Integrating Factor for First Order ODE/Function of One Variable

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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:

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

is a function of $x$ only.

Then:

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

is an integrating factor for $(1)$.


Similarly, suppose that:

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

is a function of $y$ only.

Then:

$\map \mu y = e^{\int -\map h y \rd y}$

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 $x$ or $y$ only

Suppose that $\mu$ is a function of $x$ only.

Then:

$\dfrac {\partial \mu} {\partial x} = \dfrac {d \mu} {d x}, \dfrac {\partial \mu} {\partial y} = 0$

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

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

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


Similarly, if $\mu$ is a function of $y$ only, we find that:

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

where $\map h y$ is the function of $y$ 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

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