Problem 03
$y(2x - y + 1)~dx + x(3x - 4y + 3)~dy = 0$
Solution 03
$M~dx + N~dy = 0$
$y(2x - y + 1)~dx + x(3x - 4y + 3)~dy = 0$
$M = y(2x - y + 1) = 2xy - y^2 + y$
$N = x(3x - 4y + 3) = 3x^2 - 4xy + 3x$
$\dfrac{\partial M}{\partial y} = 2x - 2y + 1$
$\dfrac{\partial N}{\partial x} = 6x - 4y + 3$
$\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} = (2x - 2y + 1) - (6x - 4y + 3)$
$\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} = -4x + 2y - 2$