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C= l(l+1).

I know need to use the recursion formula I found to find the l = 0, 1, 2, and 3 solutions to the differential equation. Do I simply plug l in for n? If so, I get for l = 0, a2 = -Ca0/2. Is this the SOLUTION to the D.E. for

l = 0, or do I need to do something else?

Similarly, for l = 1, I get a3= a1 (2-C)/6.

Any help appreciated!