Page 57 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS 47
We now have

∞ ∞

n n
(n + 2)(n + 1)c n+2 x + c n−2 x = 0 (4.9)
n=0 n=2
which can be written as

∞
n
2c 2 + 6c 3 x + [(n + 2)(n + 1)c n+2 + c n−2 ]x = 0 (4.10)
n=2
n
Each coefficient of x must be zero in order to satisfy Eq. (4.10). Thus c 2 and c 3 must be zero
and

c n+2 =−c n−2 /(n + 2)(n + 1) (4.11)

while c 0 and c 1 remain arbitrary.
Setting n = 2, we find that c 4 =−c 0 /12 and setting n = 3, c 5 =−c 1 /20. Since c 2 and
c 3 are zero, so are c 6 , c 7 , c 10 , c 11 , etc. Also, c 8 =−c 4 /(8)(7) = c 0 /(4)(3)(8)(7) and

c 9 =−c 5 /(9)(8) = c 1 /(5)(4)(9)(8).

The first few terms of the series are

4
5
9
6
u(x) = c 0 (1 − x /12 + x /672 + ··· ) + c 1 (1 − x /20 + x /1440 + ··· ) (4.12)
The values of c 0 and c 1 may be found from appropriate boundary conditions. These are both
alternating sign series with each term smaller than the previous term at least for x ≤ 1 and it is
therefore convergent at least under these conditions.
The constants c 0 and c 1 can be determined from boundary conditions. For example if
u(0) = 0, c 0 + c 1 = 0, so c 1 =−c 0 .If u(1) = 1,

c 0 [−1/12 + 1/20 + 1/672 − 1/1440 + ··· ] = 1

Example 4.2. Find a series solution in the form of Eq. (4.3) of the differential equation

u + xu + u = x 2 (4.13)

valid near x = 0.
Assuming a solution in the form of (4.3), differentiating and inserting into (4.13),

∞ ∞ ∞
n−2 n n 2
(n − 1)nc n x + nc n x + c n x − x = 0 (4.14)
n=0 n=0 n=0