Question

If $y={\left({\tan }^{-1}x\right)}^2$, then ${(x^2+1)}^2\frac{d^{2}y}{d{x}^2}+2x(x^2+1)\frac{dy}{dx}=$

• A. $4$
• B. $2$
• C. $1$
• D. $0$

Answer 1

Answer: (B)

$2$

Given
Differential equation,
$(x^2+1{)}^2\frac{{\partial }^{2}y}{\partial x^2}+2x(x^2+1)\frac{\partial y}{\partial x}$ .......(1)
$\Rightarrow y=({\tan }^{-1}x{)}^2$
Differentiate above equation with respect to $x$.
$\Rightarrow \frac{\partial y}{\partial x}=\frac{2\left({\tan }^{-1}x\right)}{1+x^2}$ ......(2)
Differentiate equation (2) with respect to $x$
$\Rightarrow \frac{{\partial }^{2}y}{\partial x^2}=\frac{(1+x^2)\left(\frac{2}{1+x^2}\right)-\left(2{\tan }^{-1}x\right)2x}{(1+x^2{)}^2}$
$\Rightarrow \frac{{\partial }^{2}y}{\partial x^2}=\frac{2-4x{\tan }^{-1}x}{(1+x^2{)}^2}$ ......(3)
Substitute $\frac{\partial y}{\partial x}$ and $\frac{{\partial }^{2}y}{\partial x^2}$ from equation (2) and (3) in equation (1).
$\Rightarrow (x^2+1{)}^2\frac{(2-4x{\tan }^{-1}x)}{(1+x^2{)}^2}+2x(x^2+1)\frac{2\left({\tan }^{-1}x\right)}{(1+x^2)}$
Simply the above equation.
$\Rightarrow 2-4x{\tan }^{-1}x+4x\left({\tan }^{-1}x\right)$
$\Rightarrow 2$ Ans