Module 7: Multistep Methods
  Lecture 24: Special 2nd order equations(Contd.)
 

 

Using integration by parts, we get

(7.45)

or

This gives

By comparing coefficients, we get the recurrence solution

The numerical values of for certain values of m are given in the following table:

m

0

1

2

3

4 5 6

1

-1

0