Explanation

(d)
Given, $x = \sin{t\ ,\ \ y = \cos{pt}}\$
$\ \ \frac{dx}{dt} = \cos t,\ \ \ \frac{dy}{dt} = - p\sin{pt}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$
$\therefore\ \ \ \ \frac{dy}{dx} = \ - \frac{p\sin{pt}}{\cos t}$
$\Rightarrow \ y_{1} =$ $\frac{- p\sqrt{1 - y^{2}}}{\sqrt{1 - x^{2}}}$
$\Rightarrow \ \ \ y_{1}\sqrt{1 - x^{2}} = - p\sqrt{1 - y^{2}}$
$\Rightarrow \ \ y_{1}^{2}\left( 1 - x^{2} \right) = p^{2}\left( 1 - y^{2} \right)\$
$\Rightarrow \ 2y_{1}y_{2}\left( 1 - x^{2} \right) - 2xy_{1}^{2} = \ - 2yy_{1}p^{2}\ \lbrack$differentiate]
$\Rightarrow \ \left( 1 - x^{2} \right)y_{2} - xy_{1} + p^{2}y = 0$