Question

If $y=e^{{\tan }^{-1}x}$, prove that (1 + x²)y₂ + (2x − 1)y₁ = 0.

Answer 1

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

Hence proved.