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Mathematics Question on Differential equations

If y=(tan⁡−1x)2y = \left(\tan^{-1} x\right)^{2} then (x2+1)2d2ydx2+2x(x2+1)dydx=\left(x^{2} + 1\right)^{2} \frac{d^{2}y}{dx^{2} } + 2x \left(x^{2} + 1 \right) \frac{dy}{dx} =

A

4

B

2

C

1

D

0

Answer

2

Explanation

Solution

We have,
y=(tan⁡−1x)2y=\left(\tan ^{-1} x\right)^{2}
On differentiating w.r.t. x,x, we get
dydx=2tan⁡−1x1+x2\frac{d y}{d x} =\frac{2 \tan ^{-1} x}{1+x^{2}}
⇒(1+x2)dydx=2tan⁡−1x\Rightarrow \left(1+x^{2}\right) \frac{d y}{d x} =2 \tan ^{-1} x
On squaring both sides, we get
(1+x2)2(dydx)2=4(tan⁡−1x)2\left(1+x^{2}\right)^{2}\left(\frac{d y}{d x}\right)^{2}=4\left(\tan ^{-1} x\right)^{2}
⇒(1+x2)2(dydx)2=4y[∵y=(tan⁡−1x)2]\Rightarrow \left(1+x^{2}\right)^{2}\left(\frac{d y}{d x}\right)^{2}=4 y \left[\because y=\left(\tan ^{-1} x\right)^{2}\right]
Again, differentiating w.r.t.x, we get
(1+x2)2(2dydx⋅d2ydx2)+2(1+x2)(2x)(dydx)2=4dydx\left(1+x^{2}\right)^{2}\left(2 \frac{d y}{d x} \cdot \frac{d^{2} y}{d x^{2}}\right)+2\left(1+x^{2}\right)(2 x)\left(\frac{d y}{d x}\right)^{2}=4 \frac{d y}{d x}
On dividing both sides by 2dydx2 \frac{d y}{d x},
we get
(1+x2)2(d2ydx2)+2x(1+x2)dydx=2\left(1+x^{2}\right)^{2}\left(\frac{d^{2} y}{d x^{2}}\right)+2 x\left(1+x^{2}\right) \frac{d y}{d x}=2

So, the correct option is (B): 2