If $x = \sin t, y = \cos p t$ then,

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

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

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

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

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Solution

The correct option is D
$(1 - x^{2}) y_{2} - x y_{1} + p^{2} y = 0$

Explanation for the correct answer:
Differential equation form:
Given , $x = \sin t, y = \cos p t$ $. . . (i)$
$\Rightarrow$ $\cos t = \sqrt{1 - \sin^{2} t}$
$\Rightarrow$ $\cos t = \sqrt{1 - x^{2}}$
Differentiating $x = \sin t$ with respect to $t$ we get
$\frac{d x}{d t} = \cos t$ $. . . (i i)$
$y = \cos p t$ $. . . (i i i)$
$\Rightarrow$ $\sin p t = \sqrt{1 - \cos^{2} p t}$
$\Rightarrow$ $\sin p t = \sqrt{1 - y^{2}}$
Differentiating $(i i i)$ with respect to $t$ we get
$\frac{d y}{d t} = - p \times \sin p t$ $. . . (i v)$
Using chain rule to find first order derivative of $y$ with respect to $x$ we get
$y_{1} = \frac{d y}{d x} = \frac{d y}{d t} \times \frac{d t}{d x}$
Substituting the required values we get
$\Rightarrow$ $y_{1} = \frac{- p \sin p t}{\cos t}$
$\Rightarrow$ $y_{1} = \frac{- p \sqrt{1 - y^{2}}}{\sqrt{1 - x^{2}}}$
$\Rightarrow$ $y_{1} \sqrt{1 - x^{2}} = - p \sqrt{1 - y^{2}}$
Squaring both the sides we get
$\Rightarrow$ ${(y_{1})}^{2} (1 - x^{2}) = p^{2} (1 - y^{2})$ $. . . (v)$
Differentiating $(v)$ with respect to $x$ we get
$\Rightarrow$ $2 y_{1} y_{2} (1 - x^{2}) + {y_{1}}^{2} (- 2 x) = p^{2} (- 2 y y_{1})$
$\Rightarrow$ $y_{2} (1 - x^{2}) - y_{1} x = - p^{2} y$
$\Rightarrow$ $(1 - x^{2}) y_{2} - x y_{1} + p^{2} y = 0$
Hence, option (D) is the correct answer.

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