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NDA 2019 Maths · Mathematics · Differential Equations · hard

The curve satisfying (1+x^2dy/dx=1+y^2 and passing through the point (0,1) is

Single correct option

The curve satisfying (1+x2)dydx=1+y2(1+x^2)\dfrac{dy}{dx}=1+y^2 and passing through the point (0,1)(0,1) is
  1. A. y=1−x1+xy=\dfrac{1-x}{1+x}
  2. B. y=1+xy=1+x
  3. C. y=tan⁡x+1y=\tan x+1
  4. D. y=1+x1−xy=\dfrac{1+x}{1-x}
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Answer: (D) y=1+x1−xy=\dfrac{1+x}{1-x}

Separate variables: dy1+y2=dx1+x2\frac{dy}{1+y^2}=\frac{dx}{1+x^2}, so tan⁡−1y=tan⁡−1x+C\tan^{-1}y=\tan^{-1}x+C. At (0,1)(0,1), C=π4C=\frac\pi4. Hence y=tan⁡(tan⁡−1x+π4)=x+11−xy=\tan\left(\tan^{-1}x+\frac\pi4\right)=\frac{x+1}{1-x}. The expression 1−x1+x\frac{1-x}{1+x} results from using −π4-\frac\pi4.
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Original question written in the NDA 2019 Maths pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake