Which solution for this DE is preferred?

The second one is certainly not a solution. There is a solution in terms of exponentials, but that is not it.

 

The first of your solutions can be arrived at from the second. I would say that the second is the fundamental.

 

No, the second is not correct.

 

No you don't. You get roots of +k and -k. Try plugging your second solution into the DE and see if it works. You'll see that it doesn't.

 

The characteristic equation for the D.E. you give, [itex]y''- k^2y= 0[/itex] is [itex]r^2- k= 0[/itex] which is equivalent to [itex]r^2= k^2[/itex] and has roots k and -k. You, apparently, miswrote the equation as [itex]y''- k^2y'= 0[/itex], which has characteristic equation [itex]r^2- k^2r= r(r- k^2)= 0[/itex] and has roots 0 and [itex]k^2[/itex].