Higher Engineering Mathematics, Sixth Edition

Power series methods of solving ordinary differential equations 511

Now try the following exercise

Exercise 197 Further problemson Bessel’s

equation and Bessel’s functions

1. Determine the power series solution of Bes-

sel’s equation:x^2

d^2 y

dx^2

+x

dy

dx

+(x^2 −v^2 )y= 0

whenv=2, up to and includingthe term in[ x^6.

y=Ax^2

{

1 −

x^2

12

+

x^4

384

−···

}]

2. Find the power series solution of the
   Bessel function:x^2 y′′+xy′+

(

x^2 −v^2

)

y= 0

in terms of the Bessel functionJ 3 (x)when

v=3. Give the answer up to and including the

term inx^7.

⎡

⎢

⎢

⎣

y=AJ 3 (x)=

(x

2

) 3 { 1

 4

−

x^2

22  5

+

x^4

25  6

−···

}

⎤

⎥

⎥

⎦

3. Evaluate the Bessel functionsJ 0 (x)andJ 1 (x)
   whenx=1, correct to 3 decimal places.
   [J 0 (x)= 0. 765 ,J 1 (x)= 0 .440]

52.7 Legendre’s equation and

Legendre polynomials

Another important differential equation in physics
and engineering applications is Legendre’s equation

of the form:( 1 −x^2 )

d^2 y

dx^2

− 2 x

dy

dx

+k(k+ 1 )y=0or

( 1 −x^2 )y′′− 2 xy′+k(k+ 1 )y=0wherek is a real
constant.

Problem 12. Determine the general power series

solution of Legendre’s equation.

To solve Legendre’s equation
( 1 −x^2 )y′′− 2 xy′+k(k+ 1 )y=0 using the Frobenius
method:

(i) Let a trial solution be of the form

y=xc

{

a 0 +a 1 x+a 2 x^2 +a 3 x^3

+···+arxr+···

}

(43)

wherea 0 =0,

i.e. y=a 0 xc+a 1 xc+^1 +a 2 xc+^2 +a 3 xc+^3

+···+arxc+r+··· (44)

(ii) Differentiating equation (44) gives:

y′=a 0 cxc−^1 +a 1 (c+ 1 )xc

+a 2 (c+ 2 )xc+^1 +···

+ar(c+r)xc+r−^1 +···

andy′′=a 0 c(c− 1 )xc−^2 +a 1 c(c+ 1 )xc−^1

+a 2 (c+ 1 )(c+ 2 )xc+···

+ar(c+r− 1 )(c+r)xc+r−^2 +···

(iii) Substitutingy,y′andy′′into each term of the

given equation:(

1 −x^2

)

y′′− 2 xy′+k(k+ 1 )y=0gives:

a 0 c(c− 1 )xc−^2 +a 1 c(c+ 1 )xc−^1

+a 2 (c+ 1 )(c+ 2 )xc+···

+ar(c+r− 1 )(c+r)xc+r−^2 +···

−a 0 c(c− 1 )xc−a 1 c(c+ 1 )xc+^1

−a 2 (c+ 1 )(c+ 2 )xc+^2 −···

−ar(c+r− 1 )(c+r)xc+r−···− 2 a 0 cxc

− 2 a 1 (c+ 1 )xc+^1 − 2 a 2 (c+ 2 )xc+^2 −···

− 2 ar(c+r)xc+r−···+k^2 a 0 xc

+k^2 a 1 xc+^1 +k^2 a 2 xc+^2 +···+k^2 arxc+r

+ ···+ka 0 xc+ka 1 xc+^1 +···

+karxc+r+···= 0 (45)

(iv) Theindicial equationis obtained by equating the

coefficient of the lowest power ofx(i.e.xc−^2 )to

zero. Hence,a 0 c(c− 1 )=0 from which,c= 0 or

c= 1 sincea 0 =0.

For the term inxc−^1 ,i.e.a 1 c(c+ 1 )=0 With

c=1,a 1 =0; however, whenc=0,a 1 is inde-

terminate, since any value ofa 1 combined with

the zero value ofcwould make the product zero.

For the term inxc+r,

ar+ 2 (c+r+ 1 )(c+r+ 2 )−ar(c+r− 1 )

(c+r)− 2 ar(c+r)+k^2 ar+kar= 0