Higher Engineering Mathematics, Sixth Edition

Power series methods of solving ordinary differential equations 509

+

(x

2

)−v{ 1

( 1 −v)

−

x^2

22 (1!)( 2 −v)

+

x^4

24 (2!)( 3 −v)

−···

}

From Problem 10 above,whenc=+v,

a 2 =

−a 0

22 (v+ 1 )

If we leta 0 =

1

2 v(v+ 1 )

then

a 2 =

− 1

22 (v+ 1 ) 2 v(v+ 1 )

=

− 1

2 v+^2 (v+ 1 )(v+ 1 )

=

− 1

2 v+^2 (v+ 2 )

from equation (41)

Similarly, a 4 =

a 2

v^2 −(c+ 4 )^2

from equation (37)

=

a 2

(v−c− 4 )(v+c+ 4 )

=

a 2

− 4 ( 2 v+ 4 )

sincec=v

=

−a 2

23 (v+ 2 )

=

− 1

23 (v+ 2 )

− 1

2 v+^2 (v+ 2 )

=

1

2 v+^4 (2!)(v+ 3 )

since(v+ 2 )(v+ 2 )=(v+ 3 )

and a 6 =

− 1

2 v+^6 (3!)(v+ 4 )

and so on.

Therecurrence relationis:

ar=

(− 1 )r/^2

2 v+r

(r

2

!

)



(

v+

r

2

+ 1

)

And if we letr= 2 k,then

a 2 k=

(−1)k

2 v+^2 k(k!)(v+k+1)

(42)

fork=1, 2, 3,...

Hence, it is possible to write the new form for equation
(38) as:

y=Axv

{

1

2 v(v+ 1 )

−

x^2

2 v+^2 (1!)(v+ 2 )

+

x^4

2 v+^4 (2!)(v+ 3 )

−···

}

This is calledthe Bessel function of the first order kind,

of orderv,and is denoted byJv(x),

i.e. Jv(x)=

(x

2

)v{ 1

(v+1)

−

x^2

22 (1!)(v+2)

+

x^4

24 (2!)(v+3)

−···

}

providedvis not a negative integer.

For the second solution,whenc=−v, replacing v

by−vin equation (42) above gives:

a 2 k=

(− 1 )k

22 k−v(k!)(k−v+ 1 )

from which, when k= 0 ,a 0 =

(− 1 )^0

2 −v(0!)( 1 −v)

=

1

2 −v( 1 −v)

since 0!=1 (see page 495)

whenk=1,a 2 =

(− 1 )^1

22 −v(1!)( 1 −v+ 1 )

=

− 1

22 −v(1!)( 2 −v)

whenk=2,a 4 =

(− 1 )^2

24 −v(2!)( 2 −v+ 1 )

=

1

24 −v(2!)( 3 −v)

whenk=3,a 6 =

(− 1 )^3

26 −v(3!)( 3 −v+ 1 )

=

1

26 −v(3!)( 4 −v)

and so on.

Hence,y=Bx−v

{

1

2 −v( 1 −v)

−

x^2

22 −v(1!)( 2 −v)

+

x^4

24 −v(2!)( 3 −v)

−···

}

i.e. J−v(x)=

(x

2

)−v{ 1

(1−v)

−

x^2

22 (1!)(2−v)

+

x^4

24 (2!)(3−v)

−···

}

providedvis not a positive integer.

Jv(x)andJ−v(x)are two independent solutions of the

Bessel equation; the complete solution is:

y=AJv(x)+BJ−v(x)whereAandBare constants