The document solves the Bessel equation with order v = 1/3. It finds two partial solutions by expanding the general solution as a power series and solving the characteristic equation. The first partial solution is a power series with terms of x^(2k+1/3) and the second partial solution is a power series with terms of x^(2k-1/3). The general solution is the linear combination of these two partial solutions.

1. Example. Solution of the Bessel
Equation with v = 1
3
August 18, 2006
Solve the Bessel equation
x2
y00
+ xy0
+ x2 1
9
y = 0 (1)
x0 = 0
0.0.1 Step 1.
x = 0 is the singular point.
Here P(x) = 1
x ; Q(x) = 1 1
9x2
p (x) = xP (x) = 1; q (x) = x2
Q (x) = x2 1
9 :
Hence by the second theorem of the existence 9 a solution y (x) of (35)
expanded in the generalized power series in the vicinity of x0 = 0.
Let y (x) be the solution of di¤erential equation (??) expanded in the gen-
eralized power series of x:
y (x) = x
1X
n=0
anxn
; a0 6= 0; jxj < 1 (2)
0.0.2 Step 2.
xy0
(x) = x
1X
n=0
(n + ) anxn
(3)
x2
y00
(x) = x
1X
n=0
(n + ) (n + 1) anxn
(4)
1

2. 0.0.3 Step 3.
x2
y00
+ xy0
+ x2 1
9
y = (5)
= x
1X
n=0
(n + ) (n + 1) anxn
+ (6)
+x
1X
n=0
(n + ) anxn
+ x2 1
9
x
1X
n=0
anxn
= 0
x2
1X
n=0
anxn
=
1X
n=0
anxn+2
=
1X
n=2
an 2xn
1X
n=0
(n + ) (n + 1) anxn
+ (7)
+
1X
n=0
(n + ) anxn
+
1X
n=2
an 2xn 1
9
1X
n=0
anxn
= 0
( 1) +
1
9
a0 + (1 + ) + (1 + )
1
9
a1x +
+
1X
n=2
(n + ) (n + 1) + (n + )
1
9
an + an 2 xn
= 0; a0 6= 0:
2 1
9
a0 = 0; a0 6= 0
(1 + )
2 1
9
a1 = 0
(n + ) (n + 1) + (n + )
1
9
an + an 2 = 0; (8)
an =
1
(n + )
2 1
9
an 2; n = 2; 3::::;(9)
Characteristic equation is:
2 1
9
= 0
, Solution is : = 1
3 ; = 1
3
= 1
3 :
a1 = 0 ! a2k+1 = 0; k = 0; 1; 2; :::
2

3. an =
1
n + 1
3
2 1
9
an 2;
n = 2; 3; :::; a0 6= 0:
n + 1
3
2 1
9 = n2
+ 2
3 n; n = 2k ) n2
+ 2
3 n = 4k k + 1
3
a2k =
1
4k k + 1
3
a2(k 1);
k = 1; 2; 3; :::; a0 6= 0:
a2k = 1
4k(k+ 1
3 )
a2(k 1) = 1
4k(k+ 1
3 )
1
4(k 1)(k 1+ 1
3 )
a2(k 2) =
= 1
4k(k+ 1
3 )
1
4(k 1)(k 1+ 1
3 )
1
4(k 2)(k 2+ 1
3 )
a2(k 3) = ::: =
= 1
4k(k+ 1
3 )
1
4(k 1)(k 1+ 1
3 )
1
4(k 2)(k 2+ 1
3 )
::: 1
4(k l)(k l+ 1
3 )
1
4 2 (2+ 1
3 )
1
4 1 (1+ 1
3 )
a0 =
= ( 1)k
22kk!(1+ 1
3 )(2+ 1
3 ):::(k 1+ 1
3 )(k+ 1
3 )
a0 = ( 1)k
22kk!
kQ
l=1
(l+ 1
3 )
a0; k = 1; 2; :::; a0 6= 0:
a2k =
( 1)
k
22kk!
kQ
l=1
l + 1
3
a0; k = 0; 1; 2; :::; a0 6= 0; (10)
a2k+1 = 0; k = 0; 1; 2; :::
0.0.4 First partial solution.
The …rst partial solution y1 (x) of equation (1) is:
y1 (x) = a0
1X
k=0
( 1)
k
x2k+ 1
3
22kk!
kQ
l=1
l + 1
3
(11)
0.0.5 Second partial solution.
1 = 1
3 ; 2 = 1
3 ; 1 2 = 2
3 =2 Z
The second partial solution y2 (x) of equation (1) is:
y2 (x) = b0
1X
k=0
( 1)
k
x2k 1
3
22kk!
kQ
l=1
l 1
3
(12)
3

4. 0.0.6 General solution.
The general solution y (x) of equation (1) is:
y (x) = C1
1X
k=0
( 1)
k
x2k+ 1
3
22kk!
kQ
l=1
l + 1
3
+ C2
1X
k=0
( 1)
k
x2k 1
3
22kk!
kQ
l=1
l 1
3
(13)
4