# Problem 03 | Exact Equations

**Problem 03**

$(2xy - 3x^2) \, dx + (x^2 + y) \, dy = 0$

**Solution 03**

$M = 2xy - 3x^2$

$N = x^2 + y$

Test for exactness

$\dfrac{\partial M}{\partial y} = 2x$

$\dfrac{\partial N}{\partial x} = 2x$

Exact!

Let

$\dfrac{\partial F}{\partial x} = M$

$\dfrac{\partial F}{\partial x} = 2xy - 3x^2$

$\partial F = (2xy - 3x^2) \, \partial x$

Integrate partially in x, holding y as constant

$\displaystyle \int \partial F = \int (2xy - 3x^2) \, \partial x$

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

Differentiate partially in y, holding x as constant

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

Let

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

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

$f'(y) = y$

Integrate partially in y, holding x as constant

$\displaystyle \int f'(y) = \int y \, \partial y$

$f(y) = \frac{1}{2}y^2$

Substitute f(y) to Equation (1)

$F = x^2y - x^3 + \frac{1}{2}y^2$

Equate F to c

$F = c$

$x^2y - x^3 + \frac{1}{2}y^2 = c$ *answer*