Question

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

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

Answer 1

Answer: (A)

Let $x = \sin\: t$ and $y = \sin\: pt$ ...(1)
Differentiate both sides w.r.t 't'
$\therefore \:\: \frac{dx}{dt} = \cos t$ and $\frac{dy}{dt} = p \cos pt $
$\therefore \:\:\: \frac{dy}{dx} =\frac{dy}{dt} \times\frac{dt}{dx}= \frac{p \cos pt}{\cos t} $
$= \frac{p\sqrt{1 -\sin^{2} pt}}{\sqrt{1-\sin^{2} t}} \left( \because \:\: \sin^{2} \theta +\cos^{2} \theta =1\right)$
$ \Rightarrow \frac{dy}{dx} =p \frac{\sqrt{\left(1 -y^{2}\right)}}{\sqrt{\left(1-x^{2}\right)}} $ (from (1))
On squaring both sides, we get
$\left(\frac{dy}{dx}\right)^{2} =p^{2} \frac{\left(1 -y^{2}\right)}{\left(1 -x^{2}\right)}$
$ \Rightarrow \left(\frac{dy}{dx}\right)^{2} \left(1-x^{2}\right)=p^{2} \left(1-y^{2}\right)$
Differentiating this equation, we get
$2\left(\frac{dy}{dx}\right)\left(\frac{d^{2}y}{dx^{2}}\right)\left(1-x^{2}\right)+\left(\frac{dy}{dx}\right)^{2}\left(-2x\right) =p^{2}\left(-2y\right) \frac{dy}{dx}$
Cancelled out 2. $\frac{dy}{dx}$ on both the sides,
$\left(1-x^{2}\right) \frac{d^{2}y}{dx} -x \frac{dy}{dx} +p^{2}y =0$