Integrating Factor for First Order ODE/Preliminary Work

Theorem
Let the first order ordinary differential equation:
 * $(1): \quad \map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$

be non-homogeneous and not exact.

Let $\map \mu x$ be an integrating factor for $(1)$.

Let:
 * $\map P {x, y} := \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$

Then:
 * $\dfrac 1 \mu = \dfrac {\map P {x, y} } {N \dfrac {\partial \mu} {\partial x} - M \dfrac {\partial \mu} {\partial y} }$

Proof
Let us for ease of manipulation express $(1)$ in the form of differentials:
 * $(2): \quad \map M {x, y} \rd x + \map N {x, y} \rd y = 0$

Let $\mu$ be an integrating factor for $(2)$.

Then, by definition:
 * $\mu \, \map M {x, y} \rd x + \mu \, \map N {x, y} \rd y = 0$

is an exact differential equation.

By Solution to Exact Differential Equation:
 * $\dfrac {\map \partial {\mu M} } {\partial y} = \dfrac {\map \partial {\mu N} } {\partial x}$

Evaluating this, using the Product Rule:
 * $\mu \dfrac {\partial M} {\partial y} + M \dfrac {\partial \mu} {\partial y} = \mu \dfrac {\partial N} {\partial x} + N \dfrac {\partial \mu} {\partial x}$

which leads us to:
 * $\dfrac 1 \mu \paren {N \dfrac {\partial \mu} {\partial x} - M \dfrac {\partial \mu} {\partial y} } = \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$

Let us use $\map P {x, y}$ for $\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$.

Then:
 * $(3): \quad \dfrac 1 \mu = \dfrac {\map P {x, y} } {N \dfrac {\partial \mu} {\partial x} - M \dfrac {\partial \mu} {\partial y}}$