# xy''+(1-x)y'+y=0

• August 26th 2005, 06:39 AM
Ene Dene
xy''+(1-x)y'+y=0
I'm supposed to solve differential equation:

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

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

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.
• September 6th 2005, 06:17 AM
Ene Dene
I know the solution now... so no need to answer now, this one is correct.