Question

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

• A. $(1-x^2)y_2+x{y}_1+p^{2}y=0$
• B. $(1-x^2)y_2+x{y}_1-p^{2}y=0$
• C. $(1+x^2)y_2-x{y}_1+p^{2}y=0$
• D. $(1-x^2)y_2-x{y}_1+p^{2}y=0$

Answer 1

The correct option is D $(1-x^2)y_2-x{y}_1+p^{2}y=0$

Given that
$x=\sin t,y=\cos pt$
$\frac{dx}{dt}=\cos t,\frac{dy}{dt}=-p\sin pt$
$\frac{dy}{dx}=-\frac{p\sin pt}{\cos t}$

$\Rightarrow y_1=\frac{-p\sqrt{1-y^2}}{\sqrt{1-x^2}}$

$\Rightarrow y_1\sqrt{1-x^2}=-p\sqrt{1-y^2}$

On squaring both sides, we get
$y_1^2(1-x^2)=p^2(1-y^2)$
Again differentiating
$2y_1{y}_2(1-x^2)-2x{y}_1^2=-2y{y}_1{p}^2$
or $(1-x^2)y_2-x{y}_1+p^{2}y=0$