You are correct that are all values of for which
Now, suppose for some other value of , i.e., . What happens as varies?
For what values of k does the function y = sin(kt) satisfy the differential equation 16y'' = -y? I have the k = 1/4 and -1/4 solutions, but don't know what the general solution of sin(kt) = 0 would be. I guess i can't say that sin(kt)= 0 iff kt = arcsin(0) = 0, therefore k = 0/t = 0? That was my first guess but it's wrong. Then i tried, sin(kt) = 0 iff kt = npi for nonneg integer n. then k = n(pi)/t, but i guess that's also wrong?
Well, i thought no matter what, if sin(kt) = 0, then kt = n(pi) for nonnegative n and so, k= n(pi)/t and the nonzero k's are those where n is positive. So included in the solution set {nPi/t, 1/4, -1/4} which also covers the case k= 0. Maybe i'm thinking about it the wrong way though.