Higher Engineering Mathematics

POWER SERIES METHODS OF SOLVING ORDINARY DIFFERENTIAL EQUATIONS 501

I

Thus, whenr=1,

a 1 =

a 0

1(2+1)

=

a 0

1 × 3

whenr=2,

a 2 =

a 1

2(4+1)

=

a 1

(2×5)

=

a 0

(1×3)(2×5)

or

a 0

(1×2)×(3×5)

whenr=3,

a 3 =

a 2

3(6+1)

=

a 2

3 × 7

=

a 0

(1× 2 ×3)×(3× 5 ×7)

whenr=4,

a 4 =

a 3

4(8+1)

=

a 3

4 × 9

=

a 0

(1× 2 × 3 ×4)×(3× 5 × 7 ×9)

and so on.

From equation (23), the trial solution was:

y=xc

{

a 0 +a 1 x+a 2 x^2 +a 3 x^3 +···

+arxr+···

}

Substitutingc=1 and the above values of
a 1 ,a 2 ,a 3 , ... into the trial solution gives:

y=x^1

{

a 0 +

a 0

(1×3)

x+

a 0

(1×2)×(3×5)

x^2

+

a 0

(1× 2 ×3)×(3× 5 ×7)

x^3

+

a 0

(1× 2 × 3 ×4)×(3× 5 × 7 ×9)

x^4

+ ···

}

i.e.y=a 0 x^1

{

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)

+ ···

}

(26)

(b) Withc=

1

2

ar=

ar− 1

2(c+r−1)(c+r)−(c+r)+ 1

from equation (25)

i.e.ar=

ar− 1

2

(

1

2

+r− 1

)(

1

2

+r

)

−

(

1

2

+r

)

+ 1

=

ar− 1

2

(

r−

1

2

)(

r+

1

2

)

−

1

2

−r+ 1

=

ar− 1

2

(

r^2 −

1

4

)

−

1

2

−r+ 1

=

ar− 1

2 r^2 −

1

2

−

1

2

−r+ 1

=

ar− 1

2 r^2 −r

=

ar− 1

r(2r−1)

Thus, whenr=1,a 1 =

a 0

1(2−1)

=

a 0

1 × 1

whenr=2,a 2 =

a 1

2(4−1)

=

a 1

(2×3)

=

a 0

(2×3)

whenr=3,a 3 =

a 2

3(6−1)

=

a 2

3 × 5

=

a 0

(2×3)×(3×5)

whenr=4,a 4 =

a 3

4(8−1)

=

a 3

4 × 7

=

a 0

(2× 3 ×4)×(3× 5 ×7)

and so on.

From equation (23), the trial solution was:

y=xc

{

a 0 +a 1 x+a 2 x^2 +a 3 x^3 +···

+arxr+···

}

Substitutingc=

1

2

and the above values of

a 1 ,a 2 ,a 3 , ... into the trial solution gives:

y=x

1

2

{

a 0 +a 0 x+

a 0

(2×3)

x^2 +

a 0

(2×3)×(3×5)

x^3

+

a 0

(2× 3 ×4)×(3× 5 ×7)

x^4 + ···

}