Question

If y = (tan⁻¹ x)², then prove that (1 + x²)² y₂ + 2x(1 + x²)y₁ = 2.

Answer 1

Here,
$$\begin{gathered} y={\left({\tan }^{-1}x\right)}^2 \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get} \\ {y}_1=\frac{2{\tan }^{-1}x}{1+x^2} \\ \mathrm{Differentiating}\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get} \\ {y}_2=\frac{2-4x{\tan }^{-1}x}{{\left(1+x^2\right)}^2} \\ \Rightarrow y_2=\frac{2}{{\left(1+x^2\right)}^2}-\frac{2{\tan }^{-1}x\times 2x}{{\left(1+x^2\right)}^2} \\ \Rightarrow y_2=\frac{2}{{\left(1+x^2\right)}^2}-\frac{2x{y}_1}{\left(1+x^2\right)} \\ \Rightarrow {\left(1+x^2\right)}^2{y}_2=2-2x\left(1+x^2\right)y_1 \\ \Rightarrow {\left(1+x^2\right)}^2{y}_2+2x\left(1+x^2\right)y_1=2 \end{gathered}$$

Hence proved.