If $x =$ cost and $y = sin 4 t$ then $(1 - x^{2}) y_{2} - x y_{1} =$

4 y

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- 4 y

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16 y

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- 16 y

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Solution

The correct option is D $- 16 y$
$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{4 c o s 4 t}{- s i n t}$
$\frac{d}{d t} (\frac{d y}{d x}) \frac{d t}{d x} = \frac{d^{2} y}{d x^{2}}$

$\frac{d}{d t} (\frac{- 4 c o s 4 t}{s i n t}) \frac{1}{- s i n t} = \frac{d^{2} y}{d x^{2}}$

$\frac{+ 4 (- 4 s i n 4 t s i n t - c o s 4 t c o s t)}{(s i n^{2} t)} \frac{1}{s i n t} = y_{2}$

$4 (- 4 s i n 4 t - \frac{c o s 4 t}{s i n t} c o s t) = s i n^{2} t y_{2}$
$- 16 s i n 4 t + c o s t y 1 = s i n^{2} t y_{2}$
$- 16 y = (1 - x^{2}) y_{2} - x y_{1}$

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