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Higher Order Derivatives

If x=sin t,...

Question

If $x = sin t, y = sin k t$ , then $(1 - x^{2}) y_{2} - x y_{1} =$

A

k y

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B

- k y

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C

k^{2} y

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D

- k^{2} y

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Solution

The correct option is C $- k^{2} y$
$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{k c o s k t}{c o s t}$
$\frac{d^{2} y}{d x^{2}} = \frac{d}{d t} (\frac{d y}{d x}) \frac{d t}{d x} = k (\frac{- k s i n k . t c o s t + s i n t . c o s k t}{c o s^{2} t}) \frac{1}{c o s t}$
$y_{2} c o s^{2} t = - k^{2} s i n k t + s i n t \frac{c o s k t}{c o s t} (- k)$
$(1 - x^{2}) y_{2} = - k^{2} y + x y_{1}$
$(1 - x^{2}) y_{2} - x y_{1} = - k^{2} y$

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Similar questions

x = sin t

y = sin k t

(1 - x^{2}) y_{2} - x y_{1}

Q. If x=sin t,y=sin kt,show that

(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + k^{2} y = 0

x = sin t, y = sin k t

(1 - x^{2}) y_{2} - x y_{1} + A y = 0

A

x = sin t, y = sin k t

,then show that

(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - k y = 0

x = sin t

y = sin p t,

(1 - x^{2}) y_{2} - x y_{1} + p^{2} y = 0

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