The general solution of the differential equation 1 + x 2 + y 2 + x 2 y 2 + x y d y d x = 0 (where C is a…

1. $\sqrt{1+y^2}+\sqrt{1+x^2}=\frac{1}{2}{\log }_e\left(\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right)+C$
2. $\sqrt{1+y^2}-\sqrt{1+x^2}=\frac{1}{2}{\log }_e\left(\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right)+C$
3. $\sqrt{1+y^2}+\sqrt{1+x^2}=\frac{1}{2}{\log }_e\left(\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}-1}\right)+C$
4. $\sqrt{1+y^2}-\sqrt{1+x^2}=\frac{1}{2}{\log }_e\left(\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}-1}\right)+C$

Solution

Asked in: JEE Main 2020 (06 Sep Shift 1)

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