$(x + y) \, dx + (x - y) \, dy = 0$
Test for exactness
$M = x + y$ ; $\dfrac{\partial M}{\partial y} = 1$
$N = x - y$ ; $\dfrac{\partial N}{\partial x} = 1$
$\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}$ ; thus, exact!
Step 1: Let
$\dfrac{\partial F}{\partial x} = M$
$\dfrac{\partial F}{\partial x} = x + y$
Step 2: Integrate partially with respect to x, holding y as constant
$\partial F = (x + y) \, \partial x$
$\displaystyle \int \partial F = \int (x + y) \, \partial x$
$F = \frac{1}{2}x^2 + xy + f(y)$ → Equation (1)
Step 3: Differentiate Equation (1) partially with respect to y, holding x as constant
$\dfrac{\partial F}{\partial y} = x + f'(y)$
Step 4: Equate the result of Step 3 to N and collect similar terms. Let
$\dfrac{\partial F}{\partial y} = N$
$x + f'(y) = x - y$
$f'(y) = -y$
Step 5: Integrate partially the result in Step 4 with respect to y, holding x as constant
$\displaystyle \int f'(y) = - \int y \, dy$
$f(y) = -\frac{1}{2}y^2$
Step 6: Substitute f(y) to Equation (1)
$F = \frac{1}{2}x^2 + xy - \frac{1}{2}y^2$
Equate F to ½c
$F = \frac{1}{2}c$
$\frac{1}{2}x^2 + xy - \frac{1}{2}y^2 = \frac{1}{2}c$
$x^2 + 2xy - y^2 = c$ answer
