Question

Solve: $\sqrt{1+x^2}\sqrt{1+y^2}dx+xydy=0$

Answer 1

$\sqrt{1+x^2}\sqrt{1+y^2}dx=-xydy\int \left(\frac{\sqrt{1+x^2}}{x}\right)dx=\int \left(\frac{-y}{\sqrt{1+y^2}}\right)dylet1+x^2=t^{2}or{x}^2=(t^2-1)2xdx=2tdtordx=\left(\frac{tdt}{x}\right)and1+y^2=z^{2}2ydy=2zdzydy=zdz\therefore \int \left(\frac{t}{x}\right)\left(\frac{tdt}{x}\right)=-\int \left(\frac{zdz}{z}\right)\int \left(\frac{t^2}{t^2-1}\right)dt=-z+C\int dt+\int \left(\frac{1}{t^2-1}\right)dt=-z+Ct+\left(\frac{1}{2}\right)log\left|\right(\frac{t-1}{t+1}\left)\right|=-z+C\sqrt{1+x^2}+\left(\frac{1}{2}\right)log\left|\right(\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\left)\right|=-\sqrt{1+y^2}+C$