Exact Differential Equations and Integrating Factors

TL;DR

flowchart TD
	ODE[Given an ODE that might be exact]
	IsExact{Determine if exact}

	ODE --> IsExact

	Solve[Apply solution method for exact differential equations]
	CheckR{Check R and R*}

	IsExact -->|If exact| Solve
	IsExact -->|If not exact| CheckR

	DetermineFactor[Find integrating factor]
	fail[Try other solution methods]

	CheckR -->|"If single-variable function R(x) or R*(y) exists"| DetermineFactor
	CheckR --->|If no single-variable integrating factor can be found| fail
	DetermineFactor --> Solve

Exact Differential Equations

A first-order ordinary differential equation $M(x,y)+N(x,y)y’=0$ can be written as:

\[M(x,y)dx+N(x,y)dy=0 \tag{1}\]

If

\[\exists u(x,y): \frac{\partial u}{\partial x}=M(x,y) \land \frac{\partial u}{\partial y}=N(x,y) \tag{2}\]

then

\[M(x,y)dx+N(x,y)dy=\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy=du \tag{3}\]

and the ODE $M(x,y)dx+N(x,y)dy=0$ is called an exact differential equation. In this case, the ODE can be written as:

\[du=0\]

Integrating this gives the general solution in the form:

\[u(x,y)=c \tag{4}\]

Identifying Exact Differential Equations

Consider a closed region in the xy-plane bounded by a closed curve that does not intersect itself, where $M$ and $N$ and their first-order partial derivatives are continuous. Looking at condition (2) again:

\[\begin{align*} \frac {\partial u}{\partial x}&=M(x,y) \tag{2a} \\ \frac {\partial u}{\partial y}&=N(x,y) \tag{2b} \end{align*}\]

Partially differentiating these equations:

\[\begin{align*} \frac {\partial M}{\partial y} &= \frac{\partial^2 u}{\partial y \partial x} \\ \frac {\partial N}{\partial x} &= \frac{\partial^2 u}{\partial x \partial y} \end{align*}\]

Given the assumed continuity, these two second-order partial derivatives are equal.

\[\therefore \frac {\partial M}{\partial y}=\frac {\partial N}{\partial x} \tag{5}\]

Thus, condition (5) is a necessary condition for ODE (1) to be an exact differential equation, and while not proven here, it is also a sufficient condition. Therefore, we can determine if an ODE is exact by checking if this condition is satisfied.

Solving Exact Differential Equations

Integrating equation (2a) with respect to x, treating y as a constant:

\[u = \int M(x,y) dx + k(y) \tag{6}\]

Here, $k(y)$ acts as an integration constant as y was treated as a constant. Now, treating x as a constant and differentiating equation (6) with respect to y to find $\partial u/\partial y$:

\[\frac{\partial u}{\partial y} = \frac{\partial}{\partial y}\int M(x,y) dx + \frac{dk}{dy}\]

Comparing this with equation (2b), we can find $dk/dy$:

\[\frac{\partial}{\partial y}\int M(x,y) dx + \frac{dk}{dy} = N(x,y)\] \[\frac{dk}{dy} = N(x,y) - \frac{\partial}{\partial y}\int M(x,y) dx\]

Finally, integrate this equation to determine $k(y)$, substitute it into equation (6) to find the implicit solution $u(x,y)=c$:

\[k(y) = \int N(x,y)dy - \int \left(\frac{\partial}{\partial y}\int Mdx\right)dy + c^*\] \[\int M(x,y)dx + \int N(x,y)dy - \int \left(\frac{\partial}{\partial y}\int Mdx\right)dy = c\]

It’s more important to understand the process of solving rather than memorizing this general solution formula as a rule.

Integrating Factors

Consider an inexact differential equation:

\[P(x,y)dx+Q(x,y)dy = 0 \quad \left( \frac {\partial P}{\partial y} \neq \frac {\partial Q}{\partial x} \right) \tag{7}\]

If

\[\exists F(x,y): \frac {\partial}{\partial y}(FP) = \frac {\partial}{\partial x}(FQ) \tag{8}\]

then multiplying the given ODE (7) by function $F$ yields the following exact differential equation:

\[FP\ dx+FQ\ dy = 0 \tag{9}\]

The function $F(x,y)$ is called an integrating factor of equation (7).

Method for Finding Integrating Factors

Applying the product rule to equation (8) and using subscript notation for partial derivatives:

\[F_y P + FP_y = F_x Q + FQ_x\]

In many practical cases, an integrating factor exists that depends on only one variable. If $F=F(x)$, then $F_y=0$ and $F_x=F’=dF/dx$, giving:

\[FP_y = F'Q + FQ_x\]

Dividing both sides by $FQ$ and rearranging:

\[\begin{align*} \frac{1}{F} \frac{dF}{dx} &= \frac{P_y}{Q} - \frac{Q_x}{Q} \\ &= \frac{1}{Q}\left(\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x} \right) \end{align*} \tag{10}\]

Therefore:

For a given ODE (7), if the right side of equation (10), $R$, is a function of x only, then equation (7) has an integrating factor $F=F(x)$.

\[F(x)=e^{\int R(x)dx}, \quad \text{where }R=\frac{1}{Q}\left(\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x} \right) \tag{11}\]

Similarly, if $F^=F^(y)$, instead of equation (10) we get:

\[\frac{1}{F^*} \frac{dF^*}{dy} = \frac{1}{P}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right) \tag{12}\]

Therefore:

For a given ODE (7), if the right side of equation (12), $R^$, is a function of y only, then equation (7) has an integrating factor $F^=F^*(y)$.

\[F^*(y)=e^{\int R^*(y)dy}, \quad \text{where }R^*=\frac{1}{P}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right) \tag{13}\]