Power series methods of solving ordinary differential equations 501
(iv) The sum of these three terms forms the left-hand
side of the equation. Since the right-hand side is
zero, the coefficients of each power of x can be
equated to zero.
For example, the coefficient of x c−1 is equated
to zero giving: 3a 0 c(c − 1) + a 0 c = 0
or a 0 c[3c − 3 + 1] = a 0 c(3c − 2) = 0
(18)
The coefficient of x c is equated to zero giving:
3a 1 c(c + 1) + a 1 (c + 1) − a 0 = 0
i.e. a 1 (3c
2
+ 3c + c + 1) − a 0
= a 1 (3c
2
+ 4c + 1) − a 0 = 0
or
a 1 (3c + 1)(c + 1)− a 0 = 0
(19)
In each of series (a), (b) and (c) an x c term is
involved, after which, a general relationship can
be obtained for x c+r , where r ≥ 0.
In series (a) and (b), terms in x c+r−1 are present;
replacing r by (r + 1) will give the corresponding
terms in x
c+r , which occurs in all three equations, i.e.
in series (a), 3a r+1 (c + r)(c + r + 1)x c+r
in series (b), a r+1 (c + r + 1)x c+r
in series (c), −a r x c+r
Equating the total coefficients of x c+r to zero
gives:
3a r+1 (c + r)(c + r + 1) + a r+1 (c + r + 1)
− a r = 0
which simplifies to:
a r+1 {(c + r + 1)(3c + 3r +1)} − a r = 0
(20)
Equation (18), which was formed from the coefficients of the lowest power of x, i.e. x
c−1 , is called
the indicial equation, from which, the value of
c is obtained. From equation (18), since a 0 = 0,
then c = 0 or c =
2
3
(a) When c = 0:
From equation (19), if c = 0, a 1 (1 × 1) − a 0 = 0,
i.e. a 1 = a 0
From equation (20), if c = 0,
a r+1 (r + 1)(3r + 1) − a r = 0,
i.e. a r+1 =
a r
(r + 1)(3r + 1)
r ≥ 0
Thus, when r = 1, a 2 =
a 1
(2 × 4)
=
a 0
(2 × 4)
since a 1 = a 0
when r = 2, a 3 =
a 2
(3 × 7)
=
a 0
(2 × 4)(3 × 7)
or
a 0
(2 × 3)(4 × 7)
when r = 3, a 4 =
a 3
(4 × 10)
=
a 0
(2 × 3 × 4)(4 × 7 × 10)
and so on.
From equation (16), the trial solution was:
y = x
c
{a 0 + a 1 x + a 2 x
2
+ a 3 x
3
+· · ·+ a r x
r
+ · · ·}
Substituting c = 0 and the above values of a 1 , a 2 , a 3 , …
into the trial solution gives:
y = x
0
a 0 + a 0 x +
a 0
(2 × 4)
x
2
+
a 0
(2 × 3)(4 × 7)
x
3
+
a 0
(2 × 3 × 4)(4 × 7 × 10)
x
4
+ · · ·
i.e. y = a 0
1 + x +
x 2
(2 × 4)
+
x 3
(2 × 3)(4 × 7)
+
x 4
(2 × 3 × 4)(4 × 7 × 10)
+ · · ·
(21)
(b) When c =
2
3
:
From equation (19), if c =
2
3
, a 1 (3)
5
3
− a 0 = 0, i.e.
a 1 =
a 0
5
From equation (20), if c =
2
3
a r+1
2
3
+ r + 1
(2 + 3r + 1) − a r = 0,
i.e. a r+1
r +
5
3
(3r + 3) − a r
= a r+1 (3r
2
+ 8r + 5) − a r = 0,
i.e. a r+1 =
a r
(r + 1)(3r + 5)
r ≥ 0
(iv) The sum of these three terms forms the left-hand
side of the equation. Since the right-hand side is
zero, the coefficients of each power of x can be
equated to zero.
For example, the coefficient of x c−1 is equated
to zero giving: 3a 0 c(c − 1) + a 0 c = 0
or a 0 c[3c − 3 + 1] = a 0 c(3c − 2) = 0
(18)
The coefficient of x c is equated to zero giving:
3a 1 c(c + 1) + a 1 (c + 1) − a 0 = 0
i.e. a 1 (3c
2
+ 3c + c + 1) − a 0
= a 1 (3c
2
+ 4c + 1) − a 0 = 0
or
a 1 (3c + 1)(c + 1)− a 0 = 0
(19)
In each of series (a), (b) and (c) an x c term is
involved, after which, a general relationship can
be obtained for x c+r , where r ≥ 0.
In series (a) and (b), terms in x c+r−1 are present;
replacing r by (r + 1) will give the corresponding
terms in x
c+r , which occurs in all three equations, i.e.
in series (a), 3a r+1 (c + r)(c + r + 1)x c+r
in series (b), a r+1 (c + r + 1)x c+r
in series (c), −a r x c+r
Equating the total coefficients of x c+r to zero
gives:
3a r+1 (c + r)(c + r + 1) + a r+1 (c + r + 1)
− a r = 0
which simplifies to:
a r+1 {(c + r + 1)(3c + 3r +1)} − a r = 0
(20)
Equation (18), which was formed from the coefficients of the lowest power of x, i.e. x
c−1 , is called
the indicial equation, from which, the value of
c is obtained. From equation (18), since a 0 = 0,
then c = 0 or c =
2
3
(a) When c = 0:
From equation (19), if c = 0, a 1 (1 × 1) − a 0 = 0,
i.e. a 1 = a 0
From equation (20), if c = 0,
a r+1 (r + 1)(3r + 1) − a r = 0,
i.e. a r+1 =
a r
(r + 1)(3r + 1)
r ≥ 0
Thus, when r = 1, a 2 =
a 1
(2 × 4)
=
a 0
(2 × 4)
since a 1 = a 0
when r = 2, a 3 =
a 2
(3 × 7)
=
a 0
(2 × 4)(3 × 7)
or
a 0
(2 × 3)(4 × 7)
when r = 3, a 4 =
a 3
(4 × 10)
=
a 0
(2 × 3 × 4)(4 × 7 × 10)
and so on.
From equation (16), the trial solution was:
y = x
c
{a 0 + a 1 x + a 2 x
2
+ a 3 x
3
+· · ·+ a r x
r
+ · · ·}
Substituting c = 0 and the above values of a 1 , a 2 , a 3 , …
into the trial solution gives:
y = x
0
a 0 + a 0 x +
a 0
(2 × 4)
x
2
+
a 0
(2 × 3)(4 × 7)
x
3
+
a 0
(2 × 3 × 4)(4 × 7 × 10)
x
4
+ · · ·
i.e. y = a 0
1 + x +
x 2
(2 × 4)
+
x 3
(2 × 3)(4 × 7)
+
x 4
(2 × 3 × 4)(4 × 7 × 10)
+ · · ·
(21)
(b) When c =
2
3
:
From equation (19), if c =
2
3
, a 1 (3)
5
3
− a 0 = 0, i.e.
a 1 =
a 0
5
From equation (20), if c =
2
3
a r+1
2
3
+ r + 1
(2 + 3r + 1) − a r = 0,
i.e. a r+1
r +
5
3
(3r + 3) − a r
= a r+1 (3r
2
+ 8r + 5) − a r = 0,
i.e. a r+1 =
a r
(r + 1)(3r + 5)
r ≥ 0