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Section 6.4 Integrating Factor Method

Let’s now bring everything together into a clear, repeatable process. First with a recap, then a step-by-step guide to the Integrating Factor Method and finally, a series of examples that show how to apply this method to solve first-order linear differential equations.

Subsection Method Steps

Before jumping into the official method, let’s recap the story so far ...
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All first-order linear differential equations have a standard form.
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Once in standard form, the equation can be multiplied by an integrating factor to “complete a product rule” on the left-hand side.
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A “Complete product rule” is a sum of two terms that can be grouped inside a single derivative via a reversed product rule.
More concretely, the process was broken down in a previous example as follows:
Standard Form:
\begin{equation*} \frac{dy}{dx} + 2 y = 5 \end{equation*}
\(\leftarrow\) Find & Multiply by \(e^{2x}\)
Complete Product Rule:
\begin{equation*} e^{2x}\frac{dy}{dx} + 2e^{2x} y = 5e^{2x} \end{equation*}
\(\leftarrow\) Reverse Product Rule
Grouped Derivative:
\begin{equation*} \frac{d}{dx} \left[e^{2x} y \right] = 5e^{2x} \end{equation*}
\(\leftarrow\) Direct Integration
This outlines the complete process of the integrating factor method.

✍🏻 Method 1. Integrating Factor (IF) Method.

Given a first-order linear equation in standard form:
\begin{equation} y' + P(x) y = Q(x),\tag{6.4} \end{equation}
the general solution can be found through the following three-step process:
Step 1: Find the Integrating Factor
Identify \(P(x)\) and compute the integrating factor:
\begin{equation*} \mu = e\vphantom{\large|}^{\textstyle\int P(x)\ dx}\text{.} \end{equation*}
Step 2: Multiply by \(\mu\) to Complete the Product Rule
Multiply both sides of the equation (6.4) by \(\mu\) and reverse the product rule on the left side.
\begin{align} \mu(x)\frac{dy}{dx} + P(x) \mu(x) y \amp = Q(x) \mu(x)\tag{6.5}\\ \frac{d}{dx}\left[\mu(x) \cdot y\right] \amp = Q(x) \mu(x)\text{.}\tag{6.6} \end{align}
Step 3: Solve Using Direct Integration
Integrate both sides of the equation (6.6) and solve for \(y\) to find the general solution.

Checkpoint 76. .

(a) 📖❓ Checking When the Method Applies.

Which type of equations can always be solved using the IF method?
  • First-order linear equations.
  • Correct! That’s the specific class for which this method is designed.
  • First-order separable equations.
  • Incorrect. The IF method applies to first-order linear equations, not all separable equations.
  • Any linear equation.
  • Incorrect. This method only applies to first-order equations.
  • Any differential equation that contains \(y\text{.}\)
  • Too broad. The method is specific to first-order linear equations.

(b) 📖❓ Before Finding the Integrating Factor.

Suppose you’re given the equation:
\begin{equation*} 2xy' + 4x^2 y = \ln x\text{.} \end{equation*}
What should you do before computing an integrating factor?
  • Rewrite the equation in standard form by dividing through by \(2x\text{.}\)
  • Correct! Standard form is required before identifying \(P(x)\text{.}\)
  • Group the left side as a single derivative using the product rule.
  • Not quite. Grouping follows multiplication by the integrating factor.
  • Compute the integrating factor.
  • Incorrect. The integrating factor depends on \(P(x)\text{,}\) which must first be identified in standard form.
  • Integrate both sides of the equation.
  • Incorrect. Integration follows multiplication by the integrating factor.

(c) 📖❓ Multiplying by the Integrating Factor.

Consider multiplying this differential equation by its integrating factor.
\begin{equation*} y' + 2xy = 4x^2\text{.} \end{equation*}
Which of the following statements are true? (Select all that apply.)
  • The \(y\) and \(y'\) terms can now be grouped as a single derivative.
  • Correct! Multiplying by the integrating factor allows us to rewrite the left-hand side as a single derivative, completing the product rule.
  • The integrating factor is \(x^2\text{.}\)
  • Incorrect. \(x^2\) comes from \(\int P(x)\ dx\text{,}\) so it is part of the integrating factor, but not the entire integrating factor.
  • The solution is equal to the integrating factor.
  • Incorrect. The solution is not equal to the integrating factor; rather, the integrating factor is used to find the solution.
  • The integrating factor is \(e^{x^2}\text{.}\)
  • Correct! The integrating factor is \(e^{\int 2x\ dx} = e^{x^2}\text{.}\)

(d) 📖❓ Purpose of the Integrating Factor.

Which statement describes the purpose of multiplying both sides of a differential equation by an integrating factor?
  • To eliminate the dependent variable.
  • To convert the equation into a separable form.
  • To solve for the constant of integration.
  • To transform the equation into a direct integration problem.

Subsection Full Solutions Using the Method

With the three-step method in hand, we’re ready to solve some full equations from start to finish. These examples show how the integrating factor method handles a variety of situations, from clean equations already in standard form to messy ones that need rearranging.

🌌 Example 77. Already in Standard Form.

Find the general solution to the equation
\begin{equation*} y' + 6y = 1\text{.} \end{equation*}
Solution.
It is already in standard form, so we proceed with the integrating factor method.
Step 1. Identify \(P(x)\) and compute \(\mu\)
📝: 📌 Note.
\begin{equation*} y' + \os{P}{\boxed{6}}\ y = 1\quad \rightarrow \quad \mu(x) = e^{\int 6\ dx} = e^{6x} \end{equation*}
Step 2. Multiply both sides by \(e^{6x}\) to complete the product rule on the left.
\begin{align*} e^{6x}y' + 6e^{6x}y \amp = e^{6x} \\ \frac{d}{dx}\left[e^{6x}\cdot y\right] \amp = e^{6x} \end{align*}
Step 3, Finally, we integrate:
\begin{align*} \int \frac{d}{dx}\left[e^{6x}\cdot y\right]\ dx \amp = \int e^{6x}\ dx \\ e^{6x}\cdot y \amp = \frac{1}{6} e^{6x} + c \\ y \amp = \frac{1}{6} + ce^{-6x}. \end{align*}

🌌 Example 78. Rearranging to Standard Form.

Solve the equation
\begin{equation*} x^2y' = 5x^3 + 2x^3y \text{.} \end{equation*}
Solution.
This equation is first-order and linear, but not in standard form, so we divide both sides by \(x^2\) and move the \(y\) to the same side as the \(y'\) term:
\begin{equation*} y' = 5x + 2x y \quad \Rightarrow \quad y' - 2x y = 5x\text{.} \end{equation*}
Now the IF method applies.
Step 1. Identify \(P(x)\) and compute \(\mu\text{:}\)
\begin{equation*} y' + \os{P}{\boxed{-2x}} y = 5x \quad \rightarrow \quad \mu = e^{\int -2x\ dx} = e^{-x^2}\text{.} \end{equation*}
Step 2. Multiply by \(e^{-x^2}\) to complete the product rule on the left:
\begin{align*} e^{-x^2}y' - 2x e^{-x^2}y \amp = 5xe^{-x^2} \\ \frac{d}{dx}\left[e^{-x^2}\cdot y\right] \amp = 5xe^{-x^2} \text{.} \end{align*}
Step 3. Here, the integral on the right is computed with \(u\)-substitution:
\begin{align*} \int \frac{d}{dx}\left[e^{-x^2}\cdot y\right]dx \amp = \int 5xe^{-x^2}dx \\ e^{-x^2}\cdot y \amp = -\frac52 \int e^{u}du \\ e^{-x^2}\cdot y \amp = -\frac52 e^{u} + c \\ e^{-x^2}\cdot y \amp = -\frac52 e^{-x^2} + c \\ y \amp = -\frac52 + ce^{x^2} \end{align*}
\begin{gather*} u = -x^2 \end{gather*}
Solution. Verify the Solution
We have found the general solution, but it is worth noting that we can verify it by substituting it back into the original equation.
\begin{gather*} \ul{LHS} \\ x^2y' \\ x^2 \cdot \frac{d}{dx}\left( -\frac{5}{2} + ce^{x^2} \right) \\ x^2 \cdot \left( 0 + 2cxe^{x^2} \right) \\ 2c x^3 e^{x^2} \quad \end{gather*}
\begin{gather*} \ul{RHS} \\ 5x^3 + 2x^3y \\ 5x^3 + 2x^3\left( -\frac{5}{2} + ce^{x^2} \right) \\ 5x^3 - 5x^3+ 2c x^3 e^{x^2} \\ 2c x^3 e^{x^2} \quad \end{gather*}
The same result on the left & right confirms our solution.

🌌 Example 79. Requires Integration by Parts.

Solve the equation:
\begin{equation*} t^2 y' + 2ty = t^3 \ln t \text{.} \end{equation*}
Solution.
Put the equation in standard form, identify \(P(t)\text{,}\) and compute \(\mu\text{:}\)
\begin{equation*} y' + \us{P}{\boxed{\frac{2}{t}}}\ y = t \ln t \quad \implies \quad \mu(t) = e^{\large \int (\sfrac{2}{t})\, dt} = e^{2 \ln t} = t^2 \end{equation*}
Multiply by \(\mu(t)=t^2\) and reverse the product rule on the left-hand side.
\begin{align*} t^2 y' + 2t y \amp = t^3 \ln t \\ \frac{d}{dt} \left[t^2 y\right] \amp = t^3 \ln t \end{align*}
Now, integrate both sides and use integration by parts on the right.
\begin{align*} \int \frac{d}{dt} \big[t^2 y\big] dt \amp = \int t^3 \ln t\ dt\\ t^2 y = \us{\textcolor{BurntOrange}{ u}}{\us{\textcolor{BurntOrange} \uparrow}{\ul{\vphantom{\frac11}\ln t}}} \cdot \us{\textcolor{BurntOrange}{ v}}{\us{\textcolor{BurntOrange} \uparrow}{\ul{\frac14 t^4}}} \amp\ - \int \us{\textcolor{BurntOrange}{ v}}{\us{\textcolor{BurntOrange} \uparrow}{\ul{\frac14 t^4}}} \cdot \us{\textcolor{BurntOrange}{du}}{\us{\textcolor{BurntOrange} \uparrow}{\ul{\frac{1}{t}\ dt}}}\\ t^2 y = \frac14 t^4 \ln t \amp\ - \frac14 \int t^3\ dt \\ t^2 y = \frac{t^4}{4} \ln t \amp\ - \frac{t^4}{16} + c \\ y = \frac{t^2}{4} \ln t \amp\ - \frac{t^2}{16} + \frac{c}{t^2} \end{align*}
\begin{align*} \small\textcolor{BurntOrange}{ u = \ln t,} \amp \ \ \small\textcolor{BurntOrange}{v = \frac14t^4} \\ \small\textcolor{BurntOrange}{ du = \frac{1}{t}\ dt,} \amp \ \ \small\textcolor{BurntOrange}{dv = t^3\ dt} \end{align*}
Finally, solve for \(y\text{:}\)
\begin{gather*} y = \frac{t^2}{4} \ln t - \frac{t^2}{16} + \frac{c}{t^2} \end{gather*}

🌌 Example 80. An Initial-Value Problem.

Solve the initial value problem.
\begin{equation*} A = \sec\theta A' - \frac{\theta e^{\large\theta^2 + \sin\theta}}{\cos \theta}, \hspace{1cm} A(0) = -\frac{1}{2}\text{.} \end{equation*}
Solution.
Put the equation in standard form by replacing \(\sec\theta\) with \(\frac{1}{\cos\theta}\) and moving the \(A'\) term to the left:
\begin{equation*} \frac{1}{\cos\theta} A' - A = \frac{\theta e^{\large\theta^2 + \sin\theta}}{\cos \theta}\text{.} \end{equation*}
Multiplying this by \(\cos\theta\) allows us to identify \(P\) and compute \(\mu\text{:}\)
\begin{equation*} A' + \us{P}{\boxed{- \cos\theta}}\ A = \theta e^{\large \theta^2 + \sin\theta} \quad \implies \quad \mu = e\vphantom{\large|}^{\large\int -\cos\theta d\theta} = e^{\large -\sin\theta} \end{equation*}
Multiply by \(\mu=e^{\large -\sin\theta}\) and complete the product rule on the left-hand side.
\begin{gather*} e^{\large -\sin\theta}A' - \cos\theta e^{\large -\sin\theta}A = \theta e^{\large\theta^2} \\ \frac{d}{d\theta}\left[ e^{\large -\sin\theta}A \right] = \theta e^{\large\theta^2} \text{.} \end{gather*}
Integrate both sides and use \(u\)-substitution:
\begin{gather*} \int \frac{d}{d\theta}\left[ e^{\large -\sin\theta}A \right] d\theta = \int \theta e^{\large\theta^2} d\theta\\ e^{\large -\sin\theta}A = \frac{1}{2}e^{\large\theta^2} + c\\ A = \frac{1}{2}e^{\large\theta^2 + \sin\theta} + ce^{\large\sin\theta} \end{gather*}
\begin{align*} u \amp = \theta^2 \\ du \amp = 2\theta\ d\theta \end{align*}
Finally, we apply the initial condition to determine the constant \(c\text{.}\)
\begin{align*} -\frac{1}{2} = A(0) \amp = \frac{1}{2}e^{\large 0^2 + \sin(0)} + ce^{\large\sin(0)} \\ \amp = \frac{1}{2}e^0 + ce^0 \\ \amp = \frac{1}{2} + c\quad \implies \quad c = -1 \end{align*}
We can then write the particular solution to the initial value problem.
\begin{equation*} A(\theta) = \frac{1}{2}e^{\large\theta^2 + \sin\theta} - e^{\large\sin\theta} \end{equation*}
These examples demonstrate the versatility of the Integrating Factor Method. Regardless of the complexity of the given equation, this three-step approach provides a reliable solution strategy for any first-order linear differential equation.
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