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

$(1-x^2)y_2-x{y}_1+p^{2}y=0$

Explanation for the correct option.

Step 1: Find $y_1$.

$\begin{array}{rcl}x& =& \sin t\\ & \Rightarrow & \frac{dx}{dt}=\cos t\\ y& =& \cos pt\\ & \Rightarrow & \frac{dy}{dt}=-p\sin pt\end{array}$

$\begin{array}{rcl}{y}_1& =& \frac{dy}{dx}\\ & =& \frac{dy}{dt}\times \frac{dt}{dx}\\ & =& \frac{-p\sin pt}{\cos t}\end{array}$

We have, $x=\sin t$, so $\cos t=\sqrt{1-x^2}$ and $y=\cos pt$, so $\sin pt=\sqrt{1-y^2}$.

$\begin{array}{rcl}\Rightarrow y_1& =& \frac{-p\sqrt{1-y^2}}{\sqrt{1-x^2}}\\ & \Rightarrow & y_1\sqrt{1-x^2}=-p\sqrt{1-y^2}\end{array}$

By squaring both sides we get

${\left(y_1\right)}^2\left(1-x^2\right)=p^2\left(1-y^2\right)....\left(1\right)$

Step 2: Find $y_2$.

$y_2=\frac{d^{2}y}{d{x}^2}$

By differentiating $\left(1\right)$ with respect to $x$, we get

$\begin{array}{rcl}-2x{\left(y_1\right)}^2+\left(1-x^2\right)2y_1{y}_2& =& p^2\left(-2y{y}_1\right)\\ \Rightarrow 2y_1{y}_2\left(1-x^2\right)-2x{\left(y_1\right)}^2& =& -2y{y}_1{p}^2\\ & \Rightarrow & y_2\left(1-x^2\right)-x\left(y_1\right)+p^{2}y=0\end{array}$

Hence, option D is correct.