Mathematical Methods for Physicists: A concise introduction - Site Map

ORDINARY DIFFERENTIAL EQUATIONS
Case 1 Distinct roots not di€ering by an integer
This is the simplest case. Let r 1 and r 2 be the roots of the indicial equation (2.29).
If we insert r ˆ r 1 into the recurrence relation and determine the coecients
a 1 , a 2 ; ... successively, as before, then we obtain a solution
y 1 …x† ˆx r 1
…a 0 ‡ a 1 x ‡ a 2 x 2 ‡†:
Similarly, by inserting the second root r ˆ r 2 into the recurrence relation, we will
obtain a second solution
y 2 …x† ˆx r 2
…a 0
* ‡ a 1
*x ‡ a 2
*x 2 ‡†:
Linear independence of y 1 and y 2 follows from the fact that y 1 =y 2 is not constant
because r 1 r 2 is not an integer.
Case 2 Double roots
The indicial equation (2.29) has a double root r if, and only if,
…g 0 1† 2 4h 0 ˆ 0, and then r ˆ…1 g 0 †=2. We may determine a ®rst solution
y 1 …x† ˆx r …a 0 ‡ a 1 x ‡ a 2 x 2 ‡† r ˆ 1g
0
…2:30†
2
as before. To ®nd another solution we may apply the method of variation of
parameters, that is, we replace constant c in the solution cy 1 …x† by a function
u…x† to be determined, such that
y 2 …x† ˆu…x†y 1 …x†
…2:31†
is a solution of Eq. (2.27). Inserting y 2 and the derivatives
y2 0 ˆ u 0 y 1 ‡ uy1 0 y2 00 ˆ u 00 y 1 ‡ 2u 0 y1 0 ‡ uy1
00
into the di€erential equation (2.27) we obtain
x 2 …u 00 y 1 ‡ 2u 0 y1 0 ‡ uy1 00 †‡xg…u 0 y 1 ‡ uy1†‡huy 0 1 ˆ 0
or
x 2 y 1 u 00 ‡ 2x 2 y1u 0 0 ‡ xgy 1 u 0 ‡…x 2 y1 00 ‡ xgy1 0 ‡ hy 1 †u ˆ 0:
Since y 1 is a solution of Eq. (2.27), the quantity inside the bracket vanishes; and
the last equation reduces to
x 2 y 1 u 00 ‡ 2x 2 y1u 0 0 ‡ xgy 1 u 0 ˆ 0:
Dividing by x 2 y 1 and inserting the power series for g we obtain
u 00 ‡ 2 y 1
0 ‡ g
0
y 1 x ‡ u 0 ˆ 0:
Here and in the following the dots designate terms which are constants or involve
positive powers of x. Now from Eq. (2.30) it follows that
y1
0 ˆ xr1 ‰ra 0 ‡…r ‡ 1†a 1 x ‡Š
y 1 x r ˆ 1 ra 0 ‡…r ‡ 1†a 1 x ‡
ˆ r
‰a 0 ‡ a 1 x ‡Š x a 0 ‡ a 1 x ‡ x ‡:
90