Solve the differential quation: dy/dx = -(2xy^2+1)/(2x^2y)

Given the differential
equation:

dy/dx = -(2xy^2+1) /
(2x^2y)

First we will cross
multiply.

==> (2x^2 y ) dy = -(2xy^2 + 1)
dx

Now we will integrate both
sides.

==> Int 2x^2y dy = Int -(2xy^2 + 1)
dx

==> 2x^2 y^2 /2 = -(2x^2*y^2/2 + x) +
C

==> x^2 y^2 = - x^2*y^2 - x +
C

==> 2x^2 *y^2 + x =
C

==> Then the equation is given by
:

2x^2y^2 + x + C =
0