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Exploring Differential Equations An Interactive, Student-Centered Approach

Geoffrey W. Cox, Ph.D.

Section 6.3 The Integrating Factor

In the last section, we explored how rewriting the left-hand side of a differential equation as a single derivative (i.e., “completing the product rule”) made it easier to solve. But here’s the catch: most equations aren’t initially set up for this.
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For example, take the equation
\begin{equation*} \frac{dy}{dx} + 2y = 5. \end{equation*}
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It’s not a complete product rule, but it can be if we multiply both sides by \(e^{2x}\text{:}\)
\begin{equation*} e^{2x} \frac{dy}{dx} + 2e^{2x} y = 5e^{2x}\text{.} \end{equation*}
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Now, we can reverse the product rule on the left, giving us
\begin{equation*} \frac{d}{dx} \left[ e^{2x} y \right] = 5e^{2x}\text{,} \end{equation*}
which is just one integration away from the solution. The function, \(e^{2x}\text{,}\) that we multiplied onto the equation is known as the integrating factor.
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But, where did \(e^{2x}\) come from? Can we always find such a function? How?
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Subsection Finding the Right Function to Multiply

To figure out where \(e^{2x}\) came from, let’s return to our example:
\begin{equation*} \frac{dy}{dx} + 2y = 5. \end{equation*}
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To complete the product rule, we need to multiply the equation by a function. We don’t yet know what that function is, so let’s call it \(\mu(x)\) and multiply both sides of the equation by it:
\begin{equation*} \mu(x)\ \frac{dy}{dx} + 2\mu(x)\ y = 5\mu(x)\text{.} \end{equation*}
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If \(\mu(x)\) completes a product rule on the left, then it should match the form \(fg'+f'g\text{.}\) Setting the labels: \(g = y\text{,}\) so \(g' = dy/dx\text{,}\) and looking for functions \(f\) and \(f'\) that complete the pattern, we match as follows:
\begin{equation*} \underset{\Large f}{ \underset{\uparrow}{\ul{\mu(x)}}}\ \underset{\Large g'}{ \underset{\uparrow}{\ul{\vphantom{|}\frac{dy}{dx}}}} + \underset{\Large f^\prime}{ \underset{\uparrow}{\ul{2\mu(x)}}}\ \underset{\Large g}{ \underset{\uparrow}{\ul{\vphantom{|}y}}}\text{.} \end{equation*}
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This suggests that
\begin{equation*} f = \mu(x) \quad\text{and}\quad f' = 2\mu(x)\text{,} \end{equation*}
and taking the derivative of the first of these means we also have \(f' = \mu'(x)\text{.}\)
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Setting the two versions of \(f'\) equal, an interesting equation emerges:
\begin{equation} \mu'(x) = 2\mu(x)\text{.}\tag{6.2} \end{equation}
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That’s a differential equation. And since it’s separable, we know how to solve it:
\begin{align*} \frac{1}{\mu} \frac{d\mu}{dx} = 2 \quad \Rightarrow \quad \int \frac{1}{\mu}\ d\mu \amp = \int 2\ dx \\ \ln|\mu| \amp = 2x + c_1 \\ |\mu| \amp = e^{2x + c_1} \\ \mu \amp = ce^{2x} \text{.} \end{align*}
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📝: 📌 \(\pm c \to c\text{?}\)

When selecting an integrating factor, we are free to choose any nonzero value for \(c\text{.}\) However, we usually choose \(c = 1\) to keep things simple, so, our integrating factor is:
\begin{equation*} \mu(x) = e^{2x}. \end{equation*}
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This is the function we were looking for. It completes the product rule, allowing us to write the left-hand side of the equation as a single derivative and solve by integration.
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Checkpoint 68. Reading Questions.

(a) 📖❓ Integrating Factor Dependence.
In the example above, we derived the integrating factor for the equation
\begin{equation*} y' + 2y = 5\text{.} \end{equation*}
Based on this derivation, select the correct ending to the statement:
Of the values in the equation, the integrating factor depends on .
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  • only the value \(2\text{.}\)
  • Correct! The integrating factor depends only on the coefficient of \(y\text{,}\) which is \(2\text{.}\)
  • only the value \(5\text{.}\)
  • Incorrect. The integrating factor does not depend on the free term \(5\text{,}\) but rather on the coefficient of \(y\text{,}\) which is \(2\text{.}\)
  • both \(2\) and \(5\text{.}\)
  • Incorrect. The integrating factor does not depend on the free term \(5\text{,}\) but rather only on the coefficient of \(y\text{,}\) which is \(2\text{.}\)
  • neither \(2\) nor \(5\text{.}\)
  • Incorrect. The integrating factor does depend on the coefficient of \(y\text{,}\) which is \(2\text{.}\)
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(b) 📖❓ Integrating Factor Equation.
To solve the differential equation
\begin{equation*} y' + \frac{1}{x}y = x\text{,} \end{equation*}
you multiply both sides by the integrating factor, \(\mu(x)\text{,}\) to get
\begin{equation*} \mu(x) y' + \frac{\mu(x)}{x}y = \mu(x) x. \end{equation*}
Which separable differential equation do you solve to find \(\mu(x)\text{?}\)
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  • \(\quad\ds\mu'(x) = \frac{1}{x}\mu(x) \)
  • Correct!
  • \(\quad\ds\mu'(x) = x\mu(x) \)
  • Incorrect
  • \(\quad\ds\mu'(x) + \frac{\mu(x)}{x} = x \)
  • Incorrect. This is not a separable equation.
  • \(\quad\ds\mu'(x) = \frac{1}{x}y\mu(x) \)
  • Incorrect. This equation has too many variables.
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Subsection A Formula for Any Equation

In the last example, we figured out the integrating factor by treating the left-hand side as a product rule and labeling the parts. Let’s now extend that reasoning to any first-order linear differential equation.
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We’ll start with defining the standard form for these equations:
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📙 Definition 69. Standard Form (First-Order, Linear).

A differential equation is said to be in standard form if it can be written as:
\begin{equation*} \frac{dy}{dx} + P(x)\,y = Q(x), \end{equation*}
where \(P(x)\) and \(Q(x)\) are known functions of the independent variable \(x\text{,}\) and \(y\) is the unknown function (dependent variable) we want to find.
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Multiplying both sides of the general standard form by a function \(\mu\) leads us to the following formula for the integrating factor:
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✳️ Integrating Factor: General Formula.

For a first-order linear equation written in standard form
\begin{equation*} y' + P(x)\ y = Q(x) \end{equation*}
the integrating factor is given by \(\ \ds\mu(x) = e\vphantom{\sum}^{{\Large\int} P(x)\ dx} \text{.}\)
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Subsection Try Out the Formula

With the integrating factor formula in hand, let’s practice finding a few with some examples. Remember, the formula assumes the equation is in standard form. Once in this form, it is easy to identify \(P(x)\) as the coefficient of \(y\) and apply it to the formula.
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Find the integrating factor for the following differential equations:
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🌌 Example 72.

\(\quad\dfrac{dy}{dx} + 4y = x\)
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Solution.
The equation is in standard form and the coefficient of \(y\) is \(4\text{,}\) so \(P(x) = 4\) and the integrating factor is:
\begin{equation*} \mu(x) = e^{\large\int 4\ dx} = e^{4x}. \end{equation*}
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🌌 Example 73.

\(x\dfrac{dy}{dx} + 2y = x^4, \quad x > 0\)
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Solution.
This equation is not in standard form because of \(x\) in the coefficient of \(dy/dx\text{.}\) Dividing through by \(x\) (since \(x \neq 0\)) puts it in standard form:
\begin{equation*} \frac{dy}{dx} + \frac{2}{x}y = x^3. \end{equation*}
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Now, we can easily see that \(P(x) = 2/x\text{.}\) Plugging this into the formula:
\begin{equation*} \mu(x) = e^{\large\int \frac{2}{x}\ dx} = e^{2\ln |x|} = e^{\ln x^2} = x^2. \end{equation*}
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🌌 Example 74.

\(z^2\ \frac{dR}{dz} = \ln z - (1 - 3z)R,\quad z > 0\)
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Solution.
The variables in this equation are \(z\) and \(R\text{,}\) so the formula adjusts accordingly:
\begin{equation*} \mu(z) = e^{\large\int P(z)\, dz}\text{.} \end{equation*}
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Also, this equation is not in standard form, so move the \(R\) term to the left:
\begin{equation*} z^2\,\frac{dR}{dz} + (1 - 3z)R = \ln z\text{.} \end{equation*}
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Then divide through by \(z^2\) and note \(P(z)\) as the coefficient of \(R\text{:}\)
\begin{equation*} \frac{dR}{dz} + \os{\large P(z)}{\boxed{\left(\frac{1 - 3z}{z^2}\right)}}\ R = \frac{\ln z}{z^2} \end{equation*}
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Using \(P(z)\text{,}\) calculate the integral of \(P(z)\) used in the formula:
\begin{equation*} \int P(z)\ dz = \int \frac{1 - 3z}{z^2}\ dz \end{equation*}
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To integrate this, we can split the fraction and integrate each term:
\begin{gather*} \int \left( \frac{1}{z^2} - \frac{3}{z} \right) dz = -\frac{1}{z} - 3\ln|z|, \quad z > 0 \end{gather*}
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📝: 📌 Note.
Thus, the integrating factor is simplified and given as follows:
\begin{equation*} \mu(z) = e^{-\sfrac{1}{z} - 3\ln z} = e^{-\sfrac{1}{z}} \cdot e^{-3\ln z} = e^{-\sfrac{1}{z}} z^{-3}. \end{equation*}
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Checkpoint 75. 📖❓ Reading Questions.

(a) Integrating Factor Dependence.
Suppose you want to compute the integrating factor for the differential equation
\begin{equation*} -4\frac{dy}{dx} + 6xy = 15.5\text{.} \end{equation*}
After rewriting it in standard form, which parts of the equation does the integrating factor depend on?
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  • \(\quad 6x\)
  • Correct! The integrating factor depends on the coefficient of \(y\text{,}\) which is \(6x\text{.}\)
  • \(\quad -4\)
  • Correct! To put the equation in standard form, we divide by \(-4\text{,}\) so the integrating factor will depend on this coefficient.
  • \(\quad 15.5\)
  • Incorrect. The integrating factor does not depend on the free term.
  • \(\quad y\)
  • Incorrect. The integrating factor does not depend on \(y\text{,}\) but it does depend on the coefficient of \(y\text{.}\)
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(b) Give the Integrating Factor.
What is the integrating factor for the equation
\begin{equation*} y' + 3y = e^x\text{?} \end{equation*}
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  • \(\quad\ds\mu(x) = e^{e^{3x}}\)
  • Incorrect. This is not the correct expression for the integrating factor.
  • \(\quad\mu(x) = e^{3x}\)
  • Correct! The integrating factor is \(e^{3x}\text{.}\)
  • \(\quad\ds\mu(x) = e^{x^2/2}\)
  • Incorrect. The correct integrating factor should involve the integral of the coefficient of \(y\text{,}\) which is \(3\text{,}\) not \(x\text{.}\)
  • \(\quad\ds\mu(x) = e^{x}\)
  • Incorrect. This is not the correct form for the integrating factor.
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(c) Find the Integrating Factor (WebWorK Here For Mathquill?).
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(d) Fill-In the Integrating Factor 1 (WebWorK Here For Mathquill?).
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(e) Fill-In the Integrating Factor 2 (WebWorK Here For Mathquill?).
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