Question

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

Answer 1

$y=(ta{n}^{-1}x{)}^2$ ________ (1)

On diff. wrt x we get

$\frac{dy}{dx}=2ta{n}^{-1}x\times \frac{1}{x^2+1}$ ________ (2)

again diff. wrt we get

$\Rightarrow \frac{d^{2}y}{d{x}^2}=2\left[\frac{\frac{1}{1+x^2}\times (1+x^2)-ta{n}^{-1}x.2x}{(1+x^2{)}^2}\right]$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=2\left[\frac{1-2xta{n}^{-1}x}{(1+x^2{)}^2}\right]$

$\Rightarrow (1+x^2{)}^2\frac{d^{2}y}{d{x}^2}=2-4xta{n}^{-1}x$

$(1+x^2)y_2+4xta{n}^{-1}x=2$

From equation (2)

$\left[\right(1+x^2)y_2+2x(1+x^2)y_1=2]$