Higher Engineering Mathematics

510 DIFFERENTIAL EQUATIONS

=

−a 2 (k+3)(k−2)

(3)(4)

=

−(k+3)(k−2)

(3)(4)

.

a 0 [−k(k+1)]

(1)(2)

=

a 0 k(k+1)(k+3)(k−2)

4!

Forr=3,

a 5 =

a 3 [(3)(4)−k(k+1)]

(4)(5)

=

−a 3 [k^2 +k−12]

(4)(5)

=

−a 3 (k+4)(k−3)

(4)(5)

=

−(k+4)(k−3)

(4)(5)

.

−a 1 (k−1)(k+2)

(2)(3)

=

a 1 (k−1)(k−3)(k+2)(k+4)

5!

and so on.

Substituting values into equation (43) gives:

y=x^0

{

a 0 +a 1 x−

a 0 k(k+1)

2!

x^2

−

a 1 (k−1)(k+2)

3!

x^3

+

a 0 k(k+1)(k−2)(k+3)

4!

x^4

+

a 1 (k−1)(k−3)(k+2)(k+4)

5!

x^5

+ ···

}

i.e.y=a 0

{

1 −

k(k+1)

2!

x^2

+

k(k+ 1 )(k−2)(k+3)

4!

x^4 −···

}

+a 1

{

x−

(k−1)(k+2)

3!

x^3

+

(k−1)(k−3)(k+2)(k+4)

5!

x^5 −···

}

(47)

From page 503, it was stated that if two solutions
of the indicial equation differ by an integer, as in
this case, wherec=0 and 1, and if one coefficient is
indeterminate, as with whenc=0, then the complete
solution is always given by using this value of c.
Using the second value ofc, i.e.c=1 in this problem,
will give a series which is one of the series in the first
solution. (This may be checked forc=1 and where
a 1 =0; the result will be the first part of equation
(47) above).

Legendre’s polynomials

(A polynomial is an expression of the form:

f(x)=a+bx+cx^2 +dx^3 + ···). Whenkin equa-

tion (47) above is an integer, say, n, one of the

solution series terminates after a finite number of

terms. For example, ifk=2, then the first series

terminates after the term inx^2. The resulting poly-

nomial inx, denoted byPn(x), is called aLegendre

polynomial. Constantsa 0 anda 1 are chosen so

thaty=1 whenx=1. This is demonstrated in the

following worked problems.

Problem 13. Determine the Legendre polyno-

mialP 2 (x).

Since inP 2 (x),n=k=2, then from the first part of

equation (47), i.e. the even powers ofx:

y=a 0

{

1 −

2(3)

2!

x^2 + 0

}

=a 0 { 1 − 3 x^2 }

a 0 is chosen to makey=1 whenx= 1

i.e. 1=a 0 { 1 −3(1)^2 }=− 2 a 0 , from which,a 0 =−

1

2

Hence, P 2 (x)=−

1

2

(

1 − 3 x^2

)

=

1

2

( 3 x^2 − 1 )

Problem 14. Determine the Legendre poly-

nomialP 3 (x).

Since inP 3 (x),n=k=3, then from the second part

of equation (47), i.e. the odd powers ofx:

y=a 1

{

x−

(k−1)(k+2)

3!

x^3

+

(k−1)(k−3)(k+2)(k+4)

5!

x^5 − ···

}

i.e.y=a 1

{

x−

(2)(5)

3!

x^3 +

(2)(0)(5)(7)

5!

x^5

}

=a 1

{

x−

5

3

x^3 + 0

}

a 1 is chosen to makey=1 whenx=1.

i.e. 1=a 1

{

1 −

5

3

}

=a 1

(

−

2

3

)

from which,a 1 =−

3

2

Hence,P 3 (x)=−

3

2

(

x−

5

3

x^3

)

orP 3 (x)=

1

2

(5x^3 − 3 x)