Question

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

• A. $(1-x^2)y_2 + xy_1 + p^2y = 0$
• B. $(1-x^2)y_2 + xy_1 - p^2y = 0$
• C. $(1+x^2)y_2 - xy_1 + p^2y = 0$
• D. $(1-x^2)y_2 - xy_1 + p^2y = 0$

Answer 1

Answer: (D)

Given that, $x=\sin t, y=\cos \,p t$
$\frac{d x}{d t} =\cos t, \frac{d y}{d t}=-p \sin p t $
$\frac{d y}{d x} =-\frac{p \sin p t}{\cos t} $
$\Rightarrow \, y_{1} =\frac{-p \sqrt{1-y^{2}}}{\sqrt{1-x^{2}}} $
$\Rightarrow \,y_{1} \sqrt{1-x^{2}} =-p \sqrt{1-y^{2}}$
On squaring both sides, we get
$y_{1}^{2}\left(1-x^{2}\right)=p^{2}\left(1-y^{2}\right)$
Again differentiating,
$2 y_{1} y_{2}\left(1-x^{2}\right)-2 x y_{1}^{2}=-2 y y_{1} p^{2}$
or $\left(1-x^{2}\right) y_{2}-x y_{1}+p^{2} y=0$