Integrating Factor Product Derivative: Collapse the Left Side

On this page: The Principle | Conditions | Failure Modes | EE Questions | Retrieval Practice | Practice Ground | Solve a Problem | Related Guides | FAQ

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The Principle

The move: after multiplying $y^{\prime}+p(x)y=q(x)$ by an integrating factor $\mu$, rewrite the two left-side terms as the derivative of the product $\mu y$.

The invariant: the multiplied equation remains equivalent; only the left side is repackaged using the product rule and the defining relation $\mu^{\prime}=p\mu$.

Pattern:

$$\mu^{\prime}=p\mu \quad \Longrightarrow \quad \mu y^{\prime}+p\mu y=(\mu y)^{\prime}$$

Legal route	Illegal route
$\mu^{\prime}=p\mu,\quad \mu y^{\prime}+p\mu y \Longrightarrow (\mu y)^{\prime}$	$\mu^{\prime}\neq p\mu,\quad \mu y^{\prime}+p\mu y \not\Longrightarrow (\mu y)^{\prime}$

The illegal route is tempting because the symbols look close to the product rule. The exact product rule is $(\mu y)^{\prime}=\mu y^{\prime}+\mu^{\prime}y$, so $p\mu y$ can replace $\mu^{\prime}y$ only when $\mu^{\prime}=p\mu$.

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Conditions of Applicability

Condition: $y and \mu differentiable$; $\mu^{\prime}=p\mu$

Before applying, check: confirm that the multiplier was built by Integrating Factor Definition, so its derivative relation matches the coefficient $p$ from First-Order Linear Standard Form.

If the condition is violated: the product derivative would be $(\mu y)^{\prime}=\mu y^{\prime}+\mu^{\prime}y$, not $\mu y^{\prime}+p\mu y$, so integrating the collapsed form solves a different equation.

• Both $y$ and $\mu$ must be differentiable on the working interval.
• The $p$ in $p\mu y$ must be the same coefficient used in $\mu^{\prime}=p\mu$.
• The equation should already be multiplied through by $\mu$ before this identity is used.

Want the complete framework behind this guide? Read Masterful Learning.

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Common Failure Modes

Failure mode: collapse $\mu y^{\prime}+p\mu y$ without checking $\mu^{\prime}=p\mu$ -> the derivative term uses the wrong coefficient.

Debug: expand $(\mu y)^{\prime}$ as $\mu y^{\prime}+\mu^{\prime}y$ and compare the second term with $p\mu y$.

Failure mode: use the visible coefficient before normalizing the equation -> the multiplier relation is built from the wrong $p$.

Debug: rewrite the equation as $y^{\prime}+p(x)y=q(x)$ before choosing $\mu$ or applying the product-derivative identity.

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Elaborative Encoding

Use these questions to build deep understanding. (See Elaborative Encoding for the full method.)

Within the Principle

• Why does the second product-rule term $\mu^{\prime}y$ become $p\mu y$ only after the integrating-factor condition is checked?
• What is preserved when the left side changes from two terms to one product derivative?

For the Principle

• What fast test would tell you that a proposed multiplier cannot license the product-derivative rewrite?
• Why does normalizing into first-order linear standard form have to happen before this step?

Between Principles

• How does this identity use Integrating Factor Definition after First-Order Linear Standard Form has identified $p$?

Generate an Example

• Write one eligible pair $(p,\mu)$ with $\mu^{\prime}=p\mu$ and one near miss where the proposed $\mu$ does not match $p$.

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Retrieval Practice

Answer from memory, then click to reveal and check. (See Retrieval Practice for the full method.)

State the move in one sentence: _____Rewrite mu y prime plus p mu y as the derivative of mu y when mu prime equals p times mu.

Write the canonical pattern: _____$\mu^{\prime}=p\mu \Rightarrow \mu y^{\prime}+p\mu y=(\mu y)^{\prime}$

State the canonical condition: _____$y and \mu differentiable; \mu^{\prime}=p\mu$

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Practice Ground

Use these exercises to build move-selection fluency. (See Self-Explanation for how to learn from worked examples.)

Procedure Walkthrough

Starting from $y^{\prime}+\frac{2}{x}y=x^3$ on $x>0$, multiply by $\mu=x^2$ and collapse the left side.

Step	Expression	Operation
0	$y^{\prime}+\frac{2}{x}y=x^3$	Start from first-order linear standard form with $p(x)=2/x$.
1	$\mu=x^2,\quad \mu^{\prime}=2x=\frac{2}{x}\mu$	Check the integrating-factor relation on $x>0$.
2	$x^2y^{\prime}+2xy=x^5$	Multiply the whole equation by $\mu=x^2$.
3	$(x^2y)^{\prime}=x^5$	Collapse $x^2y^{\prime}+2xy$ into one product derivative.

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Drills

Forward Step

Apply the product-derivative identity once. Assume $y$ and $\mu$ are differentiable and $\mu^{\prime}=3\mu$.

$$\mu y^{\prime}+3\mu y$$

Reveal

Since $p=3$ and $\mu^{\prime}=3\mu$,

$$\mu y^{\prime}+3\mu y=(\mu y)^{\prime}.$$

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Apply the product-derivative identity once. Assume $x>0$, $p(x)=1/x$, and $\mu=x$.

$$xy^{\prime}+y$$

Reveal

Here $\mu^{\prime}=1=(1/x)x=p\mu$, so

$$xy^{\prime}+y=(xy)^{\prime}.$$

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Reject or complete the route choice. Assume $p(x)=2$ and a student chooses $\mu=x^2$.

$$x^2y^{\prime}+2x^2y$$

Reveal

Reject the collapse as written. If $\mu=x^2$, then $\mu^{\prime}=2x$, but $p\mu=2x^2$. The product derivative is

$$(x^2y)^{\prime}=x^2y^{\prime}+2xy,$$

not $x^2y^{\prime}+2x^2y$.

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Apply the identity after multiplication. Assume $\mu=e^{-x}$ for $y^{\prime}-y=\sin x$.

$$e^{-x}y^{\prime}-e^{-x}y$$

Reveal

The equation has $p=-1$, and $\mu^{\prime}=-e^{-x}=p\mu$. Therefore

$$e^{-x}y^{\prime}-e^{-x}y=(e^{-x}y)^{\prime}.$$

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Which expression is eligible for the collapse if $\mu^{\prime}=4x^3$ and $\mu=x^4$?

A. $x^4y^{\prime}+4x^3y$
B. $x^4y^{\prime}+4x^4y$

Reveal

A is eligible because it matches $\mu y^{\prime}+\mu^{\prime}y$.

B would require $\mu^{\prime}=4x^4$, which is false for $\mu=x^4$.

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Action Label

What was done between these two steps? Assume $x>0$.

$$x^3y^{\prime}+3x^2y=x^6 \quad \Longrightarrow \quad (x^3y)^{\prime}=x^6$$

Reveal

The integrating-factor product derivative was applied. With $\mu=x^3$, $\mu^{\prime}=3x^2$, so $x^3y^{\prime}+3x^2y=(x^3y)^{\prime}$.

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What condition licenses this transition?

$$e^{2x}y^{\prime}+2e^{2x}y \quad \Longrightarrow \quad (e^{2x}y)^{\prime}$$

Reveal

The condition is $y$ and $\mu=e^{2x}$ differentiable and $\mu^{\prime}=p\mu$ with $p=2$. Since $(e^{2x})^{\prime}=2e^{2x}$, the collapse is legal.

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A student claims this is a product derivative. What is wrong?

$$x^2y^{\prime}+xy \quad \Longrightarrow \quad (x^2y)^{\prime}$$

Reveal

The product rule gives $(x^2y)^{\prime}=x^2y^{\prime}+2xy$. The second term is missing a factor of $2$, so the proposed collapse is not legal.

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Name the move in this chain.

$$\mu y^{\prime}+\mu^{\prime}y=\mu q \quad \Longrightarrow \quad (\mu y)^{\prime}=\mu q$$

Reveal

This is the product-rule form of the integrating-factor product derivative. Once the second term is $\mu^{\prime}y$, the left side is exactly $(\mu y)^{\prime}$.

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Transition Identification

Where does the integrating-factor identity enter this worked chain?

$$y^{\prime}+2y=\cos x \quad \Longrightarrow \quad e^{2x}y^{\prime}+2e^{2x}y=e^{2x}\cos x \quad \Longrightarrow \quad (e^{2x}y)^{\prime}=e^{2x}\cos x$$

Reveal

It enters in the second transition, when $e^{2x}y^{\prime}+2e^{2x}y$ is collapsed to $(e^{2x}y)^{\prime}$. The first transition is multiplication by the integrating factor.

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What is missing from this worked chain?

$$y^{\prime}+\frac{1}{x}y=x \quad \Longrightarrow \quad xy^{\prime}+y=x^2 \quad \Longrightarrow \quad (xy)^{\prime}=x^2$$

Reveal

The chain should state the working interval and the multiplier check. On an interval with $x\neq 0$, $\mu=x$ has $\mu^{\prime}=1=(1/x)x=p\mu$, so the collapse is legal.

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Solve a Problem

Apply what you’ve learned with Problem Solving.

Problem: Starting from $y^{\prime}-\frac{3}{x}y=x^2$ on $x>0$, use $\mu=x^{-3}$ to reach product-derivative form.

Full solution

Step	Expression	Move
0	$y^{\prime}-\frac{3}{x}y=x^2$	Identify $p(x)=-3/x$ in first-order linear standard form.
1	$\mu=x^{-3},\quad \mu^{\prime}=-3x^{-4}=-\frac{3}{x}\mu$	Check $\mu^{\prime}=p\mu$ on $x>0$.
2	$x^{-3}y^{\prime}-3x^{-4}y=x^{-1}$	Multiply the whole equation by $\mu$.
3	$(x^{-3}y)^{\prime}=x^{-1}$	Collapse the left side with the product-derivative identity.
4	$\int (x^{-3}y)^{\prime}\,dx=\int x^{-1}\,dx$	The equation is ready for integration.

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Related Guides

• Differential Equations Subdomain - See where integrating factors sit in the first-order ODE sequence.
• First-Order Linear Standard Form - Normalize the equation before identifying $p$.
• Integrating Factor Definition - Build the multiplier whose derivative relation licenses this identity.
• Differential Equation Solution Condition - Check any final candidate against the original differential equation.
• Principle Structures - Treat the name, condition, and canonical pattern as separate recall targets.

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FAQ

What is Integrating Factor Product Derivative?

Integrating Factor Product Derivative is the rewrite from $\mu y^{\prime}+p\mu y$ to $(\mu y)^{\prime}$ when $\mu^{\prime}=p\mu$. It is the product-rule identity that turns a multiplied first-order linear equation into a form that can be integrated.

When is the product-derivative collapse valid?

It is valid when $y$ and $\mu$ are differentiable and the multiplier satisfies $\mu^{\prime}=p\mu$ on the working interval. Without that exact relation, the second term in the product rule does not match the equation’s left side.

Why does this identity use p times mu instead of mu prime?

The product rule gives $(\mu y)^{\prime}=\mu y^{\prime}+\mu^{\prime}y$. The integrating-factor condition says $\mu^{\prime}=p\mu$, so $\mu^{\prime}y$ can be written as $p\mu y$.

Is this the same as finding the integrating factor?

No. Integrating Factor Definition finds or verifies $\mu$. Integrating Factor Product Derivative is the later rewrite that uses that verified $\mu$ to collapse the left side.

What goes wrong if the equation was not normalized first?

The coefficient used as $p$ may be wrong. For example, in $2y^{\prime}+6y=x$, standard form is $y^{\prime}+3y=x/2$, so $p=3$, not $6$.

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How This Fits in Unisium

Within the differential equations subdomain, Unisium treats this as a route-selection move: normalize the first-order linear equation, build or verify $\mu$, then collapse the left side only when $\mu^{\prime}=p\mu$ is true. The Unisium Study System pairs that condition check with retrieval practice, self-explanation, and compact problem-solving chains so the product-rule identity becomes fluent without hiding the legality check.

Ready to practice differential equations with structure? Check access and join the Unisium waitlist or explore the complete framework in Masterful Learning.

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