Integrating Factor for First Order ODE/Function of Sum of Variables

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.

Suppose that:

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

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

Then:

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

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 for Derivatives:

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

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:

$\dfrac {\partial z} {\partial x} = 1 = \dfrac {\partial z} {\partial y}$

Thus:

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

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

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

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:

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

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 $3$