Higher Engineering Mathematics, Sixth Edition

502 Higher Engineering Mathematics

Thus, whenr=1,a 2 =

a 1

( 2 × 8 )

=

a 0

( 2 × 5 × 8 )

sincea 1 =

a 0

5

whenr=2,a 3 =

a 2

( 3 × 11 )

=

a 0

( 2 × 3 )( 5 × 8 × 11 )

whenr=3,a 4 =

a 3

( 4 × 14 )

=

a 0

( 2 × 3 × 4 )( 5 × 8 × 11 × 14 )

and so on.

From equation (16), the trial solution was:

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

Substitutingc=

2

3

and the above values of a 1 , a 2 ,

a 3 , ... into the trial solution gives:

y=x

2

3

{

a 0 +

(a

0

5

)

x+

(

a 0

2 × 5 × 8

)

x^2

+

(

a 0

( 2 × 3 )( 5 × 8 × 11 )

)

x^3

+

(

a 0

( 2 × 3 × 4 )( 5 × 8 × 11 × 14 )

)

x^4 +···

}

i.e.y=a 0 x

2

3

{

1 +

x

5

+

x^2

( 2 × 5 × 8 )

+

x^3

( 2 × 3 )( 5 × 8 × 11 )

+

x^4

( 2 × 3 × 4 )( 5 × 8 × 11 × 14 )

+ ···

}

(22)

Sincea 0 is an arbitrary (non-zero) constant ineach

solution, its value could well be different.

Leta 0 =Ainequation(21), anda 0 =Binequation(22).

Also, if the first solution is denoted byu(x)and the

second byv(x), then the general solution of the given

differential equation isy=u(x)+v(x). Hence,

y=A

{

1 +x+

x^2

(2×4)

+

x^3

(2×3)(4×7)

+

x^4

(2× 3 ×4)(4× 7 ×10)

+···

}

+Bx

2

3

{

1 +

x

5

+

x^2

(2× 5 ×8)

+

x^3

(2×3)(5× 8 ×11)

+

x^4

(2× 3 ×4)(5× 8 × 11 ×14)

+···

}

Problem 8. Use the Frobenius method to

determine the general power series solution of the

differential equation:

2 x^2

d^2 y

dx^2

−x

dy

dx

+( 1 −x)y=0.

The differential equation may be rewritten as:

2 x^2 y′′−xy′+( 1 −x)y=0.

(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+···} (23)

wherea 0 = 0 ,

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

+ ···+arxc+r+··· (24)

(ii) Differentiating equation (24) 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′and y′′ into each term of

the given equation 2x^2 y′′−xy′+( 1 −x)y= 0

gives:

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

+ 2 a 2 (c+ 1 )(c+ 2 )xc+^2 +···

+ 2 ar(c+r− 1 )(c+r)xc+r+···

(a)

−xy′=−a 0 cxc−a 1 (c+ 1 )xc+^1

−a 2 (c+ 2 )xc+^2 −···

−ar(c+r)xc+r−··· (b)

( 1 −x)y=( 1 −x)(a 0 xc+a 1 xc+^1 +a 2 xc+^2

+a 3 xc+^3 +···+arxc+r+···)

=a 0 xc+a 1 xc+^1 +a 2 xc+^2 +a 3 xc+^3

+ ···+arxc+r+···