Let p=y/x, then p'=dp/dx=y'/x-y/x^2.
y"+y'/x-y/x^2=0=y"+p' so y"=-p'.
Integrating: y'=-y/x.
So dy/dx=-y/x; dy/y=-dx/x and ln(y)=-ln(x)+C, ln(y)+ln(x)=C or xy=A, a constant (A=e^C).
y(1)=1 so A=1 and xy=1. y(r)=1/r=n.
CHECK
xy=1 so y=1/x=x^-1; y'=-x^-2=-1/x^2; y"=2x^-3.
So y"+y'/x-y/x^2=(2x^-3)-(x^-3)-(x^-3)=0. OK.