1. ## simple differential equation

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?

2. You are correct that $\left\{\frac{1}{4},\,0,\,-\frac{1}{4}\right\}$ are all values of $k$ for which

$16y''=-y.$

Now, suppose $\sin\,(kt)=0$ for some other value of $k$, i.e., $k\ne 0$. What happens as $t$ varies?

3. 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.

4. You're looking for a k that satisfies the equation for all $t$. Your suggestion of $k = \frac{n\pi}{t}$ requires k to be a function of t, whereas in this problem you're looking for a constant value for k.