Question

If $x=\sin t$ and $y=\sin pt$, then $(1-x^2)\frac{d^{2}x}{d{x}^2}-x\frac{dy}{dx}$ is equal to

• A. $-y$
• B. $y$
• C. $py$
• D. $-p^{2}y$

Answer 1

The correct option is D $-p^{2}y$

For $x=\sin t$ and $y=\sin pt$.

$\Rightarrow \frac{dx}{dt}=\cos t$.....(1) and $\frac{dy}{dt}=p\cos pt$.......(2)

Now (2) $÷$ (1) gives

$\frac{dy}{dx}=p\frac{\cos pt}{\cos t}$

And $x\frac{dy}{dx}=p\frac{\cos pt.\sin t}{\cos t}$.......(3).

Now $\frac{d^{2}y}{d{x}^2}$

$=\frac{d}{dt}\left(\frac{dy}{dx}\right).\frac{dt}{dx}$

$=p\frac{-p\cos t.\sin pt+\cos t.\sin t}{{\cos }^{3}t}$

Again $(1-x^2)\frac{d^2}{d{x}^2}=$$p\frac{-p\cos t.\sin pt+\cos pt.\sin t}{\cos t}$........(4).

Using (3) and (4) we get,

$(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=-p^2\sin pt=-p^{2}y$.