# Integrating Factor for First Order ODE

## 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.

By Existence of Integrating Factor, if $(1)$ has a general solution, there exists an integrating factor $\map \mu {x, y}$ such that:

- $\displaystyle \map \mu {x, y} \paren {\map M {x, y} + \map N {x, y} \frac {\d y} {\d x} } = 0$

is an exact differential equation.

Unfortunately, there is no systematic method of finding such a $\map \mu {x, y}$ for all such equations $(1)$.^{[1]}

However, there are certain types of first order ODE for which an integrating factor *can* be found procedurally.

### Function of One Variable: $x$ or $y$ only

Suppose that:

- $g \left({x}\right) = \dfrac {\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x} } {N \left({x, y}\right)}$

is a function of $x$ only.

Then:

- $\mu \left({x}\right) = e^{\int g \left({x}\right) \mathrm d x}$

is an integrating factor for $(1)$.

Similarly, suppose that:

- $h \left({y}\right) = \dfrac {\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x} } {M \left({x, y}\right)}$

is a function of $y$ only.

Then:

- $\mu \left({y}\right) = e^{\int -h \left({y}\right) \mathrm d y}$

is an integrating factor for $(1)$.

### Function of $x + y$

Suppose that:

- $g \left({z}\right) = \dfrac {\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x} } {N \left({x, y}\right) - M \left({x, y}\right)}$

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

Then:

- $\mu \left({x + y}\right) = \mu \left({z}\right) = e^{\int g \left({z}\right) \mathrm d z}$

is an integrating factor for $(1)$.

### Function of $x y$

- $g \left({z}\right) = \dfrac {\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}} {N y - M x}$

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

Then:

- $\mu \left({x y}\right) = \mu \left({z}\right) = e^{\int g \left({z}\right) \mathrm d z}$

is an integrating factor for $(1)$.

## Proof

We have one of these:

- Integrating Factor for First Order ODE: Function of One Variable: $x$ or $y$ only
- Integrating Factor for First Order ODE: Function of $x + y$
- Integrating Factor for First Order ODE: Function of $x y$

### Conclusion of Proof

We have an equation of the form:

- $\dfrac 1 \mu \dfrac {\mathrm d \mu} {\mathrm d w} = f \left({w}\right)$

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

- $\dfrac {\mathrm d \left({\ln \mu}\right)}{\mathrm d w} = f \left({w}\right)$

Thus:

- $\displaystyle \ln \mu = \int f \left({w}\right) \mathrm d w$

and so:

- $\mu = e^{\int f \left({w}\right) \mathrm d w}$

Hence the results as stated.

$\blacksquare$

## Technique for finding an Integrating Factor

Suppose you have a first order ODE which is in (or can be manipulated into) the form:

- $M \left({x, y}\right) + N \left({x, y}\right) \dfrac {\mathrm d y} {\mathrm d x} = 0$

and it is not homogeneous, exact or linear.

Then what you can do is evaluate:

- $\dfrac {\partial M}{\partial y} - \dfrac {\partial N}{\partial x}$

and see what you get when you divide it by each of $N$, $M$, $N - M$ and $N y - M x$ in turn.

Then examine what you get to see if you have a function in $x$ only, $y$ only, $x + y$ or $xy$ respectively.

Suppose this has been achieved.

Then from Integrating Factor for First Order ODE, you have found an integrating factor and can solve the equation by using the technique defined in Solution to Exact Differential Equation.

## Examples

### $y \rd x + \paren {x^2 y - x} \rd y = 0$

The first order ODE:

- $(1): \quad y \rd x + \paren {x^2 y - x} \rd y = 0$

has the general solution:

- $\dfrac {y^2} 2 - \dfrac y x = C$

### $\paren {3 x^2 - y^2} \rd y - 2 x y \rd x = 0$

The first order ODE:

- $(1): \quad \paren {3 x^2 - y^2} \rd y - 2 x y \rd x = 0$

has the general solution:

- $\dfrac 1 y - \dfrac {x^2} {y^3} = C$

### $\paren {x y - 1} \rd x + \paren {x^2 - x y} \rd y = 0$

The first order ODE:

- $(1): \quad \paren {x y - 1} \rd x + \paren {x^2 - x y} \rd y = 0$

has the general solution:

- $x y - \ln x - \dfrac {y^2} 2 + C$

### $y \rd x + x \rd y + 3 x^3 y^4 \rd y = 0$

The first order ODE:

- $(1): \quad y \rd x + x \rd y + 3 x^3 y^4 \rd y = 0$

has the general solution:

- $-\dfrac 1 {2 x^2 y^2} + \dfrac {3 y^2} 2 = C$

## References

- ↑ "In general this is quite difficult."
- -- 1972: George F. Simmons:
*Differential Equations*: $\S 2.9$: Integrating Factors

- -- 1972: George F. Simmons:

## Sources

- 1972: George F. Simmons:
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