Solve the following differential equation: `sqrt(1+x^2+y^2+x^2\ y^2)\ +x y(dy)/(dx)=0`

Explanation

Note that `f(x,y) = sqrt(1+x^(2)+y^(2)+x^(2)y^(2)) = sqrt((1+x^(2))(1+y^(2)))`
`sqrt(1+x^(2)) . Sqrt(1+y^(2)) = g(x)h(y)`
Thus, f(x,y) can be written as a product of two functions, one of x alone and the othr of y alone.
`sqrt(1+x^(2)+y^(2)+x^(2)y^(2))+ xy(dy)/(dx) = 0`
`rArr sqrt(1+ x^(2))sqrt(1+y^(2)) + xy (dy)/(dx) = 0`
`rArr (sqrt(1+x^(2)))/(x) dx + (y)/(sqrt(1+y^(2)))dy = 0`
Integration yields
`int (1+x^(2))/(xsqrt(1+x^(2)))dx+(1)/(2)int (d(1+y^(2)))/(sqrt(1+y^(2))) = k`
` rArr int (1)/(xsqrt(1+ x^(2)))dx+ int (x)/(sqrt(1+x^(2))) dx + (1)/(2) xx 2sqrt(1+y^(2)) = k`
`rArr int (-(1)/(2)dt)/(((1)/(2))sqrt(1+((1)/(2))^(2)))+ sqrt(1+ x^(2)) + sqrt(1+y^(2)) = k` [setting x = 1/t in the first integral].
`rArr - int (dt)/(sqrt(1+ t^(2))) + sqrt(1+x^(2)) + sqrt(1+y^(2)) = k`
`rarr -ln|t + sqrt(1+t^(2))| + sqrt(1+ x^(2)) + sqrt(1+ y^(2)) = k`
`rArr -ln |(1)/(x)+sqrt(1+(1)/(x^(2)))| + sqrt(1+x^(2)) + sqrt(1+y^(2)) = k`
`rArr sqrt(1+ x^(2)) - ln|(x+sqrt(1+x^(2)))/(x)| + sqrt(1+ y^(2)) = k`
which is the general solution of the differential equation.