$(6x + y^2) \, dx + y(2x - 3y) \, dy = 0$
$M = 6x + y^2$
$N = y(2x - 3y) = 2xy - 3y^2$
Test for exactness
$\dfrac{\partial M}{\partial y} = 2y$
$\dfrac{\partial N}{\partial x} = 2y$
Exact!
Let
$\dfrac{\partial F}{\partial x} = M$
$\dfrac{\partial F}{\partial x} = 6x + y^2$
$\partial F = (6x + y^2) \, \partial x$
Integrate partially in x, holding y as constant
$\displaystyle \int \partial F = \int (6x + y^2) \, \partial x$
$F = 3x^2 + xy^2 + f(y)$ → Equation (1)
Differentiate partially in y, holding x as constant
$\dfrac{\partial F}{\partial y} = 2xy + f'(y)$
Let
$\dfrac{\partial F}{\partial y} = N$
$2xy + f'(y) = 2xy - 3y^2$
$f'(y) = -3y^2$
Integrate partially in y, holding x as constant
$\displaystyle \int f'(y) = -3 \int y^2 \, \partial y$
$f(y) = -y^3$
Substitute f(y) to Equation (1)
$F = 3x^2 + xy^2 - y^3$
Equate F to c
$F = c$
$3x^2 + xy^2 - y^3 = c$ answer
