DE-BOOK The Integrating Factor

Section 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

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\begin{equation*} \frac{dy}{dx} + 2y = 5. \end{equation*}

As it is, it’s not a complete product rule, but it can be if we multiply both sides by \(e^{2x}\text{:}\)

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\begin{equation*} e^{2x} \frac{dy}{dx} + 2e^{2x} y = 5e^{2x}\text{.} \end{equation*}

Now, we can reverse product rule on the left, giving us

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\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. And the natural question is:

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Where did \(e^{2x}\) come from? Can we always find such a function? How?

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This section answers these questions and prepares you for next solution technique, the integrating factor method.

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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:

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\begin{equation*} \frac{dy}{dx} + 2y = 5. \end{equation*}

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:

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\begin{equation*} \mu(x)\,\frac{dy}{dx} + 2\mu(x)\,y = 5\mu(x). \end{equation*}

If \(\mu(x)\) completes a product rule on the left, then if 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:

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\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*}

This suggests that

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\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:

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\begin{equation} \mu'(x) = 2\mu(x)\text{.}\tag{17} \end{equation}

That’s a differential equation. And since it’s separable, we know exactly how to solve it:

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📝: 📌 Note.

\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(x) \amp = ce^{2x} \text{.} \end{align*}

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:

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\begin{equation*} \mu(x) = e^{2x}. \end{equation*}

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 78. .

(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 79. Standard Form (First-Order, Linear).

A differential equation is said to be in standard form if it can be written as:

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\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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🧠 Derivation 80. Where does the Standard Form come from?

All first-order linear differential equations have a \(y'\text{,}\) \(y\text{,}\) and free term. Therefore, they can be written as

\begin{equation*} A(x)y' + B(x)y = C(x)\text{.} \end{equation*}

Dividing both sides by the leading coefficient, \(A(x)\text{,}\) and renaming the \(y\) and constant coefficients as \(P\) and \(Q\text{,}\) gives the standard form

\begin{equation*} y' + \us{P(x)}{\ub{\frac{B(x)}{A(x)}}}y = \us{Q(x)}{\ub{\frac{C(x)}{A(x)}}} \quad \implies \quad y' + P(x)y = Q(x)\text{.} \end{equation*}

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Our goal is to multiply both sides of a standard-form equation by a function \(\mu(x)\) so that the left-hand side becomes the derivative of a product.

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Just like before, we assume that multiplying by \(\mu(x)\) gives us something that matches the product rule. So we try:

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\begin{equation*} \mu(x)\,\frac{dy}{dx} + \mu(x)\,P(x)\,y = \mu(x)\,Q(x). \end{equation*}

For the left-hand side to be a product rule, we need it to match: \(fg'+f'g\text{.}\)

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As before, we set \(g = y\) and \(g' = dy/dx\text{.}\) Then the rest of the labels follow:

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\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{\mu(x)\,P(x)}}}\ \underset{\Large g}{ \underset{\uparrow}{\ul{\vphantom{|}y}}} \end{equation*}

Since \(f = \mu(x)\text{,}\) we also know \(f' = \mu'(x)\text{.}\) Setting these equal gives the condition:

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\begin{equation} \mu'(x) = P(x)\,\mu(x)\tag{18} \end{equation}

This is exactly the same kind of differential equation we solved earlier, just with a general function \(P(x)\) instead of a constant. It’s still separable, and solving it gives us the general formula for the integrating factor:

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\begin{align*} \frac{1}{\mu} \frac{d\mu}{dx} = P(x) \quad \Rightarrow \quad \int \frac{1}{\mu} \, d\mu \amp = \int P(x) \, dx \\ \ln|\mu| \amp = \int P(x)\, dx + C \\ \mu(x) \amp = ce^{\int P(x)\, dx} \end{align*}

Just like before, any nonzero constant \(c\) works, but we typically choose \(c = 1\) to simplify our computations. This gives us a general 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 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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🌌 Example 81. Constant Coefficient.

Find the integrating factor for

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\begin{equation*} \frac{dy}{dx} + 4y = x. \end{equation*}

Solution.

The equation is in standard form and the coefficient of \(y\) is \(4\text{,}\) so \(P(x) = 4\) and the integrating factor is:

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\begin{equation*} \mu(x) = e^{\large\int 4\,dx} = e^{4x}. \end{equation*}

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🌌 Example 82. Logarithmic Result.

Find the integrating factor for

\begin{equation*} x\frac{dy}{dx} + 2y = x^4, \quad x > 0\text{.} \end{equation*}

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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:

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\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 83. Non-Standard Form.

Find the integrating factor for

\begin{equation*} z^2\,\frac{dR}{dz} = \ln z - (1 - 3z)R, \quad z > 0\text{.} \end{equation*}

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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:

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\begin{equation*} z^2\,\frac{dR}{dz} + (1 - 3z)R = \ln z\text{.} \end{equation*}

Then divide through by \(z^2\) and note \(P(z)\) as the coefficient of \(R\text{:}\)

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\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*}

Using \(P(z)\text{,}\) calculate the integral of \(P(z)\) used in the formula:

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\begin{equation*} \int P(z)\ dz = \int \frac{1 - 3z}{z^2}\ dz \end{equation*}

To integrate this, we can split the fraction and integrate each term:

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📝: 📌 Note.

\begin{gather*} \int \left( \frac{1}{z^2} - \frac{3}{z} \right) dz = -\frac{1}{z} - 3\ln|z|, \quad z > 0 \end{gather*}

Thus, the integrating factor is simplified and given as follows:

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\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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Subsection 📤 Wrap-Up

🗝️ \(\textbf{Key Takeaways...}\)

The integrating factor \(\mu(x)\) is the function we multiply onto the entire equation to complete the product rule.

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For an equation in standard form, \(\ y' + P(x)y = Q(x)\ \text{,}\) a valid integrating factor, \(\mu(x)\text{,}\) can be found by solving the separable differential equation:
\begin{equation*} \mu'(x) = P(x)\mu(x)\text{.} \end{equation*}

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Solving gives the general formula:
\begin{equation*} \mu(x) = e^{\int P(x)\, dx}\text{.} \end{equation*}

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Check Your Understanding.

Checkpoint 84. 🤔💭 Separation of Variables Reading Questions.

(a) Integrating Factor Dependence.

Suppose you want to compute 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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\(\quad\ds\mu(x) = e^{e^{3x}}\)
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Incorrect. This is not the correct expression for the integrating factor.
\(\quad\mu(x) = e^{3x}\)
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Correct! The integrating factor is \(e^{3x}\text{.}\)
\(\quad\ds\mu(x) = e^{x^2/2}\)
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Incorrect. The correct integrating factor should involve the integral of the coefficient of \(y\text{,}\) which is \(3\text{,}\) not \(x\text{.}\)
\(\quad\ds\mu(x) = e^{x}\)
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Incorrect. This is not the correct form for the integrating factor.

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You have attempted of activities on this page.

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Section The Integrating Factor

Subsection Finding the Right Function to Multiply

📝: 📌 Note.

Checkpoint 78. .

(a) 📖❓ Integrating Factor Dependence.

(b) 📖❓ Integrating Factor Equation.

Subsection A Formula for Any Equation

📙 Definition 79. Standard Form (First-Order, Linear).

🧠 Derivation 80. Where does the Standard Form come from?

✳️ Integrating Factor: General Formula.

Subsection Try Out the Formula

🌌 Example 81. Constant Coefficient.

🌌 Example 82. Logarithmic Result.

🌌 Example 83. Non-Standard Form.

📝: 📌 Note.

Subsection 📤 Wrap-Up

Check Your Understanding.

Checkpoint 84. 🤔💭 Separation of Variables Reading Questions.

(a) Integrating Factor Dependence.

(b) Give the Integrating Factor.

(c) Find the Integrating Factor (WebWorK Here For Mathquill?).

(d) Fill-In the Integrating Factor 1 (WebWorK Here For Mathquill?).

(e) Fill-In the Integrating Factor 2 (WebWorK Here For Mathquill?).