The D.E whose solution is $y = C_{1} cos (4 x + C_{2})$ is:

y_{3} - 16 y_{1} = y

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y_{1} - 4 y_{1} + 4 = 0

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y_{2} + 16 y = 0

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y_{2} = 16 y

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Solution

The correct option is D $y_{2} + 16 y = 0$
$y = C_{1} cos (4 x + C_{2})$
$y_{1} = - 4 C_{1} sin (4 x + C_{2})$
$y_{2} = - 16 C_{1} cos (4 x + C_{2})$
$\Rightarrow y_{2} = - 16 y$
$\Rightarrow y_{2} + 16 y = 0$

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