Integrating Factors

Author: John J Weber III, PhD Corresponding 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$$M(x,y)dx+N(x,y)dy=0$ is not exact, but the equation $\mu (x,y)M(x,y)dx+\mu (x,y)N(x,y)dy=0$$\mu(x,y)M(x,y)dx+\mu(x,y)N(x,y)dy=0$ for $\mu (x,y)\ne 0$$\mu(x,y)\ne 0$ is exact, then the function $\mu (x,y)$$\mu(x,y)$ is an integrating factor of the equation.

Derivation

We know from the Test for Exactness that $\mu (x,y)M(x,y)dx+\mu (x,y)N(x,y)dy=0$$\mu(x,y)M(x,y)dx+\mu(x,y)N(x,y)dy=0$ is an exact equation if and only if ${\displaystyle {\displaystyle \frac{\partial [\mu (x,y)M(x,y)]}{\partial y}}={\displaystyle \frac{\partial [\mu (x,y)N(x,y)]}{\partial x}}}$$\displaystyle\dfrac{\partial\left[\mu(x,y)M(x,y)\right]}{\partial y}=\dfrac{\partial\left[\mu(x,y)N(x,y)\right]}{\partial x}$

Using the product rule for derivatives:

${\displaystyle M(x,y)\frac{\partial \mu (x,y)}{\partial y}+\mu (x,y)\frac{\partial M(x,y)}{\partial y}=N(x,y)\frac{\partial \mu (x,y)}{\partial x}+\mu (x,y)\frac{\partial N(x,y)}{\partial x}}$$\displaystyle {\color{gsured}{M(x,y)\frac{\partial\mu(x,y)}{\partial y}}}+{\color{gsured!50!red}{\mu(x,y)}}\frac{\partial M(x,y)}{\partial y}={\color{gsured!70!gsublue}{N(x,y)\frac{\partial\mu(x,y)}{\partial x}}}+{\color{gsured!50!red}{\mu(x,y)}}\frac{\partial N(x,y)}{\partial x}$

Rewriting to isolate for $\mu (x,y)$$\mu(x,y)$:

${\displaystyle M(x,y)\frac{\partial \mu (x,y)}{\partial y}-N(x,y)\frac{\partial \mu (x,y)}{\partial x}=(\frac{\partial N(x,y)}{\partial x}-\frac{\partial M(x,y)}{\partial y})\mu (x,y)}$$\displaystyle {\color{gsured}{M(x,y)\frac{\partial\mu(x,y)}{\partial y}}}-{\color{gsured!70!gsublue}{N(x,y)\frac{\partial\mu(x,y)}{\partial x}}}=\left(\frac{\partial N(x,y)}{\partial x}-\frac{\partial M(x,y)}{\partial y}\right){\color{gsured!50!red}{\mu(x,y)}}$

Solving for $\mu (x)$$\mu(x)$

Suppose the integrating is a function of only $x$$x$, i.e., $\mu (x,y)=\mu (x)$$\mu(x,y)=\mu(x)$:

${\displaystyle M(x,y)\frac{\partial \mu (x)}{\partial y}-N(x,y)\frac{\partial \mu (x)}{\partial x}=(\frac{\partial N(x,y)}{\partial x}-\frac{\partial M(x,y)}{\partial y})\mu (x)}$$\displaystyle {\color{gsured}{M(x,y)\frac{\partial\mu(x)}{\partial y}}}-{\color{gsured!70!gsublue}{N(x,y)\frac{\partial\mu(x)}{\partial x}}}=\left(\frac{\partial N(x,y)}{\partial x}-\frac{\partial M(x,y)}{\partial y}\right){\color{gsured!50!red}{\mu(x)}}$

We know from Calculus III: ${\displaystyle \frac{\partial \mu (x)}{\partial y}=0}$$\displaystyle{\color{gsured}{\frac{\partial\mu(x)}{\partial y}=0}}$ and ${\displaystyle \frac{\partial \mu (x)}{\partial x}=\frac{d\mu (x)}{dx}}$$\displaystyle{\color{gsured!70!gsublue}{\frac{\partial\mu(x)}{\partial x}}}={\color{gsured!70!gsublue}{\frac{d\mu(x)}{dx}}}$. Thus,

${\displaystyle 0-N(x,y)\frac{d\mu (x)}{dx}=(\frac{\partial N(x,y)}{\partial x}-\frac{\partial M(x,y)}{\partial y})\mu (x)}$$\displaystyle {\color{gsured}{0}}-{\color{gsured!70!gsublue}{N(x,y)\frac{d\mu(x)}{dx}}}=\left(\frac{\partial N(x,y)}{\partial x}-\frac{\partial M(x,y)}{\partial y}\right){\color{gsured!50!red}{\mu(x)}}$

Solve for ${\displaystyle \frac{d\mu (x)}{dx}}$$\displaystyle{\color{gsured!70!gsublue}{\frac{d\mu(x)}{dx}}}$:

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

Rewriting the expression in the parenthesis:

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

which is separable if and only if ${\displaystyle \left({\displaystyle \frac{{\displaystyle \frac{\partial M(x,y)}{\partial y}}-{\displaystyle \frac{\partial N(x,y)}{\partial x}}}{N(x,y)}}\right)}$$\displaystyle\left(\dfrac{\dfrac{\partial M(x,y)}{\partial y}-\dfrac{\partial N(x,y)}{\partial x}}{{N(x,y)}}\right)$ is a function of only $x$$x$. That is,

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

and

${\displaystyle \mu (x)=e^{\int \left({\displaystyle \frac{{\displaystyle \frac{\partial M(x,y)}{\partial y}}-{\displaystyle \frac{\partial N(x,y)}{\partial x}}}{N(x,y)}}\right)dx}}$$\displaystyle \mu(x)=e^{{\int{\left(\dfrac{\dfrac{\partial M(x,y)}{\partial y}-\dfrac{\partial N(x,y)}{\partial x}}{N(x,y)}\right)\,dx}}}$

Solving for $\mu (y)$$\mu(y)$

Suppose the integrating is a function of only $x$$x$, i.e., $\mu (x,y)=\mu (y)$$\mu(x,y)=\mu(y)$:

${\displaystyle M(x,y)\frac{\partial \mu (y)}{\partial y}-N(x,y)\frac{\partial \mu (y)}{\partial x}=(\frac{\partial N(x,y)}{\partial x}-\frac{\partial M(x,y)}{\partial y})\mu (y)}$$\displaystyle {\color{gsured}{M(x,y)\frac{\partial\mu(y)}{\partial y}}}-{\color{gsured!70!gsublue}{N(x,y)\frac{\partial\mu(y)}{\partial x}}}=\left(\frac{\partial N(x,y)}{\partial x}-\frac{\partial M(x,y)}{\partial y}\right){\color{gsured!50!red}{\mu(y)}}$

We know from Calculus III: ${\displaystyle \frac{\partial \mu (y)}{\partial x}=0}$$\displaystyle{\color{gsured!70!gsublue}{\frac{\partial\mu(y)}{\partial x}=0}}$ and ${\displaystyle \frac{\partial \mu (y)}{\partial y}=\frac{d\mu (y)}{dy}}$$\displaystyle{\color{gsured}{\frac{\partial\mu(y)}{\partial y}}}={\color{gsured}{\frac{d\mu(y)}{dy}}}$. Thus,

${\displaystyle M(x,y)\frac{d\mu (y)}{dy}-0=(\frac{\partial N(x,y)}{\partial x}-\frac{\partial M(x,y)}{\partial y})\mu (y)}$$\displaystyle {\color{gsured}{M(x,y)\frac{d\mu(y)}{dy}}}-{\color{gsured!70!gsublue}{0}}=\left(\frac{\partial N(x,y)}{\partial x}-\frac{\partial M(x,y)}{\partial y}\right){\color{gsured!50!red}{\mu(y)}}$

Solve for ${\displaystyle \frac{d\mu (y)}{dy}}$$\displaystyle{\color{gsured}{\frac{d\mu(y)}{dy}}}$:

${\displaystyle {\displaystyle \frac{d\mu (y)}{dy}}=\left({\displaystyle \frac{{\displaystyle \frac{\partial N(x,y)}{\partial x}}-{\displaystyle \frac{\partial M(x,y)}{\partial y}}}{M(x,y)}}\right)\mu (y)}$$\displaystyle {\dfrac{d\mu(y)}{dy}}=\left(\dfrac{\dfrac{\partial N(x,y)}{\partial x}-\dfrac{\partial M(x,y)}{\partial y}}{{M(x,y)}}\right){\mu(y)}$

which is separable if and only if ${\displaystyle \left({\displaystyle \frac{{\displaystyle \frac{\partial N(x,y)}{\partial x}}-{\displaystyle \frac{\partial M(x,y)}{\partial y}}}{M(x,y)}}\right)}$$\displaystyle\left(\dfrac{\dfrac{\partial N(x,y)}{\partial x}-\dfrac{\partial M(x,y)}{\partial y}}{{M(x,y)}}\right)$ is a function of only $y$$y$. That is,

${\displaystyle {\displaystyle \frac{d\mu (y)}{\mu (y)}}=\left({\displaystyle \frac{{\displaystyle \frac{\partial N(x,y)}{\partial x}}-{\displaystyle \frac{\partial M(x,y)}{\partial y}}}{M(x,y)}}\right)dy}$$\displaystyle \dfrac{{d\mu(y)}}{{\mu(y)}}=\left(\dfrac{\dfrac{\partial N(x,y)}{\partial x}-\dfrac{\partial M(x,y)}{\partial y}}{{M(x,y)}}\right){dy}$

and

${\displaystyle \mu (y)=e^{\int \left({\displaystyle \frac{{\displaystyle \frac{\partial N(x,y)}{\partial x}}-{\displaystyle \frac{\partial M(x,y)}{\partial y}}}{M(x,y)}}\right)dy}}$$\displaystyle \mu(y)=e^{{\int{\left(\dfrac{\dfrac{\partial N(x,y)}{\partial x}-\dfrac{\partial M(x,y)}{\partial y}}{M(x,y)}\right)\,dy}}}$

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Last Modified: Monday, 6 September 2020 13:33 EDT