I wish to solve:
y'' x^2 - xy' + (1-tx) y = 0 where t is a constant.
so if we try a Frobenius series
we arrive at the indicial equation
a_n ( n^2 + k^2 + 2nk - 2n - 2k +1) = t a_(n-1)
if we set n = 0
we have
a_0 (k^2 - 2k +1 ) = a _ (n-1)
so k = 1 since a_0 is non zero
so we have a series solution y_1 = a_0 [ x + x^2 /4 + ... (x^n) / (n!)^2 + ... ]
but what about the other solution, my notes says
y_2 = y_1 ln x + sigma b_n x^(n+ k_2)
but we how do we deal with the double root and can b_n be determined?