Integrating Factor for Exact Equation

An integrating factor that converts a First-order ODE into an exact equation if it meets certain conditions.

Theory

Consider a non-exact differential equation in the form

$$M(x,y)dx+N(x,y)dy=0$$

Then we also have

$$\frac{\partial M}{\partial y}\ne \frac{\partial N}{\partial x}$$

But, let’s assume that we can multiply both functions by some integrating factor $\mu (x,y)$ such that

$$\frac{\partial \mu M}{\partial y}=\frac{\partial \mu N}{\partial x}$$

Applying the product rule (using partial derivatives)

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

Factoring

$$\mu \left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)=N\frac{\partial \mu }{\partial x}-M\frac{\partial \mu }{\partial y}$$

While $M,N$ and its derivatives are known, it is difficult to compute $\mu$ in this form, so we have two cases.

Case 1: ${\mu }_x=\frac{du}{dx}$

or in human terms, $\mu$ is a function of only $x$.

It is then clear the ${\mu }_y=0$ and

$$\mu \left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)=N\frac{d\mu }{dx}e$$

Now, in order to proceed, we must declare a condition: the right-hand side must be a function of $x$ only. Thus, we can write

$$\frac{1}{N}\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)=f(x)$$

Then, this is a first-order separable and linear differential equation, we can write the solution using the integrating factor for product rule:

$$\mu (x)=e^{\int \frac{M_y-N_x}{N}dx}$$

Case 2: ${\mu }_y=\frac{du}{dy}$

A similar method can yield

$$\mu (y)=e^{\int \frac{N_x-M_y}{M}dy}$$

Where the condition is that $\frac{1}{M}\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)$ is a function of $y$ only.

Solution Summary

For a non-exact equation:

If $\frac{N_x-M_y}{M}$ is a function of only $y$

$$\mu (y)=e^{\int \frac{N_x-M_y}{M}dy}$$

If $\frac{M_y-N_x}{N}$ is a function of only $x$

$$\mu (x)=e^{\int \frac{M_y-N_x}{N}dx}$$

Example

For a nonlinear first-order differential equation

$$xydx+(2x^2+3y^2-20)dy=0$$

With $M=xy$ and $N=2x^2+3y^2-20$

Verifying that this is non-exact

$$\frac{\partial M}{\partial y}=x\ne \frac{\partial N}{\partial x}=4x$$

Considering case 2

$$\frac{M_y-N_x}{N}=\frac{-3x}{2x^2+3y^2-20}$$

it’s clear that this is not a pure function of $x$

Considering case 1

$$\frac{N_x-M_y}{M}=\frac{4x-x}{xy}=\frac{3}{y}$$

this is a pure function of $y$, thus we can define the integrating factor

$$I=e^{\int \frac{3}{y}dy}=e^{3\ln y}=y^3$$

After applying the integrating factor, we get

$$x{y}^{4}dx+(2x^2{y}^3+3y^5-20y^3)dy=0$$

And verifying the partial derivatives yields

$$\frac{\partial M}{\partial y}=4x{y}^3$$ $$\frac{\partial N}{\partial x}=4x{y}^3$$

So this is now an exact equation