Are they both correct? Would the first or second solution be preferred?
The second one is certainly not a solution. There is a solution in terms of exponentials, but that is not it.
Why is the second one not a solution? If you convert the original DE into an auxiliary equation, you will get roots: m1=0 and m2=k2
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.
That's weird: http://www.wolframalpha.com/input/?i=m^2-k^2m=0
Going to try plugging it in.
Edit: Oh wow. You're right. I'm an idiot. Lol
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].
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