Mathematics · 2020
JEE Main · 6 September 2020, Shift 1 · Q63
The general solution of the differential equation √(1+x^2+y^2+x^2 y^2)+x y (d y)/(d x)=0 is: (where C is a constant of integration)
The general solution of the differential equation $\displaystyle \sqrt{1+x^2+y^2+x^2 y^2}+x y \frac{\mathrm{~d} y}{\mathrm{~d} x}=0$ is : (where C is a constant of integration)
Official answer
(3)
$\displaystyle \sqrt{1+y^2}+\sqrt{1+x^2}=\frac{1}{2} \log _{\mathrm{e}}\left(\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}-1}\right)+\mathrm{C}$
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.