Page 147 - Advanced engineering mathematics

4.2 Frobenius Solutions 127

and
R(x) 3
2
(x − 2) =
P(x) x
are both analytic at 2, so 2 is a regular singular point of this differential equation.

We will not treat the case of an irregular singular point. If equation (4.7) has a regular
singular point at x 0 , there may be no power series solution about x 0 , but there will be a Frobenius
series solution, which has the form
∞

y(x) = c n (x − x 0 ) n+r
n=0
with c 0 = 0. We must solve for the coefficients c n and a number r to make this series a solution.
We will look at an example to get some feeling for how this works and then examine the method
more critically.

EXAMPLE 4.4
Zero is a regular singular point of
2

x y + 5xy + (x + 4)y = 0.

∞ n+r

Substitute y = c n x to obtain
n=0
∞ ∞

(n +r)(n +r − 1)c n x n+r−2 + 5(n +r)c n x n+r
n=0 n=0
∞ ∞

+ c n x n+r+1 + 4c n x n+r = 0.
n=0 n=0
r
Notice that the n = 0 term in the proposed series solution is c 0 x , which is not constant if c 0 = 0,
so the series for the derivatives begins with n = 0 (unlike what we saw with power series). Shift
indices in the third summation to write this equation as
∞ ∞
n+r−2 n+r
(n +r)(n +r − 1)c n x + 5(n +r)c n x
n=0 n=0
∞ ∞

+ c n−1 x n+r + 4c n x n+r = 0.
n=1 n=0
Combine terms to write
[r(r − 1) + 5r + 4]c 0 x r
∞

+ [(n +r)(n +r − 1)c n + 5(n +r)c n + c n−1 + 4c n ]x n+r = 0.
n=1
r
Since we require that c 0 = 0, the coefficient of x is zero only if
r(r − 1) + 5r + 4 = 0.
This is called the indicial equation and is used to solve for r, obtaining the repeated root r =−2.
Set the coefficient of x n+r in the series equal to zero to obtain
(n +r)(n +r − 1)c n + 5(n +r)c n + c n−1 + 4c n = 0

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October 14, 2010 14:17 THM/NEIL Page-127 27410_04_ch04_p121-136