\begin{equation*}
1. \quad \left[x^6 e^{3x}\right]' \hspace{5em} 2. \quad \frac{d}{dx}\left[e^{x^3} y\right]
\end{equation*}
Section Completing the Product Rule
In algebra, a clever rearrangement of terms can make an equation easier to solve. For instance, consider the quadratic:
\begin{equation*}
x^2 + 10x + 25 = 0.
\end{equation*}
Factoring the left-hand side gives:
\begin{equation}
(x + 5)^2 = 0.\tag{13}
\end{equation}
This grouping is helpful because taking the square root immediately reveals the solution.
Some first-order differential equations benefit from a similar idea. Instead of forming a single perfect square, we aim to group the left-hand side as a single derivative.
For example, take the equation:
\begin{equation}
\frac{dy}{dx} + 2 y = 5\text{.}\tag{14}
\end{equation}
It can be rewritten as:
\begin{equation}
\frac{d}{dx} [e^{2x} y] = 5e^{2x}\text{.}\tag{15}
\end{equation}
Just like factoring helped earlier, this new form makes solving easier, only now we integrate instead of take square roots.
Weβll call this process βcompleting the product ruleβ. Itβs based on reversing the product rule from calculus and serves as the foundation of the integrating factor method.
Subsection The Product Rule, A Quick Review
Before we reverse the product rule, letβs make sure we remember how it works in the forward direction.
In calculus, the product rule tells us how to take the derivative of two multiplied functions. If \(f(x)\) and \(g(x)\) are differentiable, then
\begin{equation*}
\frac{d}{dx} \left[ f(x) \cdot g(x) \right] = f(x) \cdot g'(x) + f'(x) \cdot g(x)\text{.}
\end{equation*}
To refresh your memory of how it works, try the following practice problems.
Checkpoint 70. πβ Remembering the Product Rule.
Subsection The Product Rule in Reverse
Now that weβve recalled how the product rule works, letβs try running it in reverse.
Suppose youβre given the expression:
\begin{equation}
\frac{1}{x} \sin x + \cos x \ln x\text{.}\tag{16}
\end{equation}
Can this be written as the derivative of a product? That is, can you find functions \(f(x)\) and \(g(x)\) so that
\begin{equation*}
\frac{d}{dx} \left[ f(x) g(x) \right] = \frac{1}{x} \sin x + \cos x \ln x?
\end{equation*}
Hereβs a tip: the product rule always combines four pieces, two functions and their derivatives. If you can spot \(f\) and \(g\) in the terms, the rest should fall into place. In this case, trying \(f(x) = \sin x\) and \(g(x) = \ln x\) works out perfectly.
Letβs get some practice with this reverse thinking.
π Example 71. Basic Reverse Product Rule.
Rewrite each expression as a derivative of a product:
\begin{equation*}
1. \quad e^{x}\cos x + e^{x}\sin x \hspace{5em} 2. \quad \frac{e^{x^2}}{x} + 2 x \ln x e^{x^2}
\end{equation*}
Solution.
Solution 1.
\begin{equation*}
e^x\
\underset{\Large f}{ \underset{\uparrow}{\ul{\vphantom{|}\cos x}}}
+
\underset{\Large g}{ \underset{\uparrow}{\ul{\vphantom{|}e^x}}}\
\sin x\text{.}
\end{equation*}
For this to work, \(f'\) should be \(\sin x\text{,}\) next to our chosen \(g\text{.}\) However,
\begin{equation*}
f' = [\cos x]' = -\sin x \ne \sin x \quad β
\end{equation*}
So this labeling doesnβt work and we instead try the labels
\begin{equation*}
\underset{\Large f}{ \underset{\uparrow}{\ul{\vphantom{|}e^x}}}\
\cos x
+
e^x\
\underset{\Large g}{ \underset{\uparrow}{\ul{\vphantom{|}\sin x}}}\text{.}
\end{equation*}
Noting that \(f' = [e^x]' = e^x\) is next to \(g\) and \(g' = [\sin x]' = \cos x\) is next to \(f\) confirms this is a valid product rule. Therefore, we can reverse the product rule as
\begin{equation*}
e^x \cos x + e^x \sin x = [e^x \sin x]'\text{.}
\end{equation*}
Solution 2.
To make this easier, letβs separate the terms as
\begin{equation*}
\frac{1}{x} e^{x^2} + \ln x (2 x e^{x^2})\text{.}
\end{equation*}
\begin{equation*}
\frac{1}{x}\
\underset{\Large f}{ \underset{\uparrow}{\ul{\vphantom{|}e^{x^2}}}}
+
\underset{\Large g}{ \underset{\uparrow}{\ul{\vphantom{|}\ln x}}}\
2 x e^{x^2}\text{,}
\end{equation*}
\begin{equation*}
f' = \left[e^{x^2}\right]' = 2 x e^{x^2} \quad β
\qquad g' = \left[\ln x\right]' = \frac{1}{x} \quad β
\text{.}
\end{equation*}
Therefore,
\begin{equation*}
\frac{e^{x^2}}{x} + 2 x \ln x e^{x^2} = \left[e^{x^2} \ln x\right]'\text{.}
\end{equation*}
Letβs now try expressions involving a variable function, like \(y(x)\text{.}\) These are the ones weβll actually encounter when solving differential equations.
π Example 72. Reversed Product Rule with a Variable.
Rewrite each expression in the form \([f(x) y(x)]'\text{.}\)
\begin{equation*}
1. \quad e^{9x}\frac{dy}{dx} + 9y e^{9x} \hspace{5em} 2. \quad e^{1/x}y' - \frac{e^{1/x}}{x^2}y
\end{equation*}
Solution.
Solution 1.
Since \(dy/dx\) and \(y\) are in separate terms, setting \(g\) to \(y\) gives you \(g'\) and \(f\) for free since
\begin{equation*}
\underset{\Large f}{ \underset{\uparrow}{\ul{e^{9x}}}}\
\underset{\Large g'}{ \underset{\uparrow}{\ul{\vphantom{|}\frac{dy}{dx}}}}
+
\underset{\Large g}{ \underset{\uparrow}{\ul{\vphantom{|}y}}}\
(9 e^{9x})\text{.}
\end{equation*}
The only thing you need to check is
\begin{equation*}
f' = [e^{9x}]' = 9e^{9x} \quad β
\end{equation*}
which is sitting right next to \(g\text{.}\) Therefore, we have
\begin{equation*}
e^{9x}\frac{dy}{dx} + 9 e^{9x} y = \left[e^{9x} y\right]'\text{.}
\end{equation*}
Solution 2.
\begin{equation*}
\underset{\Large f}{ \underset{\uparrow}{\ul{e^{1/x}}}}\
\underset{\Large g'}{ \underset{\uparrow}{\ul{\vphantom{|}y'}}}
+
\left(-\frac{e^{1/x}}{x^2}\right)\
\underset{\Large g}{ \underset{\uparrow}{\ul{\vphantom{|}y}}}\text{.}
\end{equation*}
The only thing left to do is verify the derivative of \(f\text{:}\)
\begin{equation*}
f' = \left[e^{1/x}\right]' = e^{1/x}\left[\frac{1}{x}\right]^\prime = e^{1/x}\left(-\frac{1}{x^2}\right) = -\frac{e^{1/x}}{x^2} \quad β
\text{.}
\end{equation*}
So, we can reverse the product rule as
\begin{equation*}
e^{1/x}y' - \frac{e^{1/x}}{x^2}y = \left[e^{1/x} y\right]'\text{.}
\end{equation*}
Checkpoint 73. πβ Whatβs Missing in the Product Rule.
Fill in the missing parts of each product rule below.
\(\ds\frac{d}{dx}\big[e^{3x} \) \(\big] = \) \(\ds\frac{1}{x} \cdot e^{3x} + \) \(e^{3x} \cdot \ln x \)
While reversing the product rule isnβt a standard technique, practicing it a few times helps build the intuition youβll need when working with integrating factors.
Subsection Completing the Product Rule: An Analogy and a Method
Completing the square is a classic strategy from algebra: by adding just the right constant, we can turn a messy quadratic into something easily solvable.
For example, to solve
\begin{equation*}
\os{\large\text{incomplete square}}{\qquad \ob{\ x^2 + 6x\ } = -1}\text{,}
\end{equation*}
\begin{equation*}
\ub{x^2 + 6x + 9}_{\large\text{complete square}} = -1 + 9\text{,}
\end{equation*}
which gives a perfect square:
\begin{equation*}
(x + 3)^2 = 8,
\end{equation*}
and from here we solve by taking square roots. Simple and neat.
Checkpoint 74. πβ Purpose of Adding \(9\).
In the analogy, what role does the number \(9\) play in the quadratic equation?
- The missing piece that allows the quadratic expression to be rewritten as a perfect square.
- Correct! The number 9 completes the square, making it factorable.
- The solution to the quadratic equation.
- Incorrect. The number 9 is introduced to complete the square, not as a solution.
- The coefficient of the linear term.
- Incorrect. The coefficient of the linear term in the quadratic equation is 6, not 9.
- A randomly chosen constant added to balance the equation.
- Incorrect. The number 9 is carefully selected to form a perfect square trinomial.
Now letβs look at a differential equation that behaves almost exactly the same way:
\begin{equation*}
\os{\large\text{incomplete product rule}}{\qquad \ob{\ \frac{dy}{dx} + 6 y\ } = -1}\text{.}
\end{equation*}
The left-hand side looks like it wants to be the result of a product rule. And it could be, if we multiply both sides by \(e^{6x}\text{:}\)
\begin{equation*}
\ub{e^{6x} \frac{dy}{dx} + 6e^{6x} y}_{\large\text{complete product rule}} = -e^{6x}\text{.}
\end{equation*}
That multiplication βcompletes the product ruleβ the same way adding \(9\) completed the square. Now, the left-hand side is exactly:
\begin{equation*}
\frac{d}{dx} \left[ e^{6x} y \right] = -e^{6x}.
\end{equation*}
Checkpoint 75. πβ Purpose of Multiply \(e^{6x}\).
Why do we multiply both sides of
\begin{equation*}
\frac{dy}{dx} + 6y = -1
\end{equation*}
by \(e^{6x}\text{?}\)
- To complete a product rule within this equation.
- Correct! Multiplying by \(e^{6x}\) allows us to rewrite the left-hand side as a perfect derivative.
- To eliminate the derivative term entirely.
- Incorrect. The goal is not to eliminate the derivative, but to structure the equation so it can be integrated directly.
- To make the right-hand side equal to zero.
- Incorrect. The right-hand side is transformed, but it does not become zero.
- It makes the equation linear.
- Incorrect. The equation already linear. Moreover, multiplying by any function of \(x\) will not change the equationβs linearity.
And just like we took a square root to solve a completed square, here we integrate both sides to solve a completed product rule:
\begin{align*}
\int \frac{d}{dx}\left[ e^{6x} y \right]\, dx \amp = \int -e^{6x}\, dx \\
e^{6x} y \amp = -\frac{1}{6} e^{6x} + c \\
y \amp = -\frac{1}{6} + c e^{-6x}
\end{align*}
So whatβs the big idea here? Just as completing the square gives us a clean square to work with, completing the product rule gives us a clean product rule, one we can reverse and integrate.
πΊοΈ Summary 76. Completing the Square vs Completing the Product Rule.
Hereβs how the two processes line up:
| Completing the Square | Completing the Product Rule | |
| Goal | Solve for \(x\) | Solve for \(y\) |
| Equation | \(x^2 + 6x = -1\) | \(\frac{dy}{dx} + 6y = -1\) |
| Missing Piece | \(9\) (added) | \(e^{6x}\) (multiplied) |
| Completed Form | \(x^2 + 6x + 9 = 8\) | \(e^{6x} \frac{dy}{dx} + 6e^{6x} y = -e^{6x}\) |
| Grouped Form | \((x + 3)^2 = 8\) | \(\frac{d}{dx} [e^{6x} y] = -e^{6x}\) |
| Solution Step | Take square root | Integrate |
This technique, multiplying by the right function to complete the product rule, is the heart of the integrating factor method. And it raises an important question:
How do we know what function to multiply by?
Thatβs what weβll figure out in the next section.
Subsection π€ Wrap-Up
ποΈ \(\textbf{Key Takeaways...}\)
-
Completing the square transforms a quadratic into a perfect square to make it easier to solve.
-
A similar idea applies to differential equations: we can βcomplete the product ruleβ by multiplying both sides of the equation by a carefully chosen function.
-
This multiplication turns the left-hand side into the derivative of a product, making the equation easier to solve through direct integration.
-
Completing the product rule is conceptually similar to completing the square, both introduce a missing piece that simplifies the structure of the equation.
-
This technique is the core of the integrating factor method, which is used to solve first-order linear differential equations.
-
The big question now is: how do we systematically choose the right function to multiply by? Thatβs the focus of the next section.
Check Your Understanding.
Checkpoint 77. π€π Separation of Variables Reading Questions.
(a) π€π Recognizing an Incomplete Product Rule.
Why do we say the differential equation
\begin{equation*}
\frac{dy}{dx} + 6y = -1
\end{equation*}
contains an incomplete product rule?
- It lacks a function needed to reverse a product rule.
- Correct! The equation is missing an essential function that would allow it to be rewritten using the product rule.
- Because we havenβt used the product rule yet.
- Incorrect. The issue is not that the product rule hasnβt been applied, but that the equation is not structured as a product rule.
- It has a free term of \(-1\text{.}\)
- Incorrect. While the free term is \(-1\text{,}\) the issue is related to the structure of the left-hand side.
- It is not in standard form.
- Incorrect. The equation actually is in standard form.
(b) π€π Equivalent Equations.
The differential equation,
\begin{equation*}
\frac{dQ}{dz}\ln z + \frac{Q}{z} = \tan\left(z^2\right)
\end{equation*}
is equivalent to which of the following equations? Hint: use the forward product rule in the answer choices.
- \(\quad\ds\frac{d}{dz} \left[ Q z \right] = \tan\left(z^2\right)\)
- Incorrect. This equation does not account for the logarithmic term in the original differential equation.
- \(\quad\ds\frac{dQ}{dz} = \tan\left(z^2\right)\)
- Incorrect. This equation does not account for the logarithmic term in the original differential equation.
- \(\quad\ds\frac{d}{dz} \left[ Q \ln z \right] = \tan\left(z^2\right)\)
- Correct! Applying the product rule to the left side of the equation yields the original differential equation.
- \(\quad\ds\frac{d}{dz} \left[ \frac{Q}{z} \right] = \tan\left(z^2\right)\)
- Incorrect. This equation does not account for the logarithmic term in the original differential equation.
(c) π€π Goal of Completing the Product Rule.
How does reversing the product rule help solve this equation?
\begin{equation*}
\frac{dy}{dx} \cdot e^{5x^2} + 10x e^{5x^2} \cdot y = 3x\text{.}
\end{equation*}
- It groups terms on one side of the equation into a single derivative.
- Correct! The product rule combines terms on the left side of the equation into a single derivative.
- It computes the derivative of a product of two functions.
- Incorrect. In this situation, the product rule is not used to take a derivative of a product of two functions.
- It solves the differential equation directly.
- Incorrect. The product rule can help solve some differential equations, but it does not solve them alone.
- It allows it to be easily solved by direct integration.
- Correct! Recognizing a perfect derivative lets us solve the equation by integration.
- It converts a nonlinear equation into a linear one.
- Incorrect. The equation is already linear; reversing the product rule does not change its nature.
(d) π€π Seeing the Pattern.
Given that you can multiply the equation
\begin{equation*}
\frac{dy}{dx} + 6y = -1
\end{equation*}
by the function \(e^{6x}\) to transform it into
\begin{equation*}
\frac{d}{dx}\left[e^{6x}y\right] = -e^{6x}
\end{equation*}
What function would you think to multiply onto the equation
\begin{equation*}
\frac{dy}{dx} - 8y = 3x
\end{equation*}
to accomplish the same goal?
- \(\quad\ds e^{3x}\)
- Incorrect. Review the example carefully.
- \(\quad\ds e^{8x}\)
- Incorrect. Review the example carefully.
- \(\quad\ds e^{-8x}\)
- Thatβs correct! Multiplying by \(e^{8x}\) will allow you to complete the product rule on the left side of the equation.
- \(\quad\ds \frac38 x\)
- Incorrect. Review the example carefully.
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