Author: John J Weber III, PhDCorresponding Textbook Sections:

Section 2.5 – Special Integrating Factors

Expected Educational Results

Objective 6–1: I can identify an ODE written in exact differential form.

Objective 6–2: I can determine if an ODE is an exact equation.

Objective 6–3: I can find the appropriate integrating factor to convert a DE in exact differential form into an exact equation.

Objection 6–4: I can solve ODEs in exact form by applying the appropriate integrating factor.

Integrating Factors

Definition: Integrating Factor

If the equation $M(x,y)dx+N(x,y)dy=0$ is not exact, but the equation $\mu (x,y)M(x,y)dx+\mu (x,y)N(x,y)dy=0$ for $\mu (x,y)\ne 0$is exact, then the function $\mu (x,y)$ is an integrating factor of the equation.

Method

Identify if the DE is exact:

Identify $M(x,y)$ and $N(x,y)$

Find $\frac{\partial M}{\partial y}(x,y)$ and $\frac{\partial N}{\partial x}(x,y)$

Test for exactness by verifying: $\frac{\partial M}{\partial y}(x,y)=\frac{\partial N}{\partial x}(x,y)$

If the DE is not exact, then

if $\frac{\partial M/\partial y-\partial N/\partial x}{N}$ is a function of only$x$, then the integrating factor is $\mu (x)={e}^{\int {\displaystyle \frac{\partial M/\partial y-\partial N/\partial x}{N}}{\textstyle \phantom{\rule{0.167em}{0ex}}}dx}$

if $\frac{\partial N/\partial x-\partial M/\partial y}{M}$ is a function of only$y$, then the integrating factor is $\mu (y)={e}^{\int {\displaystyle \frac{\partial N/\partial x-\partial M/\partial y}{M}}{\textstyle \phantom{\rule{0.167em}{0ex}}}dy}$

Multiply the DE with either $\mu (x)$ or $\mu (y)$.

Solve the exact equation.

State the solution curve as $F(x,y)=C$.

Identify missing solutions.

Identify solutions that are not solutions of the original DE.