Page 145 - Advanced engineering mathematics

4.1 Power Series Solutions 125

Now equation (4.5) is
∞ ∞

n n
(n + 2)(n + 1)a n+2 x + a n−2 x = 0.
n=0 n=2
We can combine the terms for n ≥ 2 in one summation. This requires that we write the n = 0 and
n = 1 terms in the last equation separately, or else we lose terms:
∞

0 n
2a 2 x + 2(3)a 3 x + [(n + 2)(n + 1)a n+2 + a n−2 ]x = 0.
n=2
The left side can be zero for all x in some interval (−h,h) only if the coefficient of each power
of x is zero:
a 2 = a 3 = 0
and
(n + 2)(n + 1)a n+2 + a n−2 = 0for n ≥ 2.
The last equation gives us
1
a n+2 =− a n−2 for n = 2,3,··· . (4.6)
(n + 2)(n + 1)
This is a recurrence relation for the coefficients of the series solution, giving us a 4 in terms of
a 0 , a 5 in terms of a 1 , and so on. Recurrence relations always give a coefficient in terms of one
or more previous coefficients, allowing us to generate as many terms of the series solution as we
want. To illustrate, use n = 2 in equation (4.6) to obtain
1 1
a 4 =− a 0 =− a 0 .
(4)(3) 12
With n = 3,
1 1
a 5 =− a 1 =− a 1 .
(5)(4) 20
In turn, we obtain
1
a 6 =− a 2 = 0
(6)(5)
because a 2 = 0,
(7)(6)
a 7 =− = 0
a 3
because a 3 = 0,
1 1 1
a 8 =− a 4 = a 0 = a 0 ,
(8)(7) (56)(12) 672
1 1 1
a 9 =− a 5 = a 1 = a 1 ,
(9)(8) (72)(20) 1440
and so on. Thus far, we have the first few terms of the series solution about 0:
1
2 3 4
y(x) = a 0 + a 1 x + 0x + 0x − a 0 x
12
1 5 6 7 1 8 1 9
− a 1 x + 0x + 0x + x + x + ···
20 672 1440

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October 14, 2010 14:17 THM/NEIL Page-125 27410_04_ch04_p121-136