# Integrating Factor for First Order ODE/Function of Product of Variables

## Theorem

$(1): \quad \map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$

be non-homogeneous and not exact.

Let $(1)$ be such that:

$\map g z = \dfrac {\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}} {N y - M x}$

is a function of $z$, where $z = x y$.

Then:

$\map \mu {x y} = \map \mu z = e^{\int \map g z \d z}$

is an integrating factor for $(1)$.

## Proof

### Preliminary Work

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:

$\mu \, \map M {x, y} \rd x + \mu \, \map N {x, y} \rd y = 0$
$\dfrac {\map \partial {\mu M} } {\partial y} = \dfrac {\map \partial {\mu N} } {\partial x}$

Evaluating this, using the Product Rule:

$\mu \dfrac {\partial M} {\partial y} + M \dfrac {\partial \mu} {\partial y} = \mu \dfrac {\partial N} {\partial x} + N \dfrac {\partial \mu} {\partial x}$

$\dfrac 1 \mu \paren {N \dfrac {\partial \mu} {\partial x} - M \dfrac {\partial \mu} {\partial y} } = \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$

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

Then:

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

### Proof for Function of x y

Suppose that $\mu$ is a function of $z = x y$.

Then:

$\dfrac {\partial z} {\partial x} = y$
$\dfrac {\partial z} {\partial y} = x$

Thus:

$\dfrac {\partial \mu} {\partial x} = \dfrac {\d \mu} {\d z} \dfrac {\partial z} {\partial x} = y \dfrac {\d \mu} {\d z}$
$\dfrac {\partial \mu} {\partial y} = \dfrac {\d \mu} {\d z} \dfrac {\partial z} {\partial y} = x \dfrac {\d \mu} {\d z}$

which, when substituting in $(3)$, leads us to:

$\dfrac 1 \mu \dfrac {\d \mu} {\d z} = \dfrac {\map P {x, y} } {N y - M x} = \map g z$

where $\map g z$ is the function of $z$ that we posited.

### Conclusion of Proof

We have an equation of the form:

$\dfrac 1 \mu \dfrac {\d \mu} {\d w} = \map f w$

which is what you get when you apply the Chain Rule for Derivatives and Derivative of Logarithm Function to:

$\dfrac {\map \d {\ln \mu} } {\d w} = \map f w$

Thus:

$\displaystyle \ln \mu = \int \map f w \rd w$

and so:

$\mu = e^{\int \map f w \rd w}$

Hence the results as stated.

$\blacksquare$