Question

If log y = tan⁻¹ x, show that (1 + x²)y₂ + (2x − 1) y₁ = 0

Answer 1

Here,
$$\begin{gathered} \log y={\tan }^{-1}x \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get} \\ \frac{1}{y}\times y_1=\frac{1}{1+x^2} \\ \Rightarrow \left(1+x^2\right)y_1=y \\ \Rightarrow \left(1+x^2\right)y_2+2x{y}_1=y_1 \\ \Rightarrow \left(1+x^2\right)y_2+2x{y}_1-y_1=0 \\ \Rightarrow \left(1+x^2\right)y_2+\left(2x-1\right)y_1=0 \end{gathered}$$

Hence proved.