Question

If $y={\left({\tan }^{-1}x\right)}^2$, show that ${(x^2+1)}^2{y}_2+2x(x^2+1)y_1=2$

Answer 1

$y=({\tan }^{-1}x{)}^2$

$\Rightarrow y_1=2\left({\tan }^{-1}x\right)\times \frac{1}{1+x^2}$
$\Rightarrow y_2=\frac{2}{(1+x^2{)}^2}+2\left({\tan }^{-1}x\right)\times \frac{-2x}{(1+x^2{)}^2}$

Now, $(x^2+1{)}^2{y}_2+2x(x^2+1)y_1=(x^2+1{)}^2(\frac{2}{(1+x^2{)}^2}-\frac{4x\left({\tan }^{-1}x\right)}{(1+x^2{)}^2})+2x(x^2+1)\left(\frac{2\left({\tan }^{-1}x\right)}{1+x^2}\right)$

$=2-4x\left({\tan }^{-1}x\right)+2x\left(2\right({\tan }^{-1}x\left)\right)$

$=2-4x\left({\tan }^{-1}x\right)+4x\left({\tan }^{-1}x\right)$

$=2$