x=cos theta, y=sin 5 theta Rightarrow (1-x^2) | vedclass

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(D) Given, $x=\cos 	heta$ and $y=\sin 5 	heta$. First, find the derivatives with respect to $	heta$: $\frac{d x}{d 	heta}=-\sin 	heta$ and $\frac

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$-25$

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(D) Given, $x=\cos 	heta$ and $y=\sin 5 	heta$. First, find the derivatives with respect to $	heta$: $\frac{d x}{d 	heta}=-\sin 	heta$ and $\frac{d y}{d 	heta}=5 \cos 5 	heta$. Then, $\frac{d y}{d x} = \frac{d y / d 	heta}{d x / d 	heta} = \frac{5 \cos 5 	heta}{-\sin 	heta} = -\frac{5 \cos 5 	heta}{\sin 	heta}$. Now, find the second derivative $\frac{d^2 y}{d x^2} = \frac{d}{d 	heta} \left( \frac{d y}{d x} ight) \cdot \frac{d 	heta}{d x}$: $\frac{d^2 y}{d x^2} = \frac{d}{d 	heta} \left( -\frac{5 \cos 5 	heta}{\sin 	heta} ight) \cdot \left( -\frac{1}{\sin 	heta} ight) = -\left( \frac{\sin 	heta (25 \sin 5 	heta) - 5 \cos 5 	heta (\cos 	heta)}{\sin^2 	heta} ight) \cdot \left( -\frac{1}{\sin 	heta} ight) = \frac{25 \sin 	heta \sin 5 	heta - 5 \cos 	heta \cos 5 	heta}{\sin^3 	heta}$. Substitute these into the expression $(1-x^2) \frac{d^2 y}{d x^2} - x \frac{d y}{d x}$: $(1-\cos^2 	heta) \left( \frac{25 \sin 	heta \sin 5 	heta - 5 \cos 	heta \cos 5 	heta}{\sin^3 	heta} ight) - \cos 	heta \left( -\frac{5 \cos 5 	heta}{\sin 	heta} ight)$ $= \sin^2 	heta \left( \frac{25 \sin 	heta \sin 5 	heta - 5 \cos 	heta \cos 5 	heta}{\sin^3 	heta} ight) + \frac{5 \cos 	heta \cos 5 	heta}{\sin 	heta}$ $= \frac{25 \sin 	heta \sin 5 	heta - 5 \cos 	heta \cos 5 	heta}{\sin 	heta} + \frac{5 \cos 	heta \cos 5 	heta}{\sin 	heta}$ $= 25 \sin 5 	heta = 25 y$. Wait, re-evaluating the sign: $\frac{d}{d 	heta} (\frac{-5 \cos 5 	heta}{\sin 	heta}) = \frac{25 \sin 5 	heta \sin 	heta + 5 \cos 5 	heta \cos 	heta}{\sin^2 	heta}$. Thus, $(1-x^2) \frac{d^2 y}{d x^2} - x \frac{d y}{d x} = \sin^2 	heta (\frac{25 \sin 5 	heta \sin 	heta + 5 \cos 5 	heta \cos 	heta}{\sin^2 	heta}) - \cos 	heta (\frac{-5 \cos 5 	heta}{\sin 	heta}) = 25 \sin 5 	heta \sin 	heta + 5 \cos 5 	heta \cos 	heta + \frac{5 \cos 	heta \cos 5 	heta}{\sin 	heta} \dots$ Actually, the standard result for $y = \sin(n \cos^{-1} x)$ is $(1-x^2) y'' - x y' + n^2 y = 0$. Here $n=5$, so $(1-x^2) y'' - x y' = -25 y$.

$x=\cos \theta, y=\sin 5 \theta \Rightarrow (1-x^2) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}$ is equal to (in $y$)

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