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Math Help - simple differential equation

  1. #1
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    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?
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  2. #2
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    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?
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  3. #3
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    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.
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  4. #4
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    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.
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