Single Choice

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

A$$4$$
B2
Correct Answer
C1
D0

Solution

Given

Differential equation,

$$(x^2+1)^2\dfrac{\partial^2 y}{\partial x^2}+2x(x^2+1)\dfrac{\partial y}{\partial x}$$ .......(1)

$$\Rightarrow y=(\tan^{-1}x)^2$$

Differentiate above equation with respect to $$x$$.

$$\Rightarrow \dfrac{\partial y}{\partial x}=\dfrac{2(\tan^{-1}x)}{1+x^2}$$ ......(2)

Differentiate equation (2) with respect to $$x$$

$$\Rightarrow \dfrac{\partial^2 y}{\partial x^2}=\dfrac{(1+x^2)(\dfrac{2}{1+x^2})-(2\tan^{-1}x)2x}{(1+x^2)^2}$$


$$\Rightarrow \dfrac{\partial^2 y}{\partial x^2}=\dfrac{2-4x\tan^{-1}x}{(1+x^2)^2}$$ ......(3)


Substitute $$\frac{\partial y}{\partial x}$$ and $$\dfrac{\partial^2 y}{\partial x^2}$$ from equation (2) and (3) in equation (1).

$$\Rightarrow (x^2+1)^2\dfrac{(2-4x\tan^{-1}x)}{(1+x^2)^2}+2x(x^2+1)\dfrac{2(\tan^{-1}x)}{(1+x^2)}$$

Simply the above equation.

$$\Rightarrow 2-4x\tan^{-1}x+4x(\tan^{-1}x)$$

$$\Rightarrow 2$$ Ans

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