Question Number 20872 by j.masanja06@gmail.com last updated on 05/Sep/17
![if y=[xtan^(−1) x]−[(1/2)ln(1+x^2 )] show that (1+x^2 )y^(′′) =1](https://www.tinkutara.com/question/Q20872.png)
$${if}\:\:{y}=\left[{xtan}^{−\mathrm{1}} {x}\right]−\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right] \\ $$$${show}\:{that}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}^{''} =\mathrm{1} \\ $$
Answered by sma3l2996 last updated on 05/Sep/17
$${y}'={tan}^{−\mathrm{1}} {x}+\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }−\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${y}''=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\Leftrightarrow\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}''=\mathrm{1} \\ $$