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The general solution of the differential equation \(\sqrt{1+{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{x}}^{2}{\mathrm{y…

Q1

The general solution of the differential equation \(\sqrt{1+{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{x}}^{2}{\mathrm{y}}^{2}}+\mathrm{xy}\frac{\mathrm{dy}}{\mathrm{dx}}=0\) is: (where C is a constant of integration)

[JEE Main 2020, 6 Sep (Shift 1)]

a

\(\sqrt{1+{\mathrm{y}}^{2}}+\sqrt{1+{\mathrm{x}}^{2}}=\frac{1}{2}{\log }_{\mathrm{e}}\left(\frac{\sqrt{1+{\mathrm{x}}^{2}}+1}{\sqrt{1+{\mathrm{x}}^{2}}-1}\right)+\mathrm{C}\)

b

\(\sqrt{1+{\mathrm{y}}^{2}}+\sqrt{1+{\mathrm{x}}^{2}}=\frac{1}{2}{\log }_{\mathrm{e}}\left(\frac{\sqrt{1+{\mathrm{x}}^{2}}-1}{\sqrt{1+{\mathrm{x}}^{2}}+1}\right)+\mathrm{C}\)

c

\(\sqrt{1+{\mathrm{y}}^{2}}-\sqrt{1+{\mathrm{x}}^{2}}=\frac{1}{2}{\log }_{\mathrm{e}}\left(\frac{\sqrt{1+{\mathrm{x}}^{2}}-1}{\sqrt{1+{\mathrm{x}}^{2}}+1}\right)+\mathrm{C}\)

d

\(\sqrt{1+{\mathrm{y}}^{2}}-\sqrt{1+{\mathrm{x}}^{2}}=\frac{1}{2}{\log }_{\mathrm{e}}\left(\frac{\sqrt{1+{\mathrm{x}}^{2}}+1}{\sqrt{1+{\mathrm{x}}^{2}}-1}\right)+\mathrm{C}\)

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