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Given the function $y = \sin(kt)$ and the differential equation $y'' + 36y = 0$, the question is to find constants that do not make the equation evaluate to zero. The answer choices are: (a) 6, (b) 36, -36, (c) -6, (d) 36, (e) -6, 6, (f) -36.

First, find the first and second derivative of $y$:

$$y' = k \cos(kt)$$

$$y'' = -k^2 \sin(kt)$$

Second, plug everything into the DE and factor:

$$-k^2 \sin(kt) + 36 \sin(kt) = 0$$

$$\sin(kt)(36-k^2) = 0$$ $$\sin(kt)(6+k)(6-k) = 0$$

So, $6$ and $-6$ would make the differential equation evaluate to zero. However, I chose answer choice (b) because plugging in $36$ or $-36$ would not make the equation evaluate to $0$, but this answer choice was wrong. Any help would be appreciated to point out where I went wrong. Thank you!

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    What is actually meant : "If $A\sin(kt)$ is not $0$ and satisfies the equation , what are the possible values of $k$ ?" – 2017-02-09

2 Answers 2

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Hint : The general solution of the differential equation is $$A\sin(-6x)+B\cos(-6x)$$

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    Given the choice in answers, e) is the most general answer, a) and c) are partial answers. – 2017-02-09
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    @LutzL Oops, you are right, because we have $-\sin(-6x)=\sin(6x)$. So, $e)$ is correct. – 2017-02-09
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I misread the question - it was asking which values of $k$ satisfied the differential equation, which would be 6 and -6.

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    Instead of writing an answer, please edit the question. – 2017-02-09