The general solution of √1+x^2+y^2+x^2y^2+xydydx=0 is (C constant):

Question

The general solution of $\sqrt{1+x^2+y^2+x^2y^2}+xy\dfrac{dy}{dx}=0$ is (C constant):

Accepted Answer

Separating: $\sqrt{1+x^2}\cdot\sqrt{1+y^2}=-xy\dfrac{dy}{dx}\cdot(\dots)$; with $x=\tan\theta$ and integrating $\int\sec^3\theta/\tan\theta\,d\theta = -\int y\,dy/\sqrt{1+y^2}$, one obtains
$$\sqrt{1+y^2}+\sqrt{1+x^2} = \frac12\log_e\left|\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}-1}\right|+C.$$

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