Higher Engineering Mathematics

500 DIFFERENTIAL EQUATIONS

Leta 0 =Ain equation (21), anda 0 =Bin equation

(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) Substituting y,y′ andy′′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+···

−a 0 xc+^1 −a 1 xc+^2 −a 2 xc+^3

−a 3 xc+^4 −···−arxc+r+^1 −···

(c)

(iv) Theindicial equation, which is obtained by

equating the coefficient of the lowest power of

xto zero, gives the value(s) ofc. Equating the

total coefficients ofxc(from equations (a) to

(c)) to zero gives:

2 a 0 c(c−1)−a 0 c+a 0 = 0

i.e. a 0 [2c(c−1)−c+1]= 0

i.e. a 0 [2c^2 − 2 c−c+1]= 0

i.e. a 0 [2c^2 − 3 c+1]= 0

i.e. a 0 [(2c−1)(c−1)]= 0

from which, c= 1 orc=

1

2

The coefficient of the general term, i.e.xc+r,

gives (from equations (a) to (c)):

2 ar(c+r−1)(c+r)−ar(c+r)

+ar−ar− 1 = 0

from which,

ar[2(c+r−1)(c+r)−(c+r)+1]=ar− 1

andar=

ar− 1

2(c+r−1)(c+r)−(c+r)+ 1

(25)

(a) Withc= 1 ,ar=

ar− 1

2(r)(1+r)−(1+r)+ 1

=

ar− 1

2 r+ 2 r^2 − 1 −r+ 1

=

ar− 1

2 r^2 +r

=

ar− 1

r(2r+1)