The general solution of the differential equation

$\sqrt {1 + {x^2} + {y^2} + {x^2}{y^2}}$ + xy${{dy} \over {dx}}$ = 0 is :

(where C is a constant of integration)

$\sqrt {1 + {x^2} + {y^2} + {x^2}{y^2}}$ + xy${{dy} \over {dx}}$ = 0 <br><br>$\Rightarrow$ $\sqrt {\left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)}$ + xy${{dy} \over {dx}}$ = 0 <br><br>$\Rightarrow$ $\sqrt {\left( {1 + {x^2}} \right)} \sqrt {\left( {1 + {y^2}} \right)}$ = -xy${{dy} \over {dx}}$ <br><br>$\Rightarrow$ $$\int {{{ydy} \over {\sqrt {1 + {y^2}} }}} = - \int {{{\sqrt {1 + {x^2}} } \over x}} dx$$ .....(1) <br><br>Now put 1 + x² = u² and 1 + y² = v² <br><br>2xdx = 2udu and 2ydy = 2vdv <br><br>$\Rightarrow$ xdx = udu and ydy = vdv <br><br>Substitute these values in equation (1) <br><br>$\int {{{vdv} \over v}} = - \int {{{{u^2}du} \over {{u^2} - 1}}}$ <br><br>$\Rightarrow$ $\int {dv} = - \int {{{{u^2} - 1 + 1} \over {{u^2} - 1}}} du$ <br><br>$\Rightarrow$ v = $- \int {\left( {1 + {1 \over {{u^2} - 1}}} \right)} du$ <br><br>$\Rightarrow$ v = -u - ${1 \over 2}{\log _e}\left| {{{u - 1} \over {u + 1}}} \right|$ + C <br><br>$\Rightarrow$ $\sqrt {1 + {y^2}}$ = $- \sqrt {1 + {x^2}}$ + $${1 \over 2}{\log _e}\left| {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right|$$ + C <br><br>$\Rightarrow$ $$\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C$$