Question

$(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}+y=0$ it being given that $x=\cos 0$ then,

• A. $\frac{d^{2}y}{d{0}^2}+y=0$
• B. $\frac{d^{4}y}{d{0}^4}+y=0$
• C. $\frac{d^{3}y}{d{0}^3}+y=0$
• D. $\frac{d^{2}y}{d{0}^2}-y=0$

Answer 1

The correct option is A $\frac{d^{2}y}{d{0}^2}+y=0$

Given $x=\cos 0\Rightarrow 0={\cos }^{-1}x\therefore \frac{d0}{dx}=-\frac{1}{\sqrt{1-x^2}}\cdots \left(1\right)$

$\therefore \frac{dy}{dx}=\frac{dy}{d0}.\frac{d0}{dx}=-\frac{1}{\sqrt{1-x^2}}\frac{dy}{d0}$

$\therefore \sqrt{1-x^2}\frac{dy}{dx}=-\frac{dy}{d0}$

Differentiating both sides w.r.t.x

$\sqrt{1-x^2}\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}\frac{(-2x)}{2\sqrt{1-x^2}}=-\frac{d^{2}y}{d{0}^2}.\frac{d0}{dx}=\frac{d^{2}y}{d{0}^2}.\frac{1}{\sqrt{1-x^2}}by\left(1\right)$

$\therefore (1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=\frac{d^{2}y}{d{0}^2}$

Add $y$ to both sides, $\frac{d^{2}y}{d{0}^2}+y=0$