Question

If $y=(tan^{-1}x)^2$,show that $(x^2+1)^2y_2+2x(x^2+1)y_1=2$

Answer 1

The given relationship is,$y=(tan^{-1}x)^2$
Then,$y_1=2tan^{-1}x\frac{d}{dx}(tan^{-1}x)$
$⇒y_1=2tan^{-1}x.\frac{1}{1+x^2}$
$⇒(1+x^2)y_1=2tan^{-1}x$
Again differentiating with respect to $x$ on both sides,we obtain
$(1+x^2)y_2+2xy_1=2(\frac{1}{1+x^2})$
$⇒(1+x^2)^2y_2+2x(x^2+1)y_1=2$
Hence,proved