Question

If $x=\sin t$ and $y=\sin pt$, then the value of $\left(1-x^2\right)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}+p^{2}y$ 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
$2\left(\frac{dy}{dx}\right)\left(\frac{d^{2}y}{d{x}^2}\right)\left(1-x^2\right)+{\left(\frac{dy}{dx}\right)}^2\left(-2x\right)=p^2\left(-2y\right)\frac{dy}{dx}$
Cancelled out 2. $\frac{dy}{dx}$ on both the sides,
$\left(1-x^2\right)\frac{d^{2}y}{dx}-x\frac{dy}{dx}+p^{2}y=0$