Page 58 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

48 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
Shifting the indices as before

∞ ∞ ∞

n n n 2
(n + 2)(n + 1)c n+2 x + nc n x + c n x − x = 0 (4.15)
n=0 n=0 n=0
n
Once again, each of the coefficients of x must be zero.
Setting n = 0, we see that

n = 0:2c 2 + c 0 = 0, c 2 =−c 0 /2 (4.16)
n = 1:6c 3 + 2c 1 = 0, c 3 =−c 1 /3
n = 2:12c 4 + 3c 2 − 1 = 0, c 4 = (1 + 3c 0 /2)/12
c n
n > 2: c n+2 =
n + 2
The last of these is called a recurrence formula.
Thus,

6
4
2
u = c 0 (1 − x /2 + x /8 − x /(8)(6) + ··· )
7
5
3
+ c 1 (x − x /3 + x /(3)(5) − x /(3)(5)(7) + ··· )
2
4
+ x (1/12 − x /(12)(6) + ··· ) (4.17)
Note that the series on the third line of (4.17) is the particular solution of (4.13). The constants
c 0 and c 1 are to be evaluated using the boundary conditions.
4.1.3 Lessons: Finding Series Solutions for Differential Equations
with Ordinary Points
If x 0 is an ordinary point assume a solution in the form of Eq. (4.3) and substitute into
the differential equation. Then equate the coefficients of equal powers of x. This will give a
recurrence formula from which two series may be obtained in terms of two arbitrary constants.
These may be evaluated by using the two boundary conditions.

Problems
1. The differential equation

u + xu + xu = x

has ordinary points everywhere. Find a series solution near x = 0.
2. Find a series solution of the differential equation
2
u + (1 + x )u = x

near x = 0 and identify the particular solution.