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Section 6.2 Completing the Product Rule
In algebra, a clever rearrangement of terms can make an equation easier to solve. For instance, to solve
\begin{equation*}
\os{\large\text{incomplete square}}{\qquad \ob{\ x^2 + 6x\ } = -1}\text{,}
\end{equation*}
we complete the square by adding \(9\) (since \((6/2)^2 = 9\) ) to both sides:
\begin{equation*}
\ub{x^2 + 6x + 9}_{\large\text{complete square}} = -1 + 9\text{,}
\end{equation*}
which gives a single perfect square:
\begin{equation*}
(x + 3)^2 = 8.
\end{equation*}
This grouping is helpful because a square root immediately reveals the solution.
Some
first-order differential equations benefit from a similar idea. Instead of forming a single square, we aim to group the left-hand side as a
single derivative .
For example, take the equation:
\begin{equation*}
\os{\large\text{incomplete product rule}}{\qquad \ob{\ \frac{dy}{dx} + 6 y\ } = -1}\text{.}
\end{equation*}
As you will soon learn, multiplying both sides by \(e^{6x}\) actually “completes the product rule” in a similar way that adding \(9\) completed the square:
\begin{equation*}
\ub{e^{6x} \frac{dy}{dx} + 6e^{6x} y}_{\large\text{complete product rule}} = -e^{6x}\text{.}
\end{equation*}
Applying the product rule in reverse gives a single derivative on the left:
\begin{equation*}
\frac{d}{dx} \left[ e^{6x} y \right] = -e^{6x}\text{.}
\end{equation*}
This new form makes solving easier since direct integration can be used.
This process of 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? That’s what we’ll figure out in the next section.
🗺️ Summary 66 . Completing the Square vs Completing the Product Rule.
Here’s 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
Checkpoint 67 . 📖❓ Reading Questions.
(a) 📖❓ 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 is added to balance the equation.
Incorrect. The number 9 is carefully selected to form a perfect square trinomial.
(b) 📖❓ Purpose of Multiplying \(e^{6x}\) .
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 is already linear. Moreover, multiplying by any function of \(x\) will not change the equation’s linearity.
(c) 📖❓ 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 lacks an essential function that would enable it to be rewritten using the product rule.
Because we have not yet used the product rule.
Incorrect. The issue is not that the product rule hasn’t 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.
(d) 📖❓ 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.
(e) 📖❓ 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 differentiate a product of two functions.
It solves the differential equation directly.
Incorrect. The product rule can help solve certain differential equations, but it does not solve them on its own.
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.
(f) 📖❓ 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 the equation by
\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.
\(\quad\ds e^{-8x}\)
That’s 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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