Higher Engineering Mathematics

502 DIFFERENTIAL EQUATIONS

i.e.y=a 0 x

1

2

{

1 +x+

x^2

(2×3)

+

x^3

(2×3)×(3×5)

+

x^4

(2× 3 ×4)×(3× 5 ×7)

+ ···

}

(27)

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

each solution, its value could well be different.

Leta 0 =Ain equation (26), anda 0 =Bin equa-

tion (27). Also, if the first solution is denoted

byu(x) and the second byv(x), then the gen-

eral solution of the given differential equation

isy=u(x)+v(x),

i.e.y=Ax

{

1 +

x

( 1 × 3 )

+

x^2

( 1 × 2 )×( 3 × 5 )

+

x^3

( 1 × 2 × 3 )×( 3 × 5 × 7 )

+

x^4

( 1 × 2 × 3 × 4 )×( 3 × 5 × 7 × 9 )

+···

}

+Bx

1

2

{

1 +x+

x^2

( 2 × 3 )

+

x^3

( 2 × 3 )×( 3 × 5 )

+

x^4

( 2 × 3 × 4 )×( 3 × 5 × 7 )

+···

}

Problem 9. Use the Frobenius method to deter-

mine the general power series solution of the

differential equation:

d^2 y

dx^2

− 2 y= 0

The differential equation may be rewritten as:
y′′− 2 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+···

}

(28)

wherea 0 =0,

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

+ ···+arxc+r+··· (29)

(ii) Differentiating equation (29) 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) Replacingrby (r+2) in

ar(c+r−1)(c+r)xc+r−^2 gives:

ar+ 2 (c+r+1)(c+r+2)xc+r

Substitutingyandy′′into each term of the given

equationy′′− 2 y=0 gives:

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

+[a 2 (c+1)(c+2)− 2 a 0 ]xc+···

+[ar+ 2 (c+r+1)(c+r+2)

− 2 ar]xc+r+··· = 0 (30)

(iv) Theindicial equationis obtained by equating

the coefficient of the lowest power ofxto zero.

Hence, a 0 c(c−1)=0 from which,c= 0

or c= 1 sincea 0 = 0

For the term inxc−^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 com-

bined with the zero value ofcwould make the

product zero.

For the term inxc,

a 2 (c+1)(c+2)− 2 a 0 =0 from which,

a 2 =

2 a 0

(c+1)(c+2)

(31)

For the term inxc+r,

ar+ 2 (c+r+1)(c+r+2)− 2 ar= 0

from which,

ar+ 2 =

2 ar

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

(32)

(a)Whenc=0:a 1 is indeterminate, and from

equation (31)

a 2 =

2 a 0

(1×2)

=

2 a 0

2!

In general,ar+ 2 =

2 ar

(r+1)(r+2)

and

whenr=1,a 3 =

2 a 1

(2×3)

=

2 a 1

(1× 2 ×3)

=

2 a 1

3!