Question

If $x=\sin t\mathrm{and}y=\sin pt$, then the value of $(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}+p^{2}y=$

• A. $0$
• B. $1$
• C. $-1$
• D. $\sqrt{2}$

Answer 1

The correct option is A

$0$

Explanation for the correct option.

Step 1: Find First derivative.

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

Now,

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{dy}{dt}\times \frac{dt}{dx}\\ \Rightarrow \frac{dy}{dx}& =& \frac{p\cos pt}{\cos t}\\ & \Rightarrow & \frac{dy}{dx}\left(\cos t\right)=p\cos pt\end{array}$

By squaring both sides we get

$\begin{array}{rcl}{\left(\frac{dy}{dx}\right)}^2\left({\cos }^{2}t\right)& =& p^2{\cos }^{2}pt\\ & \Rightarrow & {\left(\frac{dy}{dx}\right)}^2\left(1-{\sin }^{2}t\right)=p^2\left(1-{\sin }^{2}pt\right)\\ \Rightarrow {\left(\frac{dy}{dx}\right)}^2\left(1-x^2\right)& =& p^2\left(1-y^2\right)\end{array}$

Step 2: Find Second derivative.

$\begin{array}{rcl}\left(1-x^2\right)2\left(\frac{dy}{dx}\right)\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2\left(-2x\right)& =& -2p^{2}y\frac{dy}{dx}\\ \Rightarrow 2\left(1-x^2\right)\left(\frac{dy}{dx}\right)\frac{d^{2}y}{d{x}^2}-2x{\left(\frac{dy}{dx}\right)}^2& =& -2p^{2}y\frac{dy}{dx}\\ \Rightarrow \left(1-x^2\right)\frac{d^{2}y}{d{x}^2}-x\left(\frac{dy}{dx}\right)& =& -p^{2}y\\ \Rightarrow \left(1-x^2\right)\frac{d^{2}y}{d{x}^2}-x\left(\frac{dy}{dx}\right)+p^{2}y& =& 0\end{array}$

Hence, option A is correct.