082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

590 Chapter 19 Ordinary differential equations I
Example19.32 By seeking a power series solution in the form
∞∑
a m
x m obtain the first three non-zero
m=0
terms inthe solution of Bessel’s equation inthe case when v = 1:
x 2d2 y
dx 2 +xdy dx + (x2 −1)y = 0. (19.28)
Solution As before weseek a power series solution inthe form
∞∑
y = a m
x m =a 0
+a 1
x+a 2
x 2 +a 3
x 3 +a 4
x 4 +a 5
x 5 ...
m=0
The strategy is to substitutey, and its first and second derivatives into Equation 19.28
and choose valuesa m
so that the equation is satisfied.yand its first two derivatives can
be written:
y=a 0
+a 1
x+a 2
x 2 +a 3
x 3 +a 4
x 4 +a 5
x 5 ...
dy
dx =a 1 +2a 2 x+3a 3 x2 +4a 4
x 3 +5a 5
x 4 ...
d 2 y
dx 2 =2a 2 +6a 3 x+12a 4 x2 +20a 5
x 3 ...
Substitution intox 2d2 y
dx 2 +xdy dx + (x2 −1)y =0gives
2a 2
x 2 +6a 3
x 3 +12a 4
x 4 +20a 5
x 5 ... + ( a 1
x +2a 2
x 2 +3a 3
x 3 +4a 4
x 4 +5a 5
x 5 ... )
Removing the brackets
+ (x 2 −1)(a 0
+a 1
x+a 2
x 2 +a 3
x 3 +a 4
x 4 +a 5
x 5 ...) =0
2a 2
x 2 +6a 3
x 3 +12a 4
x 4 +20a 5
x 5 ...+ ( a 1
x +2a 2
x 2 +3a 3
x 3 +4a 4
x 4 +5a 5
x 5 ... )
+a 0
x 2 +a 1
x 3 +a 2
x 4 +a 3
x 5 +a 4
x 6 +a 5
x 7 ...
−(a 0
+a 1
x+a 2
x 2 +a 3
x 3 +a 4
x 4 +a 5
x 5 ...) =0
We now equate coefficients on both sides of this equation starting with the constant
term:
−a 0
= 0 so that a 0
= 0.
Equating coefficients ofx:
a 1
−a 1
= 0
sothata 1
isarbitrary.
Equating coefficients ofx 2 :
2a 2
+2a 2
+a 0
−a 2
= 0 sothat 3a 2
= −a 0
, from which a 2
= 0.
Equating coefficients ofx 3 :
6a 3
+3a 3
+a 1
−a 3
= 0 sothat 8a 3
= −a 1
, from which a 3
= − 1 8 a 1 .
It is easy to verify that all subsequent terms a m
when m is even, are zero. Equating
coefficients ofx 5 :
20a 5
+5a 5
+a 3
−a 5
= 0 so that 24a 5
= −a 3