Question

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

Answer 1

Given:
$y=({\tan }^{-1}x{)}^2$ (diff wrt $x$)
$y_1=2{\tan }^{-1}x.\frac{1}{1+x^2}$
$\therefore (1+x{)}^2{y}_1=2{\tan }^{-1}x$ (diff wrt $x$)
$(1+x^2)y_2+y_1.2x=2.\frac{1}{1+x^2}\Rightarrow (1+x^2{)}^2{y}_2+2x(1+x^2)y_1=2$ (proved).