Question

If $y=({\tan }^{-1}x{)}^2$, show that $(1+x^2{)}^2\frac{d^{2}y}{d{x}^2}+2x(1+x^2)\frac{dy}{dx}-2=0$

Answer 1

Given, $y=({\tan }^{-1}x{)}^2$
Let ${\tan }^{-1}x=z$
$\therefore x=\tan z$
On differentiating both sides, we get
$dx={\sec }^{2}zdz$
$=(1+{\tan }^{2}z)dz$
$=(1+x^2)dz$
Now, $y=z^2$
$dy=2zdz$
$=2{\tan }^{-1}xdz$
$dy=2{\tan }^{-1}x.\frac{dx}{(1+x^2)}$
Hence, $\frac{dy}{dx}=\frac{2{\tan }^{-1}x}{(1+x^2)}$
$(1+x^2)\frac{dy}{dx}=2{\tan }^{-1}x$
$(1+x^2{)}^2{\left(\frac{dy}{dx}\right)}^2=4({\tan }^{-1}x{)}^2$
$(1+x^2){\left(\frac{dy}{dx}\right)}^2-4y=0$
Differentiating both sides with respect to $x$,
$(1+x^2)\left[2\frac{dy}{dx}\right]\frac{d^{2}y}{d{x}^2}+2{\left(\frac{dy}{dx}\right)}^2(1+x^2)2x--4\frac{dy}{dx}=0$
$(1+x^2{)}^2\frac{d^{2}y}{d{x}^2}+2x(1+x^2)\frac{dy}{dx}-2=0$