Section 5.1 Product Rule
The integrating factor method relies on an essential calculus concept: the product rule for derivatives. Since this rule is at the heart of the method, let’s take a moment to review it and see how it helps transform a differential equation into one that can be directly integrated.
The product rule is used to compute the derivative of a product of two functions. In the context of solving firstorder linear differential equations, we use the product rule in reverse to combine terms involving the dependent and independent variables. This allows us to rewrite the equation as the derivative of a product, making integration straightforward. Let’s begin by recalling the two directions of the product rule.
Suppose \(f\) and \(g\) are differentiable functions. The product rule states that
\begin{equation}
\underset{\xrightarrow[\LARGE\text{forward}]{} }{\ds [f \cdot g ]'} =
\underset{\xleftarrow[\LARGE\text{backward}]{} }{\ds f \cdot g' + f' \cdot g}.\tag{6}
\end{equation}
The “forward” version goes from left to right, while the “backward” version, which we’ll focus on, goes from right to left. Let’s practice applying both versions with a few examples.
Example 43.
\(\ \ \) Compute each derivative.

\(\displaystyle \left[\ln x \tan x\right]' = \)
Solution.
\begin{align*}
\big[
\overset{\color{blue}{f}}{\ln x \vphantom{\big]'}}\
\overset{\color{blue}{g}}{\tan x \vphantom{\big]'}}
\big]' \amp =
\overset{\color{blue}{f}}{\ln x \vphantom{\big]'}}\
\overset{\color{blue}{g'}}{\big[\tan x\big]'} +
\overset{\color{blue}{f'}}{\big[\ln x \big]'}
\overset{\color{blue}{g}}{\tan x \vphantom{\big]'}} \\
\amp = \ln x (\sec^2 x)+\left(\frac{1}{x}\right) \tan x.
\end{align*}

\(\displaystyle \ds \frac{d}{dx}\left[x^3 \mu\right]' = \)
Solution.
\begin{gather*}
\left[x^3 \mu\right]' = x^3 \mu' + 3x^2 \mu.
\end{gather*}
Example 44.
\(\ \ \) Rewrite each expression in the form \([f \cdot g]' \text{.}\)
\(e^x\cos x + \sin x\ e^x = \big[\ \us{?}{\ul{\qquad}} \cdot \us{?}{\ul{\qquad}}\ \big]'\) Solution.
Trial and error with different combinations of \(e^x, \sin x, \) and \(\cos x\) inside \(\big[\ \us{?}{\ul{\qquad}} \cdot \us{?}{\ul{\qquad}}\ \big]'\) is one approach, but a little thought should lead you to the answer
\begin{equation*}
e^x\cos x + \sin x\ e^x = e^x[\sin x]' + \sin x [e^x]' = [e^x\sin x]'\text{.}
\end{equation*}
\(\ds\frac{\mu}{x} + \mu' \ln x = \frac{d}{dx}\big[\ \us{?}{\ul{\qquad}} \cdot \us{?}{\ul{\qquad}}\ \big], \quad\) where
\(\mu\) is a function of
\(x\text{.}\) Solution.
It may help to first rewrite the expression as
\begin{equation*}
\mu \cdot\frac{1}{x} + \mu' \ln x\text{.}
\end{equation*}
Since \(\mu\) and \(\mu'\) appear in different terms, you know
\begin{equation*}
\frac{\mu}{x} + \mu' \ln x = \big[\ \mu \cdot \us{?}{\ul{\qquad}}\ \big]'
\end{equation*}
and since \([\ln x]' = 1/x\text{,}\) we must have
\begin{equation*}
\frac{\mu}{x} + \mu' \ln x = \big[\ \mu \cdot \ln x \ \big]'\text{.}
\end{equation*}
Finally, let’s see what this looks like on an actual differential equation.
Example 45.
\(\ \ \) Rewrite the left side of the equation in the form \(\frac{d}{dx}[f \cdot g] \text{.}\)
\begin{equation}
\frac{dy}{dx} \cdot e^{5x^2} + 10xe^{5x^2}\cdot y = 3x\text{.}\tag{7}
\end{equation}
Solution.
As before, noticing \(y\) and \(dy/dx\) in different terms of
\begin{equation*}
\frac{dy}{dx} \cdot e^{5x^2} + 10xe^{5x^2}\cdot y = 3x
\end{equation*}
immediately tells us that the left side can be rewritten as
\begin{equation*}
\frac{d}{dx}\big[\ \us{?}{\ul{\qquad}} \cdot y\ \big]' = 3x
\end{equation*}
and the observation that \([e^{5x^2}]' = 10xe^{5x^2}\) gives us the other term. Therefore, the differential equation can be rewritten as
\begin{equation*}
\frac{d}{dx}\big[\ e^{5x^2} \cdot\ y \big]' = 3x\text{.}
\end{equation*}
Do you see that direct integration can be used here to solve the differential equation?
By understanding how the product rule works in both directions, we can transform a differential equation into one that is easy to solve through integration. This transformation is the core of the integrating factor method, which we will explore next.
Reading Questions Check your Understanding
1. Fillin the missing parts of each product rule.
2. What is the main purpose of the product rule in the integrating factor method?
What is the main purpose of the product rule in the integrating factor method?
 To simplify a differential equation.
Incorrect. Simplification is a result, but the main purpose is to rewrite the equation as the derivative of a product.
 To rewrite the differential equation so that one side of the equation is the derivative of a product.
Correct! The product rule is used to express the differential equation in a form suitable for direct integration.
 To find the derivative of the solution.
Incorrect. The product rule helps us manipulate the equation into a solvable form, not to find the derivative of the solution.
 To eliminate constants from the equation.
Incorrect. The product rule does not eliminate constants; it helps combine terms involving the dependent and independent variables.
3. Which version of the product rule is used in the integrating factor method?
Which version of the product rule is used in the integrating factor method?
 The forward version.
Incorrect. The forward version computes the derivative of a product, but the backward version is used to reverse this process.
 The backward version.
Correct! The backward version is used to combine terms and express the equation as the derivative of a product.
 Neither version is used.
Incorrect. The product rule is essential in the integrating factor method.
 Both versions are equally used.
Incorrect. While both directions are important, the integrating factor method specifically uses the backward version.
4. How does the product rule help solve the equation \(\ds\frac{dy}{dx} \cdot e^{5x^2} + 10x e^{5x^2} \cdot y = 3x\text{?}\)
How does the product rule help solve the equation \(\ds\frac{dy}{dx} \cdot e^{5x^2} + 10x e^{5x^2} \cdot y = 3x\text{?}\)
 It eliminates the dependent variable, \(y\text{.}\)
Incorrect. The product rule does not eliminate \(y\text{,}\) but rather combines terms involving \(y\) and \(dy/dx\text{.}\)
 It rewrites the left side of the equation as \(\ds\frac{d}{dx}\left[e^{5x^2} \cdot y\right]\text{.}\)
Correct! The product rule helps combine terms so that the equation can be rewritten as a derivative of a product, which can then be integrated.
 It changes the equation to \(y = 3x\text{.}\)
Incorrect. The product rule does not solve for \(y\) directly; it helps rewrite the equation in a form suitable for solving.
 It differentiates both sides of the equation.
Incorrect. The product rule is used in reverse here to combine terms, not to take derivatives.
5. Which expression can be rewritten as \(\ds [\mu e^{P(x)}]'\) using the product rule?
Which expression can be rewritten as \(\ds [\mu e^{P(x)}]'\) using the product rule?
 \(\mu' e^{P(x)} + P'(x) \mu e^{P(x)} \)
Correct! Using the product rule, \([\mu e^{P(x)}]'\) expands to \(\mu' e^{P(x)} + P'(x) \mu e^{P(x)} \text{.}\)
 \(\mu e^{P(x)} + P'(x) e^{P(x)} \mu \)
Incorrect. The second term should include the derivative of \(\mu\text{,}\) not just \(P'(x) e^{P(x)}\text{.}\)
 \(r \mu e^{P(x)} + \mu e^{P(x)} \)
Incorrect. This does not account for the derivative of \(\mu\text{,}\) which is needed for the correct application of the product rule.
 \(\mu' e^{P(x)} + e^{x} \mu \)
Incorrect. The exponent of \(e\) should be \(P(x)\) in both terms, not \(x\) in the second term.
6. How can \(\ \ds\frac{dQ}{dz}\ln z + \frac{Q}{z}\ \) be rewritten using the backward product rule?
How can \(\ \ds\frac{dQ}{dz}\ln z + \frac{Q}{z}\ \) be rewritten using the backward product rule?
 \(\ds\frac{d}{dz} \left[ Q z \right]\)
Incorrect. You need to include \(\ln z \) in the product, not just \(z\text{.}\)
 \(\ds\frac{d}{dz} \left[ \ln z \right]\)
Incorrect. This only gives the derivative of \(\ln z \text{,}\) which is \(1/z \text{,}\) but doesn’t account for \(Q\) and its derivative.
 \(\ds\frac{d}{dz} \left[ Q \ln z \right]\)
Correct! Using the backward product rule, the expression \(\ln z \frac{dQ}{dz} + \frac{Q}{z} \) can be written as the derivative \(\frac{d}{dz} \left[ Q \ln z \right] \text{.}\)
 \(\ds\ln z + \frac{d}{dz} \left[ Q z \right]\)
Incorrect. This expression incorrectly separates the logarithmic and product rule components. The product rule should involve \(\ln z \) and \(Q \) together.
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