Hi,

I am having a little trouble finding the two linearly independent Frobenius solutions of the following differential equation:

2xy''+(2x+1)y'+2y=0

I have got as far as finding out the recurrence relations:

note ak is a with subscript k and a(k-1) is a with subscript (k-1)

CASE 1:

ak=-(2a(k-1))/(2k-1)

In the solutions I have it says the above equation and then hence:

ak=(-2/(2k-1))(-2/(2k-3))...(-2/1)a0

where a0 is a with subscript 0

and also

CASE 2:

ak=(ak-1)/k

The solutions again have the above line and hence:

ak=(((-1)^k)/k!)a0

However I really do not know how in the solutions they have got from each ak equation to the hence part.

Any help appreciated.