I'm supposed to solve differential equation:

$\displaystyle xy''(x)+(1-x)y'(x)+y(x)=0$

by Frobenius method.

The first solution is $\displaystyle y1=c(1-x)$, where $\displaystyle c$ is a constant. I am not sure if the second solution is correct. I got:

$\displaystyle y2=c(1-x)ln(x)+c(3x-\frac{x^2}{4}-\frac{x^3}{36}-\frac{2x^4}{576}-...)$

or if I put it in the sum:

$\displaystyle y2=c(1-x)ln(x)+c(3x-\sum_{n=0}^{\infty}\frac{n!}{((n+2)!)^2}x^{(n+2)})$

So, please if someone could check and tell me if this is OK. If not, I would appreciate if you could write the procedure how you got the second solution different from mine.