Question

If $y=e^{a{\cos }^{-1}}x$, prove that $\left(1-x^2\right)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}-a^{2}y=0$.

Answer 1

Here,
$$\begin{gathered} y=e^{a{\cos }^{-1}x} \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get} \\ \frac{\mathit{d}y}{\mathit{d}x}=-{\mathrm{e}}^{a{\cos }^{-1}x}\times \frac{a}{\sqrt{1-x^2}} \\ \mathrm{Differentiating}\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get} \\ \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}={\mathrm{e}}^{a{\cos }^{-1}x}\times \frac{a^2}{1-x^2}+\frac{2xa{e}^{a{\cos }^{-1}x}}{2{\left(1-x^2\right)}^{{\displaystyle \frac{3}{2}}}} \\ \Rightarrow \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}={\mathrm{e}}^{a{\cos }^{-1}x}\times \frac{a^2}{1-x^2}+\frac{xa{e}^{a{\cos }^{-1}x}}{\left(1-x^2\right)\sqrt{1-x^2}} \\ \Rightarrow \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}=y\times \frac{a^2}{1-x^2}-\frac{x{\displaystyle \frac{dy}{dx}}}{\left(1-x^2\right)} \\ \Rightarrow \left(1-x^2\right)\frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}=a^{2}y-x\frac{dy}{dx} \\ \Rightarrow \left(1-x^2\right)\frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}+x\frac{dy}{dx}-a^{2}y=0 \end{gathered}$$

Hence proved.