$y = a cos e c (b - x)$ , then $y y_{2} - 2 y_{1}^{2} =$

- y^{2}

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- y

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y^{2}

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y

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Solution

The correct option is B $y^{2}$
$y^{‘} = - a c o s e c (b - x) c o t (b - x) (- 1)$
$y^{‘} = y c o t (b - x)$
$y^{‘ ‘} = y^{‘} c o t (b - x) + y c o s e c^{2} (b - x)$
$y^{‘ ‘} = \frac{y^{12}}{y} + y (1 + \frac{y^{2}}{y^{2}})$

$y y^{‘ ‘} = 2 (y^{'})^{2} + y^{2}$

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y=acosec(b−x), then yy2−2y21=

The correct option is B y2y‘=−acosec(b−x)cot(b−x)(−1) y‘=ycot(b−x) y‘‘=y‘cot(b−x)+y cosec2(b−x) y‘‘=y12y+y(1+y2y2) yy‘‘=2(y′)2+y2

$y = a cos e c (b - x)$ , then $y y_{2} - 2 y_{1}^{2} =$

The correct option is B $y^{2}$
$y^{‘} = - a c o s e c (b - x) c o t (b - x) (- 1)$
$y^{‘} = y c o t (b - x)$
$y^{‘ ‘} = y^{‘} c o t (b - x) + y c o s e c^{2} (b - x)$
$y^{‘ ‘} = \frac{y^{12}}{y} + y (1 + \frac{y^{2}}{y^{2}})$

$y y^{‘ ‘} = 2 (y^{'})^{2} + y^{2}$