Higher Engineering Mathematics

POWER SERIES METHODS OF SOLVING ORDINARY DIFFERENTIAL EQUATIONS 505

I

For the term inxc+r,

ar(c+r−1)(c+r)+ar(c+r)+ar− 2

−arv^2 = 0

ar[(c+r−1)(c+r)+(c+r)−v^2 ]=−ar− 2

i.e. ar[(c+r)(c+r− 1 +1)−v^2 ]=−ar− 2

i.e. ar[(c+r)^2 −v^2 ]=−ar− 2

i.e. therecurrence relationis:

ar=

ar− 2

v^2 −(c+r)^2

for r≥ 2 (37)

For the term inxc+^1 ,

a 1 [c(c+1)+(c+1)−v^2 ] = 0

i.e. a 1 [(c+1)^2 −v^2 ]= 0

but ifc=v a 1 [(v+1)^2 −v^2 ]= 0

i.e. a 1 [2v+1]= 0

Similarly, ifc=−v a 1 [1− 2 v]= 0

The terms (2v+1)and(1− 2 v) cannot both be

zero sincevis a real constant, hencea 1 =0.

Since a 1 =0, then from equation (37)

a 3 =a 5 =a 7 =...= 0

and

a 2 =

a 0

v^2 −(c+2)^2

a 4 =

a 0

[v^2 −(c+2)^2 ][v^2 −(c+4)^2 ]

a 6 =

a 0

[v^2 −(c+2)^2 ][v^2 −(c+4)^2 ][v^2 −(c+6)^2 ]

and so on.

Whenc=+v,

a 2 =

a 0

v^2 −(v+2)^2

=

a 0

v^2 −v^2 − 4 v− 4

=

−a 0

4 + 4 v

=

−a 0

22 (v+1)

a 4 =

a 0

[

v^2 −(v+2)^2

][

v^2 −(v+4)^2

]

=

a 0

[− 22 (v+1)][− 23 (v+2)]

=

a 0

25 (v+1)(v+2)

=

a 0

24 ×2(v+1)(v+2)

a 6 =

a 0

[v^2 −(v+2)^2 ][v^2 −(v+4)^2 ][v^2 −(v+6)^2 ]

=

a 0

[2^4 ×2(v+1)(v+2)][−12(v+3)]

=

−a 0

24 ×2(v+1)(v+2)× 22 ×3(v+3)

=

−a 0

26 × 3 !(v+1)(v+2)(v+3)

and so on.

The resulting solution forc=+vis given by:

y=u=

Axv

{

1 −

x^2

22 (v+1)

+

x^4

24 × 2 !(v+1)(v+2)

−

x^6

26 × 3 !(v+1)(v+2)(v+3)

+···

}

(38)

which is valid providedvis not a negative

integer and whereAis an arbitrary constant.

Whenc=−v,

a 2 =

a 0

v^2 −(−v+2)^2

=

a 0

v^2 −(v^2 − 4 v+4)

=

−a 0

4 − 4 v

=

−a 0

22 (v− 1 )

a 4 =

a 0

[2^2 (v−1)][v^2 −(−v+4)^2 ]

=

a 0

[2^2 (v−1)][2^3 (v−2)]

=

a 0

24 ×2(v−1)(v−2)

Similarly, a 6 =

a 0

26 × 3 !(v−1)(v−2)(v−3)

Hence,

y=w=

Bx−v

{

1 +

x^2

22 (v−1)

+

x^4

24 × 2 !(v−1)(v−2)

+

x^6

26 × 3 !(v−1)(v−2)(v−3)

+···

}

which is valid providedvis not a positive

integer and whereBis an arbitrary constant.