Lecture 21
Series Solution about Singular Points
Cases of Indicial Roots
When using the method of Frobenius, we usually distinguish three cases corresponding to
the nature of the indicial roots. For the sake of discussion let us suppose that r1 and r2
are the real solutions of the indicial equation and that, when appropriate, r1 denotes the
largest root.
Case I: Roots not Differing by an Integer
If r1 and r2 are distinct and do not differ by an integer, then their exist two linearly
independent solutions of the differential equation of the form


n 0
n 0
y1   c n x n  r1 . . c0  0 , and y 2   bn x n  r2 ,
b0  0.
Example 6
2 xy   (1  x) y   y  0.
Solve
Solution

If
y   cn x n r , then
n 0

2 xy   (1  x) y   y  2 (n  r )( n  r  1)c n x n  r 1 
n 0
n 0
n 0
n
x n  r 1 


 ( n  r )c

 ( n  r )c
n
x nr   cn x nr
n 0


  (n  r )( 2n  2r  1)c n x n  r 1   (n  r  1)c n x n  r
n 0
n 0



n 1
n 0
 x r r (2r  1)c0 x 1   (n  r )( 2n  2r  1)c n x n 1   (n  r  1)c n x n
n  k 1

k n



 x r  r (2r  1)c0 x 1   [(k  r  1)(2k  2r  1)ck 1  (k  r  1)ck ]x k   0
k 0


which implies
r ( 2r  1) =0
(k  r  1)(2k  2r  1)ck 1  (k  r  1)ck  0 , k  0,1, 2,...
(1)
1
Lecture 21
For r1 
Series Solution about Singular Points
1
3
, we can divide by k  in the above equation to obtain
2
2
 ck
,
c k 1 
2(k  1)
 c0
c1 
2.1
c
 c1
c2 
 20
2.2 2 .2!
c2 c0
c3 

2.3 23.3!
cn 
In general
(1) n c0
, n  1, 2,3,...
2 n n!
Thus we have

1
2
y1  c0 x 1 
n 1
(1) n n
x , which converges for x  0 .
2 n n!
1
2
As given, the series is not meaningful for x  0 because of the presence of x .
Now for r2  0 , (1) becomes
 ck
2k  1
 c0
c1 
1
 c1 c0
c2 

3
1.3
 c 2  c0
c3 

5
1.3.5
 c3
c0
c4 

7
1.3.5.7
ck 1 
.
In general
cn 
(1) n c0
, n  1, 2, 3,...
1.3.5.7...(2n  1)
Thus second solution is



(1)n
y2  c0 1  
xn  .
 n1 1.3.5.7...(2n  1) 
On the interval ( 0,  ), the general solution is
y  C1 y1 ( x)  C2 y2 ( x).
x  .
2
Lecture 21
Series Solution about Singular Points
Method of Frobenius Cases (II and III)
When the roots of the indicial equation differ by a positive integer, we may or may not be
able to find two solutions of
a2 ( x) y   a1 ( x) y   a0 ( x) y  0
having form
(1)

y   c n ( x  x0 ) n  r
(2)
n 0
If not, then one solution corresponding to the smaller root contains a logarithmic term.
When the exponents are equal, a second solution always contains a logarithm. This latter
situation is similar to the solution of the Cauchy-Euler differencial equation when the
roots of the auxiliary equation are equal. We have the next two cases.
Case II: Roots Differing by a Positive Integer
If r1  r2  N , where N is a positive integer, then there exist two linearly independent
solutions of the form

n  r1
(3a )
y1 
cn x
, c0  0
n 0

n  r2
(3b )
y2  Cy1( x) ln x 
bn x
, b0  0
n 0
where C is a constant that could be zero.


Case III: Equal Indicial Roots:
If r1  r2 , there always exist two linearly independent solutions of (1) of the form

n  r1
(4a)
y1 
cn x
, c0  0
n 0

n r
(4b)
y2  y1( x)ln x  bn x 1
r1  r 2
n 1


3
Lecture 21
Example 7:
Series Solution about Singular Points
Solve xy   ( x  6) y   3 y  0
(1)

Solution: The assumption y   c n x n  r leads to
n 0
xy   ( x  6) y   3 y




n 0
n 0
  (n  r )( n  r  1)c n x n  r 1  6 (n  r )c n x n  r 1   (n  r )c n x n  r  3 c n x n  r
n 0
n 0


 x r  r (r  7)c0 x 1   (n  r )(n  r  7)cn x n 1   (n  r  3)cn x n 
n 1
n 0





x r  r (r  7)c0 x 1   (k  r  1)(k  r  6)ck 1  (k  r  3)ck x k   0
k 0


Thus r(r - 7)  0 so that r1  7,r 2  0, r1  r2  7, and


(k  r  1)( k  r  6) ck 1  (k  r  3)ck  0,
k  0,1, 2,3,...
For smaller root r2  0, (2)becomes
(k  1)(k  6)ck 1  (k  3)ck  0
(2)
(3)
recurrence relationbecomes
ck 1  
(k  3)
ck
(k  1)(k  6)
Since k-6=0, when, k=6, we do not divide by this term until k>6.we find
1.( 6)c1  (3)c0  0
2.( 5)c2  (2)c1  0
3.( 4)c3  (1).c2
0
4.(-3)c4  0.c3
0
5. (-2)c5  1.c4
0
6. (-1)c6  2.c5  0
7.0.c7  3.c6
0
This implies that
c4  c5  c6  0, But c0 and c7 can be chosen arbitrarily.
1
c1   c0
Hence
2
4
Lecture 21
Series Solution about Singular Points
1
1
c1 = c0
5
10
1
1
c3 = 
c0
c2  
120
12
c2 = 
(4)
and for k  7
(k  3)
ck 1 
ck
(k  1)(k  6)
4
c8 
c7
8.1
5
4.5
c9  
c8 
c7
9.2
2!8.9
6
4.5.6
c10 
c9 
c7
10.3
3!8.9.10



(1) n 1 4  5  6  (n  4)
c7 ,
n  8,9,10, 
(5)
(n  7)!8  9 10  ( n)
If we choose c7 = 0andc0  0 It follows that we obtain the polynomial solution
cn 
c0 [1 
y1 
1
1
1 3
x  x2 
x ],
2
10
120
c7  = 0andc0  0 , It follows that a second, though infinite series solution
But when
is
(1) n 1 4  5  6  (n  4) n
x ]
n 8 ( n  7)! 8  9 10  n

y2  c7 [ x 7  
(1)k 4  5  6  (k  3) k 3
(6)
x ],
x 
k 1 k !8  9 10  ( k  7)
Finally the general solution of equation (1) on the interval (0,  ) is
Y = c1 y1 ( x)  c2 y2 ( x)

1
1 2
1 3
(1)k 4  5  6  (k  3) k 7
7
[
1

x

x

x
]
= c1
+ c2 [ x  
x ]
2
10
120
n 1 k ! 8  9 10  ( k  7)
It is interesting to observe that in example 9 the larger root r 1 =7 were not used. Had we
done so, we would have obtained a series solution of the form*

=
c7 [ x 7  

y   cn x n7
(7)
n 0
Where cn are given by equation (2) with r1 =7
5
Lecture 21
Series Solution about Singular Points
(k  4)
ck ,
k  0,1, 2...
(k  8)(k  1)
Iteration of this latter recurrence relation then would yield only one solution, namely the
solution given by (6) with c0 playing the role of c7 )
When the roots of indicial equation differ by a positive integer, the second solution may
contain a logarithm.
On the other hand if we fail to find second series type solution, we can always use the
fact that
 p ( x ) dx
e 
(8)
y 2  y1 ( x)  2
dx
y1 ( x)
is a solution of the equation y   P( x) y   Q( x) y  0 ,whenever y1 is a known solution.
Note: In case 2 it is always a good idea to work with smaller roots first.
ck 1 
Example8:
Find the general solution of xy   3 y   y  0 .
Solution
The method of frobenius provide only one solution to this equation, namely,

1
1 2
1 3
2
x 
x  
y1  
xn  1 x 
(9)
3
24
360
n  0 n!( n  2)!
From (8) we obtain a second solution
 p ( x ) dx
dx
e 
y 2  y1 ( x)  2
dx = y1 ( x) 
1
1 2
1 3
y1 ( x)
x3[1  x 
x 
x  ]2
3
24
360
dx
= y1 ( x) 
2
7
1
x 3 [1  x  x 2  x 3  ]
3
36
30
1
2
1
19 3
x  ]dx
= y1 ( x)  3 [1  x  x 2 
x
3
4
270
2 1
19
 1

 y1 ( x)   2   ln x 
x  ...
270
 2 x 3x 4

1
2 19
 1

 y1 ( x) ln x  y1 ( x)   2  
x  ...
4
 2 x 3x 270

1
2 19
 1

 y  c1 y1 ( x)  c2  y1 ( x) ln x  y1 ( x )   2  
x  ...  
 2 x 3 x 270

4
(*)
(**)
6
Lecture 21
Series Solution about Singular Points
Example 9:
Find the general solution of
xy  3y  y  0
Solution :

y 2  y1 ln x   b n x n  2
(10)
n 0

2
xn
n  0 n!(n  2)!
differentiate (10) gives
y1  
(11)

y
y 2  1  y1 ln x   (n  2)b n x n 3
x
n 0

y1 2y1


y2   2 
 y1 ln x   (n  2)(n  3)b n x n  4
x
x
n 0
so that

2y
xy2  3y2  y 2  ln x  xy1  3y1  y1   2y1  1   (n  2)(n  3)b n x n 3


x n 0


3 (n  2)b n x n 3   b n x n  2
n 0

n 0

2y1
  (n  2)nb n x n 3   b n x n  2
(12)
x n 0
n 0
where we have combined the1st two summations and used the fact that
xy1  3y1  y1  0
 2y1 
Differentiate (11) we can write (12) as

4n
 n!(n  2)! x
n 0
n 1



4
x n 1   (n  2)nbn x n 3   bn x n  2
n  0 n!( n  2)!
n 0
n 0


4(n  1) n 1 
x   (n  2)nbn x n 3   bn x n  2
n  0 n!( n  2)!
n2
n 1

= 0(2)b0 x 3  (b0  b1 ) x  2  

 4(k  1)

(b0  b1 ) x 2   
 k (k  2)bk  2  bk 1 x k 1.
k  0  k !( k  2)!

(13)
Setting (13) equal to zero then gives b 1  b 0 and
7
Lecture 21
Series Solution about Singular Points
4(k  1)
 k (k  2)bk  2  bk 1  0,
k!(k  1)!
For k=0, 1, 2, …
(14)
When k=0 in equation (14) we have 2+0  2b2  b1  0 so that but
b1  2, b0  2, but b2 is arbitrary
Rewriting equation (14) as
bk 1
4(k  1)
bk  2 

k (k  2) k!(k  2)! k (k  2)
and evaluating for k=1,2,… gives
b
4
b3  2 
3 9
1
1
1
25
b4  b3 

b2 
8
32 24
288
and so on. Thus we can finally write
y 2  y1 ln x  b0 x 2  b1 x 1  b2  b3 x    
(15)
4
b
= y1 ln x  2 x 2  2 x 1  b2   2   x   
 3 9
Where b2 is arbitrary.
(16)
Equivalent Solution
At this point you may be wondering whether (*) and (16) are really equivalent. If we
choose c2  4 in equation (**), then
8
38
 2


x    
y 2  y1 ln x +   2 
3x 135
 x

1
1 3
 1

8
38
1  x  x2 
x     2


(17)
y 2  y1 ln x +  3
24
360
   2  3x  135 x     
 x

 y1 ln x  2 x 2  2 x 1 
29 19

x  ...
36 108
Which is precisely obtained what we obtained from (16). If b2 is chosen as
29
36
The next example illustrates the case when the indicial roots are equal.
8
Lecture 21
Series Solution about Singular Points
Example :10
Find the general solution of xy  y   4 y  0
(18)

Solution :The assumption y=  c n x n  r leads to
n=0



n=0
n=0
xy  y  4 y   (n+r)(n+r-1)cn x n  r 1   (n+r)cn x n  r 1  4 cn x n  r
n=0


=  (n+r) 2 c n x n  r 1  4 c n x n  r
n=0
n=0



 x r  r 2 c0 x 1   (n+r) 2 cn x n 1  4 c n x n 
n=1
n=0






 x r  r 2 c0 x 1   (k+r+1) 2 c k+1  4ck  x k  0
k=0


Therefore r 2 =0, and so the indicial roots are equal: r1  r2  0. Moreover we have
(k  r  1) 2 c k 1  4c k  0, k=0,1,2,…
(19)
Clearly the roots r1  0 will yield one solution corresponding to the coefficients defined
by the iteration of
4ck
k=0,1,2,…
ck 1 
(k  1) 2
The result is
 4n
(20)
y1  c0 
xn , x  
2
n  0 ( n!)

y2  y1 ( x) 
1
 x ) dx
e
dx
dx  y1 ( x) 
2
2
y1 ( x)
16 3


2
x 1  4 x  4 x  x    
9


 y1 (x) 
1
1472 3

1  8x  40x 2 
x  dx

x
9

1472 2
1

 y1 (x)    8  40x 
x  ...dx
9
x

9
Lecture 21
Series Solution about Singular Points
1472 3


 y1 (x) ln x  8x  20x 2 
x  
27


Thus on the interval (0,  ) the general solution of (18) is

1472 3


y  c1y1 (x)  c2  y1 (x) ln x  y1 (x)  8x  20x 2 
x  ...  
27



where y1 ( x) is defined by (20)
In caseII we can also determine y2 ( x) of example9 directly from assumption (4b)
Practice Exercises
In problem 1-12 show that the indicial roots differ by an integer. Use the method of
frobenius to obtain two linearly independent series solutions about the regular singular
point x0  0 Form the general solution on (0,  )
1. xy  2 y  xy  0
1

2. x 2 y  xy   x 2   y  0
4

3. x( x  1) y  3 y  2 y  0
3
4. y  y  2 y  0
x
5. xy  (1  x) y  y  0
6. xy  y  0
7. xy  y  y  0
8. xy  y  y  0
9. x 2 y  x( x  1) y  y  0
10. xy  y  4 xy  0
11. x 2 y  ( x  1) y  2 y  0
12. xy  y  x3 y  0
10