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

## Contents

## 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 $(1)$ 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:

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

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:

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

### 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.

$\blacksquare$

## Sources

- 1972: George F. Simmons:
*Differential Equations*... (previous) ... (next): $\S 2.9$: Integrating Factors: Problem $1$