
 x r (2r  1)c0 x   (n  r )(2n  2r  1)cn x   (n  r  1)cn x 
n 1
n 0

 
 



k  n 1
k n


1
k
 x r (2r  1)c0 x   [(k  r  1)(2k  2r  1)ck 1  (k  r  1)ck ]x 


k 0
r
which implies r(2r – 1) = 0
(k  r  1)(2k  2r  1)ck 1  (k  r  1)ck  0,
Copyright © Jones and Bartlett；滄海書局
(15)
k  0, 1, 2,  (16)
Ch5_46
Example 3 (3)
From (15), we have r1 = ½ , r2 = 0.
Foe r1 = ½ , we divide by k + 3/2 in (16) to obtain
 ck
ck 1 
,
2(k  1)
k  0, 1, 2, 
(17)
Foe r2 = 0 , (16) becomes
 ck
ck 1 
,
2k  1
Copyright © Jones and Bartlett；滄海書局
k  0, 1, 2, 
(18)
Ch5_47
Example 3 (4)
From (17)
 c0
c1 
2． 1
c0
 c1
c2 
 2
2．2 2 ．2!
 c0
 c2
c3 
 3
2．3 2 ．3!
c
c
c4  3  4 0
2．4 2 ．4!

(1) n c0
cn  n
2 n!
Copyright © Jones and Bartlett；滄海書局
From (18)
 c0
c1 
1
 c1 c0
c2 

3 1．3
 c0
 c2
c3 

5
1．3．5
c
c0
c4  3 
7
1．3．5．7

(1) n c0
cn 
1．3．5．7  (2n  1)
Ch5_48
Example 3 (5)
Thus for r1 = ½
n
n




(

1
)
(

1
)
y1 ( x)  x1/ 2 1   n x n    n x n1/ 2
 n1 2 n!  n0 2 n!
for r2 = 0
(1) n
y2 ( x )  1  
xn , | x |  
n1 1．3．5．7  ( 2n  1)

and on (0, ), the solution is y(x) = C1y1(x) + C2y2(x).
Copyright © Jones and Bartlett；滄海書局
Ch5_49
Example 4 Only One Series Solutions
Solve xy" y  0
Solution:
From xP(x) = 0, x2Q(x) = x, and the fact 0 and x are
their own power series centered at 0, we conclude a0 =
0, b0 = 0. Then form (14) we have r(r – 1) = 0, r1 = 1, r2
= 0. In other words, there is only a single series solution
(1) n n1
x 2 x3 1 4
y1 ( x)  
x  x  
x  ...
2 12 144
n0 n!( n  1)!

Copyright © Jones and Bartlett；滄海書局
Ch5_50
Three Cases
(1) If r1, r2 are distinct and do not differ by an integer,
there exists two linearly independent solutions of the
form:

y1 ( x)   cn x
n r1
n 0
Copyright © Jones and Bartlett；滄海書局

and y2 ( x)   bn x nr2
n 0
Ch5_51
(2) If r1 – r2 = N, where N is a positive integer, there
exists two linearly independent solutions of the form:

y1 ( x)   cn x nr1 , c0  0
(19)
n 0

y2 ( x)  Cy1 ( x) ln x   bn x nr2 , b0  0
(20)
n 0
Copyright © Jones and Bartlett；滄海書局
Ch5_52
(3) If r1 = r2, there exists two linearly independent
solutions of the form:

y1 ( x)   cn x nr1 , c0  0
(21)
n 0

y2 ( x)  y1 ( x) ln x   bn x nr2
(22)
n 0
Copyright © Jones and Bartlett；滄海書局
Ch5_53
Finding a Second Solution
If we already have a known solution y1, then the
second solution can be obtained by
e
y2 ( x)  y1 ( x)  2 dx
y1 ( x)
 Pdx
Copyright © Jones and Bartlett；滄海書局
(23)
Ch5_54
Example 5 Example 4 Revised—Using a
CAS
Find the general solution of xy" y  0
Solution:
From the known solution in Example 4,
1 2 1 3 1 4
y1 ( x)  x  x  x 
x 
2
12
144
we can use (23) to find y2(x). Here please use a CAS
for the complicated operations.
Copyright © Jones and Bartlett；滄海書局
Ch5_55
Example 5 (2)
e
dx
y2 ( x)  y1 ( x) 
dx  y1 ( x) 
2
2
[ y1 ( x)]
1
1
1
 x  x 2  x3 

4
x


 2

12
144
 0 dx
 y1 ( x) 
dx
 x 2  x 3  5 x 4  7 x 5  


12
12
 1 1 7 19

 y1 ( x)   2    x   dx
 x x 12 72

7
19 2
 1

 y1 ( x)    ln x  x 
x  
12 144
 x

19 2
 1 7

y2 ( x)  y1 ( x) ln x  y1 ( x)   x 
x  
 x 12 144

Copyright © Jones and Bartlett；滄海書局
Ch5_56
5.3 Special Functions
Bessel’s Equation of order v
x 2 y  xy  ( x 2  v 2 ) y  0
(1)
where v  0, and x = 0 is a regular singular point of
(1). The solutions of (1) are called Bessel functions.
Lengender’s Equation of order n
(1  x 2 ) y  2 xy  n(n  1) y  0
(2)
where n is a nonnegative integer, and x = 0 is an
ordinary point of (2). The solutions of (2) are called
Legender functions.
Copyright © Jones and Bartlett；滄海書局
Ch5_57
The Solution of Bessel’s Equation
Because x = 0 is a regular singular point, we know
there exists at least one solution of the

n r
y

c
x
form
n0 n . Then from (1),
x 2 y  xy  ( x 2  v 2 ) y




n 0
n 0
n 0
n 0
  cn (n  r )(n  r  1) x nr   cn (n  r ) x nr   cn x nr 2  v 2  cn x nr
 c0 (r 2  r  r  v 2 ) x r
x


r
 cn [(n  r )(n  r  1)  (n  r )  v ]x  x
2
n
r
n 1

n 2
x
c
n
n 0

 c0 (r 2  v 2 ) x r  x r  cn [(n  r ) 2  v 2 ]x n  x r  cn x n2
n 1
Copyright © Jones and Bartlett；滄海書局
n 0
(3)
Ch5_58
From (3) we have the indicial equation r2 – v2 = 0, r1 =
v, r2 = −v. When r1 = v, we have
(1 + 2v)c1 = 0
(k + 2)(k + 2+ 2v)ck+2 + ck = 0
or
ck 2 
 ck
,
(k  2)(k  2  2v)
k  0, 1, 2, 
(4)
The choice of c1 = 0 implies c3 = c5 = c7 = … = 0,
so for k = 0, 2, 4, …, letting k + 2 = 2n, n = 1, 2, 3, …,
we have
c2 n2
c2 n   2
(5)
2 n( n  v )
Copyright © Jones and Bartlett；滄海書局
Ch5_59
Thus
c0
c2   2
2 ．1．(1  v)
c0
c2
c4   2
 4
2 ．2(2  v) 2 ．1．2(1  v)(2  v)
c6  
c0
c4


2 2．3(3  v)
26．1．2．3(1  v)(2  v)(3  v)

(1) n c0
c2 n  2 n
,
2 n!(1  v)(2  v)  (n  v)
Copyright © Jones and Bartlett；滄海書局
n  1, 2, 3, 
(6)
Ch5_60
We choose c0 to be a specific value
c0 
1
2v (1  v)
where (1 + v) is the gamma function. See Appendix II.
There is an important relation:
(1 + ) = ()
so we can reduce the denominator of (6):
(1  v  1)  (1  v)(1  v)
(1  v  2)  (2  v)(2  v)  (2  v)(1  v)(1  v)
Copyright © Jones and Bartlett；滄海書局
Ch5_61
Hence we can write (6) as
(1) n
c2 n  2 nv
, n  0, 1, 2, ...
2 n!(1  v  n)
Copyright © Jones and Bartlett；滄海書局
Ch5_62
Bessel’s Functions of the First Kind
We define Jv(x) by
(1) n
 x
J v ( x)  
 
n0 n! (1  v  n)  2 

2 nv
(7)
and
(1) n
 x
J v ( x )  
 
n 0 n! (1  v  n)  2 

2 n v
(8)
In other words, the general solution of (1) on (0, ) is
y = c1Jv(x) + c2J-v(x), v  integer
(9)
See Fig 5.3.1.
Copyright © Jones and Bartlett；滄海書局
Ch5_63
Copyright © Jones and Bartlett；滄海書局
Ch5_64
Example 1 General Solution: v Not an
Integer
Consider the DE
x 2 y" xy'( x 2  1/4) y  0
We find v = ½, and the general solution on (0, ) is
y  c1 J1/2 ( x)  c2 J 1/2 ( x)
Copyright © Jones and Bartlett；滄海書局
Ch5_65
Bessel’s Functions of the Second Kind
If v  integer, then
cos v J v ( x)  J v ( x)
Yv ( x) 
sin v
(10)
and the function Jv(x) are linearly independent.
Another solution of (1) is y = c1Jv(x) + c2Yv(x).
As v  m, m an integer, (10) has the form 0/0. From
L’Hopital’s rule, the function
Ym ( x)  limYv ( x)
vm
and Jv(x) are linearly independent solutions of
x 2 y" xy'( x 2  m2 ) y  0
Copyright © Jones and Bartlett；滄海書局
Ch5_66
 Hence for any value of v, the general solution of (1)
is
(11)
y  c1 J v ( x)  c2Yv ( x)
Yv(x) is called the Bessel function of the second
kind of order v. Fig 5.3.2 shows y0(x) and y1(x).
Copyright © Jones and Bartlett；滄海書局
Ch5_67
Copyright © Jones and Bartlett；滄海書局
Ch5_68
Example 2 General Solution: v an Integer
Consider the DE
x 2 y" xy'( x 2  9) y  0
We find v = 3, and from (11) the general solution on
(0, ) is
y  c1 J 3 ( x)  c2Y3 ( x)
Copyright © Jones and Bartlett；滄海書局
Ch5_69
DEs Solvable in Terms of Bessel Functions
Let t = x,  > 0, in
x 2 y  xy  ( 2 x 2  v 2 ) y  0
(12)
then by the Chain Rule,
dy dy dt
dy


dx dt dx
dt
2
d 2 y d  dy  dt 
2 d y
     
2
dx
dt  dx  dx 
dt 2
Copyright © Jones and Bartlett；滄海書局
Ch5_70
Thus, (12) becomes
2
2
 t  2 d y  t  dy  2 2 
    t  v y  0
 
2
dt
 
   dt
2
d
y dy
2
2
2

y  0
t

t

t

v
2
dt
dt
The solution of the above DE is y = c1Jv(t) + c2Yv(t)
Let t = x, we have
y = c1Jv(x) + c2Yv(x)
(13)
Copyright © Jones and Bartlett；滄海書局
Ch5_71
Another equation is called the modified Bessel
equation order v,
x 2 y  xy  ( x 2  v 2 ) y  0
(14)
This time we let t = ix, then (14) becomes
2
d
y dy
t 2 2  t  (t 2  2 ) y  0
dt
dt
The solution will be Jv(ix) and Yv(ix). A real-valued
solution, called the modified Bessel function of the
first kind of order v is defined by
(15)
I ( x)  i  J (ix)
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Ch5_72
Analogous to (10), the modified Bessel function of
the second kind of order v  integer is defined by
 I  ( x)  I ( x)
(16)
K ( x) 

2
sin 
and for any integer v = n,
K n ( x)  lim K ( x)
 n
Because Iv and Kv are linearly independent on (0, ),
the general solution of (14) is
(17)
y  c1 I ( x)  c2 K ( x)
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Ch5_73
We consider another important DE:
 2 2 2 c 2 a 2  p 2 c 2 
1  2a
y 
y   b c x

 y  0, p  0
2
x
x


(18)
The general solution of (18) is
y  x a [c1 J p (bx c )  c2Yp (bx c )]
(19)
We shall not supply the details here.
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Ch5_74
Example 3 Using (18)
Find the general solution of xy  3 y  9 y  0 on (0, ).
Solution:
Writing the DE as
3
9
y  y  y  0
x
x
according to (18)
1 – 2a = 3, b2c2 = 9, 2c – 2 = −1, a2 – p2c2 = 0
then a = −1, c = ½ . In addition we take b= 6, p = 2.
From (19) the solution is
y  x 1[c1 J 2 (6 x1/ 2 )  c2Y2 (6 x1/ 2 )]
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Ch5_75
Example 4 The Aging Spring Revised
Recall the model in Sec. 3.8
mx  ke t x  0,   0
You should verify that by letting
2 k t / 2
s
e
 m
we have
2
d
x
dx
s2 2  s  s2 x  0
ds
ds
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Ch5_76
Example 4 (2)
The solution of the new equation is
x = c1J0(s) + c2Y0(s),
If we resubstitute
s
2
k t / 2
e
 m
we get the solution.
 2 k t / 2 
 2 k t / 2 
x(t )  c1 J 0 
e
e
  c2Y0 

 m

 m

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Ch5_77
Properties
(i)
J m ( x)  (1) m J m ( x)
(ii)
J m ( x)  (1) m J m ( x)
(iii)
0 , m  0
J m (0)  
1 , m  0
(iv)
lim x0 Ym ( x)  
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Ch5_78
Example 5 Derivation Using Series
Definition
Derive the formula xJ v ( x)  vJ v ( x)  xJ v1 ( x)
Solution:
It follows from (7)
(1) n (2n  v)  x 
xJ v ( x)  
 
n
!

(
1

v

n
)
2
n 0

2 nv
(1)
 x
 
n 0 n! (1  v  n )  2 

n
 v
2 nv
(1) n  x 
 
n 0 n! (1  v  n)  2 

n
 2
2 nv
2 n  v 1
(1) n
 x
 vJ v ( x)  x 
 
1 ( n  1)! (1  v  n)  2 
n


k  n 1
(1)
 x
 
k 0 k! ( 2  v  k )  2 

 vJ v ( x)  x 
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k
2 k  v 1
 vJ v ( x)  xJ v1 ( x)
Ch5_79
The result in example 5 can be written as
v
J v ( x)  J v ( x)   J v1 ( x)
x
which is a linear DE in Jv(x). Multiplying both sides
the integrating factor x-v, then
d v
(20)
[ x J ( x)]   x v J ( x)
dx
v
v 1
It can be shown
d v
[ x J v ( x)]  x v J v1 ( x)
dx
(21)
When y = 0, it follows from (14) that
J 0 ( x)   J1 ( x),
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Y0( x)  Y1 ( x)
(22)
Ch5_80
Spherical Bessel Functions
When the order v is half an odd number, that is,
1/2, 3/2, 5/2, …..
The Bessel function of the first kind Jv(x) can be
expressed as spherical Bessel function:
(1)
 x
 
n 0 n! (1  1 / 2  n)  2 

J1/ 2 ( x)  
n
2 n 1 / 2
Since (1 + ) = () and (1/2) = ½, then
 1
 (2n  1)!
1   n   2 n1

 2
 2 n!
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Ch5_81
Hence
(1)
J1/ 2 ( x)  
(2n  1)! 
n 0
n!
2 2 n1 n!

and
n
 x
 
 2
2 n 1 / 2
2  (1) n 2 n1

x

x n0 (2n  1)!
2
J1/ 2 ( x) 
sin x
x
2
J 1/ 2 ( x) 
cos x
x
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(23)
(24)
Ch5_82
The Solution of Legendre Equation
Since x = 0 is an ordinary point of (2), we use
y  n0 cn x n

After substitutions and simplifications, we obtain
n(n  1)c0  2c2  0
(n  1)(n  2)c1  6c3  0
( j  2)( j  1)c j 2  (n  j )(n  j  1)c j  0
or in the following forms:
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Ch5_83
n(n  1)
c0
2!
(n  1)(n  2)
c3  
c1
3!
(n  j )(n  j  1)
c j 2  
c j , j  2, 3, 4, 
( j  2)( j  1)
c2  
(25)
Using (25), at least |x| < 1, we obtain
 n(n  1) 2 (n  2)n(n  1)(n  3) 4
y1 ( x)  c0 1 
x 
x
2!
4!

(n  4)(n  2)n(n  1)(n  3)(n  5) 6


x  
6!

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Ch5_84
 (n  1)(n  2) 3 (n  3)(n  1)(n  2)(n  4) 5
y2 ( x)  c1  x 
x 
x
3!
5!

(n  5)(n  3)(n  1)(n  2)(n  4)(n  6) 7


x   (26)
7!

Notices: If n is an even integer, the first series
terminates, whereas y2 is an infinite series.
If n is an odd integer, the series y2 terminates with xn.
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Ch5_85
Legendre Polynomials
The following are nth order Legendre polynomials:
P0 ( x)  1,
P1 ( x)  x
1
1
P2 ( x)  (3 x 2  1),
P3 ( x)  (5 x 3  3 x)
(27)
2
2
1
1

2
P4 ( x)  (35 x  30 x  3), P5 ( x)  (63x 5  70 x 3  15 x)
8
8
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Ch5_86
They are in turn the solutions of the DEs. See Fig
5.3.5
n  0 : (1  x 2 ) y  2 xy  0
n  1 : (1  x 2 ) y  2 xy  2 y  0
n  2 : (1  x ) y  2 xy  6 y  0
n  3 : (1  x 2 ) y  2 xy  12 y  0
2

(28)

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Ch5_87
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Ch5_88
Properties
(i)
Pn ( x)  (1) n Pn ( x)
(ii)
Pn (1)  1
n
P
(

1
)

(

1
)
(iii) n
(iv) Pn (0)  0, n odd
(v)
P'n (0)  0, n even
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Ch5_89
Recurrence Relation
Without proof, we have
(k  1) Pk 1 ( x)  (2k  1) xPk ( x)  kPk 1 ( x)  0
(29)
which is valid for k = 1, 2, 3, …
Another formula by differentiation to generate
Legendre polynomials is called the Rodrigues’
formula:
1 dn 2
n
Pn ( x)  n
(
x

1
)
, n  0, 1, 2, ...
n
2 n! dx
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(30)
Ch5_90