If x=cosθ, y=sin5θ then 1-x2d2ydx2-xdydx

x=cosθ, y=sin5θ dxdθ=-sinθ dydθ=5cos5θ dydx=-5cos5θsin2θ dydx2=25 cos25θsin2θ =25(1-y2)1-x2 1-x2 dydx2=251-y2 1-x22dydx d2ydx2-2xdydx2=25-2y dydx 1-x2 d2ydx2-xdydx=-25y

If x=cosθ, y=sin5θ then 1-x2d2ydx2-xdydx

x=cosθ, y=sin5θ dxdθ=-sinθ dydθ=5cos5θ dydx=-5cos5θsin2θ dydx2=25 cos25θsin2θ =25(1-y2)1-x2 1-x2 dydx2=251-y2 1-x22dydx d2ydx2-2xdydx2=25-2y dydx 1-x2 d2ydx2-xdydx=-25y

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If $x = cosθ, y = \sin 5 θ$ then $(1 - x^{2}) \frac{d^{2} y}{{dx}^{2}} - x \frac{dy}{dx}$

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$- 5 y$

$5 y$

$25 y$

$- 25 y$

answer is D.

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Detailed Solution

$x = cosθ, y = \sin 5 θ \frac{dx}{dθ} = - sinθ \frac{dy}{dθ} = 5 \cos 5 θ \frac{dy}{dx} = \frac{- 5 \cos 5 θ}{\sin^{2} θ} {(\frac{dy}{dx})}^{2} = 25 \frac{c {os}^{2} 5 θ}{\sin^{2} θ} = \frac{25 (1 - y^{2})}{1 - x^{2}} (1 - x^{2}) {(\frac{dy}{dx})}^{2} = 25 (1 - y^{2}) (1 - x^{2}) 2 (\frac{dy}{dx}) \frac{d^{2} y}{{dx}^{2}} - 2 x {(\frac{dy}{dx})}^{2} = 25 (- 2 y \frac{dy}{dx}) (1 - x^{2}) \frac{d^{2} y}{{dx}^{2}} - x \frac{dy}{dx} = - 25 y$