Higher Engineering Mathematics

POWER SERIES METHODS OF SOLVING ORDINARY DIFFERENTIAL EQUATIONS 509

I

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=0 wherekis 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=0 gives:

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 equat-

ing the coefficient of the lowest power ofx

(i.e.xc−^2 ) to zero. Hence,a 0 c(c−1)=0 from

which,c= 0 orc= 1 sincea 0 =0.

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

Withc=1,a 1 =0; however, whenc=0,a 1 is

indeterminate, 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

from which,

ar+ 2 =

ar

[

(c+r−1)(c+r)+2(c+r)−k^2 −k

]

(c+r+1)(c+r+2)

=

ar[(c+r)(c+r+1)−k(k+1)]

(c+r+1)(c+r+2) (46)

Whenc= 0 ,

ar+ 2 =

ar[r(r+1)−k(k+1)]

(r+1)(r+2)

Forr=0,

a 2 =

a 0 [−k(k+1)]

(1)(2)

Forr=1,

a 3 =

a 1 [(1)(2)−k(k+1)]

(2)(3)

=

−a 1 [k^2 +k−2]

3!

=

−a 1 (k−1)(k+2)

3!

Forr=2,

a 4 =

a 2 [(2)(3)−k(k+1)]

(3)(4)

=

−a 2

[

k^2 +k− 6

]

(3)(4)