Power series methods of solving ordinary differential equations 505
Substituting y and y
into each term of the given
equation y − 2y = 0 gives:
y
− 2y = a 0 c(c − 1)x
c−2
+ a 1 c(c + 1)x
c−1
+ [a 2 (c+1)(c + 2)−2a 0 ]x
c
+· · ·
+ [a r+2 (c + r + 1)(c + r + 2)
− 2a r ] x
c+r
+ · · · = 0
(30)
(iv) The indicial equation is obtained by equating
the coefficient of the lowest power of x to zero.
Hence, a 0 c(c − 1) =0 from which, c = 0 or
c = 1 since a 0 = 0
For the term in x c−1 , i.e. a 1 c(c + 1) = 0
With c = 1, a 1 = 0; however, when c = 0, a 1 is
indeterminate, since any value of a 1 combined
with the zero value of c would make the product
zero.
For the term in x c ,
a 2 (c + 1)(c + 2) − 2a 0 = 0 from which,
a 2 =
2a 0
(c + 1)(c + 2)
(31)
For the term in x c+r ,
a r+2 (c + r + 1)(c + r + 2) − 2a r = 0
from which,
a r+2 =
2a r
(c + r + 1)(c + r + 2)
(32)
(a) When c = 0: a 1 is indeterminate, and from
equation (31)
a 2 =
2a 0
(1 × 2)
=
2a 0
2!
In general, a r + 2 =
2a r
(r + 1)(r + 2)
and
when r = 1, a 3 =
2a 1
(2 × 3)
=
2a 1
(1 × 2 × 3)
=
2a 1
3!
when r = 2, a 4 =
2a 2
3 × 4
=
4a 0
4!
Hence, y = x
0
a 0 + a 1 x +
2a 0
2!
x
2
+
2a 1
3!
x
3
+
4a 0
4!
x
4
+ · · ·
from equation (28)
= a 0
1 +
2x 2
2!
+
4x 4
4!
+ · · ·
+ a 1
x +
2x 3
3!
+
4x 5
5!
+ · · ·
Since a 0 and a 1 are arbitrary constants
depending on boundary conditions, let a 0 = P
and a 1 = Q, then:
y=P
1 +
2x 2
2!
+
4x 4
4!
+· · ·
+ Q
x +
2x 3
3!
+
4x 5
5!
+· · ·
(33)
(b) When c =1: a 1 = 0, and from equation (31),
a 2 =
2a 0
(2 × 3)
=
2a 0
3!
Since c = 1, a r+2 =
2a r
(c + r + 1)(c + r + 2)
=
2a r
(r + 2)(r + 3)
from equation (32) and when r = 1,
a 3 =
2a 1
(3 × 4)
= 0 since a 1 = 0
when r = 2,
a 4 =
2a 2
(4 × 5)
=
2
(4 × 5)
×
2a 0
3!
=
4a 0
5!
when r = 3,
a 5 =
2a 3
(5 × 6)
= 0
Hence, when c = 1,
y = x
1
a 0 +
2a 0
3!
x
2
+
4a 0
5!
x
4
+ · · ·
from equation (28)
i.e. y = a 0
x +
2x 3
3!
+
4x 5
5!
+ ...
Again, a 0 is an arbitrary constant; let a 0 = K ,
then
y = K
x +
2x
3
3!
+
4x
5
5!
+· · ·
However, this latter solution is not a separate solution,
for it is the same form as the second series in equation
(33). Hence, equation (33) with its two arbitrary constants P and Q gives the general solution. This is always
Substituting y and y
into each term of the given
equation y − 2y = 0 gives:
y
− 2y = a 0 c(c − 1)x
c−2
+ a 1 c(c + 1)x
c−1
+ [a 2 (c+1)(c + 2)−2a 0 ]x
c
+· · ·
+ [a r+2 (c + r + 1)(c + r + 2)
− 2a r ] x
c+r
+ · · · = 0
(30)
(iv) The indicial equation is obtained by equating
the coefficient of the lowest power of x to zero.
Hence, a 0 c(c − 1) =0 from which, c = 0 or
c = 1 since a 0 = 0
For the term in x c−1 , i.e. a 1 c(c + 1) = 0
With c = 1, a 1 = 0; however, when c = 0, a 1 is
indeterminate, since any value of a 1 combined
with the zero value of c would make the product
zero.
For the term in x c ,
a 2 (c + 1)(c + 2) − 2a 0 = 0 from which,
a 2 =
2a 0
(c + 1)(c + 2)
(31)
For the term in x c+r ,
a r+2 (c + r + 1)(c + r + 2) − 2a r = 0
from which,
a r+2 =
2a r
(c + r + 1)(c + r + 2)
(32)
(a) When c = 0: a 1 is indeterminate, and from
equation (31)
a 2 =
2a 0
(1 × 2)
=
2a 0
2!
In general, a r + 2 =
2a r
(r + 1)(r + 2)
and
when r = 1, a 3 =
2a 1
(2 × 3)
=
2a 1
(1 × 2 × 3)
=
2a 1
3!
when r = 2, a 4 =
2a 2
3 × 4
=
4a 0
4!
Hence, y = x
0
a 0 + a 1 x +
2a 0
2!
x
2
+
2a 1
3!
x
3
+
4a 0
4!
x
4
+ · · ·
from equation (28)
= a 0
1 +
2x 2
2!
+
4x 4
4!
+ · · ·
+ a 1
x +
2x 3
3!
+
4x 5
5!
+ · · ·
Since a 0 and a 1 are arbitrary constants
depending on boundary conditions, let a 0 = P
and a 1 = Q, then:
y=P
1 +
2x 2
2!
+
4x 4
4!
+· · ·
+ Q
x +
2x 3
3!
+
4x 5
5!
+· · ·
(33)
(b) When c =1: a 1 = 0, and from equation (31),
a 2 =
2a 0
(2 × 3)
=
2a 0
3!
Since c = 1, a r+2 =
2a r
(c + r + 1)(c + r + 2)
=
2a r
(r + 2)(r + 3)
from equation (32) and when r = 1,
a 3 =
2a 1
(3 × 4)
= 0 since a 1 = 0
when r = 2,
a 4 =
2a 2
(4 × 5)
=
2
(4 × 5)
×
2a 0
3!
=
4a 0
5!
when r = 3,
a 5 =
2a 3
(5 × 6)
= 0
Hence, when c = 1,
y = x
1
a 0 +
2a 0
3!
x
2
+
4a 0
5!
x
4
+ · · ·
from equation (28)
i.e. y = a 0
x +
2x 3
3!
+
4x 5
5!
+ ...
Again, a 0 is an arbitrary constant; let a 0 = K ,
then
y = K
x +
2x
3
3!
+
4x
5
5!
+· · ·
However, this latter solution is not a separate solution,
for it is the same form as the second series in equation
(33). Hence, equation (33) with its two arbitrary constants P and Q gives the general solution. This is always