Page 56 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

46 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
4.1.2 Ordinary Points and Series Solutions
If the point x 0 is an ordinary point the dependent variable has a solution in the neighborhood
of x 0 of the form
∞
n
u(x) = c n (x − x 0 ) (4.3)
n=0
We now illustrate the solution method with two examples.
Example 4.1. Find a series solution in the form of Eq. (4.3) about the point x = 0ofthe
differential equation

2
u + x u = 0 (4.4)

The point x = 0 is an ordinary point so at least near x = 0 there is a solution in the form of
the above series. Differentiating (4.3) twice and inserting it into (4.4)
∞

n−1
u = nc n x
n=0
∞

n−2
u = n(n − 1)c n x
n=0
∞ ∞
n−2 n+2
n(n − 1)c n x + x c n = 0 (4.5)
n=0 n=0
Note that the first term in the u series is zero while the first two terms in the u series are zero.

We can shift the indices in both summations so that the power of x is the same in both series
by setting n − 2 = m in the first series.
∞ ∞ ∞
n−2 m m
n(n − 1)c n x = (m + 2)(m + 1)c m+2 x = (m + 2)(m + 1)c m+2 x (4.6)
n=0 m=−2 m=0
Noting that m is a “dummy variable” and that the first two terms in the series are zero the series
can be written as

∞
n
(n + 2)(n + 1)c n+2 x (4.7)
n=0
In a similar way we can write the second term as

∞ ∞
n+2 n
c n x = c n−2 x (4.8)
n=0 n=2