thanks for your answer Mr. Fantastic...ill show ya what i did cuz i have a mistake and i don't know where it is...so let's see:

first of all i took the first solution and i evaluated it in the equation and yes it was a solution (i had already done it before but i didn't do it well)

so then i did grade reduction to get the second solution which in the exercise said it had to be:

$\displaystyle y_2= e^x\int x^Ne^{-x}dx$

well so this is what i got:

$\displaystyle y_2= e^xv(x)$ where v(x) is my unknown solution

$\displaystyle y_2'=e^xv(x)+e^xv'(x)$

$\displaystyle y_2'=e^xv(x)+2e^xv'(x)+e^xv''(x)$

then replacing on the diff eq. (which was $\displaystyle xy''-(x+N)y'+Ny=0$ with N>0)

$\displaystyle x(e^xv(x)+2e^xv'(x)+e^xv''(x))-(x+N)(e^xv(x)+e^xv'(x))+N(e^xv(x))=0$

some sums and products later...

$\displaystyle v''xe^x-v'e^x(x+N)=0$

Mr F says: I think you'll find this should be $\displaystyle {\color{red}v'' x e^x + v' e^x (x-N) = 0}$. And you can divide out the common factor of e^x.
then i suppouse that $\displaystyle u=v'(x)$ so the new equation is:

$\displaystyle u'xe^x-ue^x(x+N)=0$ which is a separable diff. equation

here is the problem i think...

$\displaystyle u'=u((x+N)/x)$

$\displaystyle du/dx=u((x+N)/x)$

$\displaystyle du/u=((x+N)/x)dx $

(while i was writing this i noticed my mistake, i was not ok because i put $\displaystyle du/u=dx/((x+N)/x))$ let's c if i'm right...)

$\displaystyle ln u=x+ln x^N$

$\displaystyle u=e^xx^N$

remember that u=v' so:

$\displaystyle v=\int e^xx^Ndx$ and finally:

$\displaystyle y_2=e^x\int e^xx^Ndx$ but i`m still missing the minus where do i get it (omg it worked pretty good)

and the other point i care is that i gotta prove that this second solution is a polynomial of N grade...