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?