Question

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

Answer 1

Here,
$$\begin{gathered} y={\mathrm{cosec}}^{-1}x \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get} \\ \frac{\mathit{d}y}{\mathit{d}x}=\frac{-1}{x\sqrt{x^2-1}} \\ \mathrm{Differentiating}\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get} \\ \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}=\frac{\sqrt{x^2-1}+{\displaystyle \frac{x^2}{\sqrt{x^2-1}}}}{x^2\left(x^2-1\right)} \\ \Rightarrow \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}=\frac{x^2-1+x^2}{x^2\left(x^2-1\right)\sqrt{x^2-1}} \\ \Rightarrow \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}=\frac{2x^2-1}{x^2\left(x^2-1\right)\sqrt{x^2-1}} \\ \Rightarrow \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}=\frac{2}{\left(x^2-1\right)\sqrt{x^2-1}}-\frac{1}{x^2\left(x^2-1\right)\sqrt{x^2-1}} \\ \Rightarrow \left(x^2-1\right)\frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}=\frac{2}{\sqrt{x^2-1}}-\frac{1}{x^2\sqrt{x^2-1}} \\ \Rightarrow \left(x^2-1\right)\frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}=-2x\frac{dy}{dx}+\frac{1}{x}\frac{dy}{dx} \\ \Rightarrow x\left(x^2-1\right)\frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}=-\left(2x^2-1\right)\frac{dy}{dx} \\ \Rightarrow x\left(x^2-1\right)\frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}+\left(2x^2-1\right)\frac{dy}{dx}=0 \end{gathered}$$

Hence proved.