How to calculate the two linearly independent solutions of the differential equation
x(x − 1)y'' + (3x−1)y'+ y = 0?
My approach was by using the Frobenius method as x=0 and x=1 are regular singular points and then finding the roots of the indicial equation and then by seeing the difference of the roots I got the solution to be 1/(1-x), and by using the method of reduction the second one turns out to be log_e(x)/(1-x), but the second solution does not seem to satisfy the equation when I put it back in the DE. Where am I going wrong here?