Question

If y = cosec⁻¹ x, x >1, then show that \[x\left( x^2 - 1 \right)\frac{d^2 y}{d x^2} + \left( 2 x^2 - 1 \right)\frac{dy}{dx} = 0\] ?

Answer 1

Here,

\[y = {cosec}^{- 1} x\]

\[\text { Differentiating w . r . t . x, we get }\]

\[\frac{d y}{d x} = \frac{- 1}{x\sqrt{x^2 - 1}}\]

\[\text { Differentiating again w . r . t . x, we get }\]

\[\frac{d^2 y}{d x^2} = \frac{\sqrt{x^2 - 1} + \frac{x^2}{\sqrt{x^2 - 1}}}{x^2 \left( x^2 - 1 \right)}\]

\[ \Rightarrow \frac{d^2 y}{d x^2} = \frac{x^2 - 1 + x^2}{x^2 \left( x^2 - 1 \right)\sqrt{x^2 - 1}}\]

\[ \Rightarrow \frac{d^2 y}{d x^2} = \frac{2 x^2 - 1}{x^2 \left( x^2 - 1 \right)\sqrt{x^2 - 1}}\]

\[ \Rightarrow \frac{d^2 y}{d x^2} = \frac{2}{\left( x^2 - 1 \right)\sqrt{x^2 - 1}} - \frac{1}{x^2 \left( x^2 - 1 \right)\sqrt{x^2 - 1}}\]

\[ \Rightarrow \left( x^2 - 1 \right)\frac{d^2 y}{d x^2} = \frac{2}{\sqrt{x^2 - 1}} - \frac{1}{x^2 \sqrt{x^2 - 1}}\]

\[ \Rightarrow \left( x^2 - 1 \right)\frac{d^2 y}{d x^2} = - 2x\frac{dy}{dx} + \frac{1}{x}\frac{dy}{dx}\]

\[ \Rightarrow x\left( x^2 - 1 \right)\frac{d^2 y}{d x^2} = - \left( 2 x^2 - 1 \right)\frac{dy}{dx}\]

\[ \Rightarrow x\left( x^2 - 1 \right)\frac{d^2 y}{d x^2} + \left( 2 x^2 - 1 \right)\frac{dy}{dx} = 0\]

Hence proved.