# Problem 04 | Exact Equations

**Problem 04**

$(y^2 - 2xy + 6x) \, dx - (x^2 - 2xy + 2) \, dy = 0$

**Solution 04**

$M = y^2 - 2xy + 6x$

$N = -x^2 + 2xy - 2$

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

$\displaystyle \int \partial F = \int (y^2 - 2xy + 6x) \, \partial x$

$F = xy^2 - x^2y + 3x^2 + f(y)$ → Equation (1)

Differentiate partially in y, holding x as constant

$\dfrac{\partial F}{\partial y} = 2xy - x^2 + f'(y)$

Let

$\dfrac{\partial F}{\partial y} = N$

$2xy - x^2 + f'(y) = -x^2 + 2xy - 2$

$f'(y) = -2$

Integrate partially in y, holding x as constant

$\displaystyle \int f'(y) = -2 \int \partial y$

$f(y) = -2y$

Substitute f(y) to Equation (1)

$F = xy^2 - x^2y + 3x^2 - 2y$

Equate F to c

$F = c$

$xy^2 - x^2y + 3x^2 - 2y = c$ *answer*