Higher Engineering Mathematics, Sixth Edition

Power series methods of solving ordinary differential equations 501

(iv) The sum of these three terms forms the left-hand

side of the equation. Since the right-hand side is

zero, the coefficients of each power of x can be

equated to zero.

For example, the coefficient ofxc−^1 is equated

to zero giving: 3a 0 c(c− 1 )+a 0 c= 0

or a 0 c[3c− 3 +1]=a 0 c(3c−2)= 0 (18)

The coefficient ofxcis equated to zero giving:

3 a 1 c(c+ 1 )+a 1 (c+ 1 )−a 0 = 0

i.e. a 1 ( 3 c^2 + 3 c+c+ 1 )−a 0

=a 1 ( 3 c^2 + 4 c+ 1 )−a 0 = 0

or a 1 (3c+1)(c+1)−a 0 = 0 (19)

In each of series (a), (b) and (c) anxcterm is

involved, after which, a general relationship can

be obtained forxc+r,wherer≥0.

In series (a) and (b), terms inxc+r−^1 are present;

replacingrby(r+ 1 )will give the corresponding

terms inxc+r, which occurs in all three equa-

tions, i.e.

in series (a), 3ar+ 1 (c+r)(c+r+ 1 )xc+r

in series (b), ar+ 1 (c+r+ 1 )xc+r

in series (c), −arxc+r

Equating the total coefficients ofxc+rto zero

gives:

3 ar+ 1 (c+r)(c+r+ 1 )+ar+ 1 (c+r+ 1 )

−ar= 0

which simplifies to:

ar+ 1 {(c+r+1)(3c+ 3 r+1)}−ar= 0 (20)

Equation (18), which was formed from the coeffi-

cients of the lowest power ofx,i.e.xc−^1 , iscalled

theindicial equation, from which, the value of

cis obtained. From equation (18), sincea 0 =0,

thenc= 0 orc=

2

3

(a) Whenc= 0 :

From equation (19), if c=0, a 1 ( 1 × 1 )−a 0 =0,
i.e.a 1 =a 0

From equation (20), ifc=0,
ar+ 1 (r+ 1 )( 3 r+ 1 )−ar=0,

i.e.ar+ 1 =

ar

(r+1)(3r+1)

r≥ 0

Thus, whenr=1,a 2 =

a 1

( 2 × 4 )

=

a 0

( 2 × 4 )

sincea 1 =a 0

whenr=2,a 3 =

a 2

( 3 × 7 )

=

a 0

( 2 × 4 )( 3 × 7 )

or

a 0

( 2 × 3 )( 4 × 7 )

whenr=3,a 4 =

a 3

( 4 × 10 )

=

a 0

( 2 × 3 × 4 )( 4 × 7 × 10 )

andsoon.

From equation (16), the trial solution was:

y=xc{a 0 +a 1 x+a 2 x^2 +a 3 x^3 +···+arxr+···}

Substitutingc=0 and the above values ofa 1 ,a 2 ,a 3 ,...

into the trial solution gives:

y=x^0

{

a 0 +a 0 x+

(

a 0

( 2 × 4 )

)

x^2

+

(

a 0

( 2 × 3 )( 4 × 7 )

)

x^3

+

(

a 0

( 2 × 3 × 4 )( 4 × 7 × 10 )

)

x^4 +···

}

i.e. y=a 0

{

1 +x+

x^2

( 2 × 4 )

+

x^3

( 2 × 3 )( 4 × 7 )

+

x^4

( 2 × 3 × 4 )( 4 × 7 × 10 )

+···

}

(21)

(b) Whenc=

2

3

:

From equation (19), ifc=

2

3

,a 1 ( 3 )

(

5

3

)

−a 0 =0, i.e.

a 1 =

a 0

5

From equation (20), ifc=

2

3

ar+ 1

(

2

3

+r+ 1

)

( 2 + 3 r+ 1 )−ar=0,

i.e.ar+ 1

(

r+

5

3

)

( 3 r+ 3 )−ar

=ar+ 1 ( 3 r^2 + 8 r+ 5 )−ar=0,

i.e.ar+ 1 =

ar

(r+1)(3r+5)

r≥ 0