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.