The general solution of the differential equation \sqrt{1+x^{2}+y^{2}+x^{2} y^{2}}+x y \frac{d y}{d x}=0 is (where {\C} is a constant of integration) [6 Sep 2020 Shift 1]

Correct Answer is : √1+y2+√1+x2 =12loge(√1+x2+1√1+x2−1)+C

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

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

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

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

JEE Mains 06-Sep-2020 Shift 1 Solved Paper

Section: Mathematics

Question : 62 of 75

Marks: +1, -0

The general solution of the differential equation √1+x2+y2+x2y2+xy

=0 is (where C is a constant of integration)

[6 Sep 2020 Shift 1]

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