Higher Engineering Mathematics

POWER SERIES METHODS OF SOLVING ORDINARY DIFFERENTIAL EQUATIONS 503

I

whenr=2,a 4 =

2 a 2

3 × 4

=

4 a 0

4!

Hence,y=x^0

{

a 0 +a 1 x+

2 a 0

2!

x^2 +

2 a 1

3!

x^3

+

4 a 0

4!

x^4 + ···

}

from equation (28)

=a 0

{

1 +

2 x^2

2!

+

4 x^4

4!

+···

}

+a 1

{

x+

2 x^3

3!

+

4 x^5

5!

+···

}

Since a 0 and a 1 are arbitrary constants
depending on boundary conditions, leta 0 =P
anda 1 =Q, then:

y=P

{

1 +

2 x^2

2!

+

4 x^4

4!

+···

}

+Q

{

x+

2 x^3

3!

+

4 x^5

5!

+···

}

(33)

(b) When c=1: a 1 =0, and from equa-
tion (31),

a 2 =

2 a 0

(2×3)

=

2 a 0

3!

Since c=1, ar+ 2 =

2 ar

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

=

2 ar

(r+2)(r+3)

from equation (32) and whenr=1,

a 3 =

2 a 1

(3×4)

=0 sincea 1 = 0

whenr=2,

a 4 =

2 a 2

(4×5)

=

2

(4×5)

×

2 a 0

3!

=

4 a 0

5!

whenr=3,

a 5 =

2 a 3

(5×6)

= 0

Hence, whenc=1,

y=x^1

{

a 0 +

2 a 0

3!

x^2 +

4 a 0

5!

x^4 +···

}

from equation (28)

i.e. y=a 0

{

x+

2 x^3

3!

+

4 x^5

5!

+...

}

Again,a 0 is an arbitrary constant; leta 0 =K,

then y=K

{

x+

2 x^3

3!

+

4 x^5

5!

+···

}

However, this latter solution is not a separate solu-

tion, for it is the same form as the second series

in equation (33). Hence, equation (33) with its two

arbitrary constantsPandQgives the general solu-

tion. This is always the case when the two values of

cdiffer by an integer (i.e. whole number). From the

above three worked problems, the following can be

deduced, and in future assumed:

(i) if two solutions of the indicial equation differ

by a quantitynot an integer, then two inde-

pendent solutions y=u(x)+v(x) results, the

general solution of which isy=Au+Bv(note:

Problem 7 hadc=0 and

2

3

and Problem 8 had

c=1 and

1

2

; in neither case did c differ by an

integer)

(ii) if two solutions of the indicial equationdodiffer

by an integer, as in Problem 9 wherec=0 and

1, and if one coefficient is indeterminate, as with

whenc=0, then the complete solution is always

given by using this value of c. Using the second

value ofc, i.e.c=1 in Problem 9, always gives

a series which is one of the series in the first

solution.

Now try the following exercise.

Exercise 197 Further problems on power

series solution by the Frobenius method

1. Produce, using Frobenius’ method, a power
   series solution for the differential equation:

2 x

d^2 y

dx^2

+

dy

dx

−y= 0

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

y=A

{

1 +x+

x^2

(2×3)

+

x^3

(2×3)(3×5)

+···

}

+Bx

1

2

{

1 +

x

(1×3)

+

x^2

(1×2)(3×5)

+

x^3

(1× 2 ×3)(3× 5 ×7)

+ ···

}

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦