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Transforming a Differential Equation by x=tan⁻¹t | JEE

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Question
If the substitution x=tan⁡−1(t)x = \tan^{-1}(t) transforms the differential equation d2ydx2+xydydx+sec⁡2x=0\dfrac{d^2y}{dx^2} + xy\dfrac{dy}{dx} + \sec^2 x = 0 into (1+t2)d2ydt2+(2t+ytan⁡−1(t))dydt=k(1 + t^2)\dfrac{d^2y}{dt^2} + \big(2t + y\tan^{-1}(t)\big)\dfrac{dy}{dt} = k, then the value of kk is:
A11
B−1-1correct
C22
D−2-2
Solution
Step 1: From x=tan⁡−1tx = \tan^{-1}t, dxdt=11+t2\dfrac{dx}{dt} = \dfrac{1}{1 + t^2}, so
dydx=(1+t2)dydt.\frac{dy}{dx} = (1 + t^2)\frac{dy}{dt}.
 Step 2: Differentiate again:
d2ydx2=ddt[(1+t2)dydt]⋅dtdx=(1+t2)[2tdydt+(1+t2)d2ydt2].\frac{d^2y}{dx^2} = \frac{d}{dt}\left[(1 + t^2)\frac{dy}{dt}\right]\cdot\frac{dt}{dx} = (1 + t^2)\left[2t\frac{dy}{dt} + (1 + t^2)\frac{d^2y}{dt^2}\right].
 Step 3: Also sec⁡2x=1+tan⁡2x=1+t2\sec^2 x = 1 + \tan^2 x = 1 + t^2 and x=tan⁡−1tx = \tan^{-1}t. Substitute into the equation:
(1+t2)[2tdydt+(1+t2)d2ydt2]+(tan⁡−1t) y (1+t2)dydt+(1+t2)=0.(1 + t^2)\left[2t\frac{dy}{dt} + (1 + t^2)\frac{d^2y}{dt^2}\right] + (\tan^{-1}t)\,y\,(1 + t^2)\frac{dy}{dt} + (1 + t^2) = 0.
 Step 4: Cancel (1+t2)(1 + t^2) throughout:
(1+t2)d2ydt2+(2t+ytan⁡−1t)dydt=−1⇒k=−1.(1 + t^2)\frac{d^2y}{dt^2} + \big(2t + y\tan^{-1}t\big)\frac{dy}{dt} = -1 \Rightarrow k = -1.
 Correct answer: (2)
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