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

588 Chapter 19 Ordinary differential equations I
As was done inSection 19.7 we seekapower 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 19.25
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 d2 y
dx + dy +xy = 0gives
2 dx
(2a 2
x +6a 3
x 2 +12a 4
x 3 +20a 5
x 4 ...) + ( a 1
+2a 2
x+3a 3
x 2 +4a 4
x 3 ... )
+ (a 0
x+a 1
x 2 +a 2
x 3 +a 3
x 4 +a 4
x 5 ...) =0
We now equate coefficients on both sides of this equation starting with the constant
term:
a 1
=0.
Equating coefficients ofx:
4a 2
+a 0
=0 sothat a 2
=− 1 4 a 0 .
Equating coefficients ofx 2 :
9a 3
+a 1
=0 sothat a 3
=− 1 9 a 1 = 0 sincea 1 = 0.
Equating coefficients ofx 3 :
16a 4
+a 2
=0 sothat a 4
=− 1 (
16 a 2 = − 1 ) (
− 1 )
a
16 4 0
= 1
64 a 0 .
Itiseasytoverifythatallsubsequenttermsa m
,whenmisodd,arezero.Youshouldverify,inasimilarway,thata
6
= − 1
2304 a 0 .Whenmiseven,alla m aremultiplesofa 0 .We
can now write down a power series that satisfies Bessel’s differential
Equation 19.25:
(
y=a 0
1 − 1 4 x2 + 1
64 x4 − 1 )
2304 x6 +...
(19.26)
We have succeeded in obtaining a power series solution of Bessel’s Equation 19.25
wherea 0
isan arbitraryconstant.
As we have already noted, a 0
is an arbitrary constant. However, historically its value
is chosen to be 1 in Equation 19.26. The resulting power series is used to define a