If $log y = log sin x - x^{2}$ then $y_{2} + 4 x y_{1} = \dots$

(4 x^{2} - 3) y

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- (4 x^{2} - 3) y

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- (4 x^{2} + 3) y

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(4 x^{2} + 3) y

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Solution

The correct option is C $- (4 x^{2} + 3) y$
Differentiating both sides w.r.t $x$
$\frac{1}{y} \frac{d y}{d x} = \frac{1}{s i n x} c o s x - 2 x$
$y_{1} = y (c o t x - 2 x)$
Differentiate again w.r.t $x$
$y_{2} = \frac{d y}{d x} (c o t x - 2 x) - y (c o s e c^{2} x + 2)$
$y_{2} = y (c o t^{2} x + 4 x^{2} - 4 x c o t x - c o s e c^{2} x - 2)$
$y_{2} = y (4 x^{2} - 4 x c o t x - 3)$
$y_{2} + 4 x y_{1} = - y (4 x^{2} + 3)$

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\frac{2 x}{3 y} = \frac{3 y}{4 z}

4 x^{2} + 9 y^{2} + 16 z^{2} = (2 x - 3 y + 4 z) (2 x + 3 y + 4 z)

(4 x^{2} + 9 y^{2}

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