Problem 04
$(y^2 - 2xy + 6x) \, dx - (x^2 - 2xy + 2) \, dy = 0$
Solution 04
$(y^2 - 2xy + 6x) \, dx - (x^2 - 2xy + 2) \, dy = 0$
$M = y^2 - 2xy + 6x$
$N = -x^2 + 2xy - 2$
Test for exactness
$\dfrac{\partial M}{\partial y} = 2y - 2x$
$\dfrac{\partial N}{\partial x} = -2x + 2y$
Exact!
Let
$\dfrac{\partial F}{\partial x} = M$
$\dfrac{\partial F}{\partial x} = y^2 - 2xy + 6x$
$\partial F = (y^2 - 2xy + 6x) \, \partial x$
Integrate partially in x, holding y as constant