Thanks Torquil.
U know the original equation was,
Y''+xy+n^2y+exp(-x)y=0,
when i used the transformation x=exp(-z/2), i got
z^2y''+zy'+(z^2-n^2)y+ylog(z)=0,
which i already post,
now according to u if z=log(u), than i am sure i will get my original equation (1), with exponential.