frobenius method

asdf1

for the frobenius method, why do you suppose that a0=1?

HallsofIvy

You don't. You assume that it is non-zero which is just a choice of "c".

The idea of Frobenius method is to write a solution in the form
[tex]\Sigma_{n=0}{\infnty}a_n x^{n+c}[/tex]
where c, while a constant, is not necessarily a positive integer.
If I, choosing c to be, say, c[sub]0[/sup], I find that a₀= 0 but a₁ is not, I could just take c= c₀+ 1 instead.

asdf1

thank you very much!!! :)