
xy''+(1x)y'+y=0
I'm supposed to solve differential equation:
$\displaystyle xy''(x)+(1x)y'(x)+y(x)=0$
by Frobenius method.
The first solution is $\displaystyle y1=c(1x)$, where $\displaystyle c$ is a constant. I am not sure if the second solution is correct. I got:
$\displaystyle y2=c(1x)ln(x)+c(3x\frac{x^2}{4}\frac{x^3}{36}\frac{2x^4}{576}...)$
or if I put it in the sum:
$\displaystyle y2=c(1x)ln(x)+c(3x\sum_{n=0}^{\infty}\frac{n!}{((n+2)!)^2}x^{(n+2)})$
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.

I know the solution now... so no need to answer now, this one is correct.