Integrating Factor: Method & Example

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  • 0:03 Integrating Factors
  • 0:47 Showing the Basic Idea
  • 3:38 Getting into the Details
  • 6:12 Solving a Mixing Problem
  • 8:07 Lesson Summary
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Lesson Transcript
Instructor: Gerald Lemay

Gerald has taught engineering, math and science and has a doctorate in electrical engineering.

,It's nice to get solutions to differential equations. When faced with a first-order linear differential equation we can use the integrating factor method to get a solution. In this lesson, we'll use examples to explain this method.

Integrating Factors

A great thing about restaurant breakfasts is the coffee service. To be more precise, it's the regular coffee refills. You might take your coffee with 2 sugars, but how do you get this concentration of sugar when your not-yet-empty cup keeps getting refilled with fresh but sugar-free coffee? This is a mixing problem: mixing fresh coffee with sweetened coffee.

Mixing problems are an application of differential equations that can be solved using the integrating factor method. Simply put, the integrating factor is a function that we multiply both sides of the differential equation by to make it easier to solve. In this lesson, we'll demonstrate how to find the integrating factor and use it to solve linear first-order differential equations.

Showing the Basic Idea

First, let's make sure we know what we're trying to do when solving the equation. Whether we're mixing fresh coffee with sweetened coffee or something else, the mixing problem often involves solving a linear first-order equation in standard form:


We're looking for a solution to the y variable. Once we find y, it can be substituted in the original differential equation. If we have a valid y solution, the differential equation's left-hand side (or LHS) must equal its right-hand side (or RHS).

Say we're given the following differential equation in standard form:


We have to identify the P and the Q. The quantity multiplying the y is the P term. Clearly, P = 4/x. The quantity on the RHS is the Q term. Thus, Q = x - 1/x. Note P and Q are functions of x.

Let's see how to verify that:


This is the solution. By the way, C is just a constant.

The LHS of this differential equation has two parts, y' and 4y/x. Remember y' is the derivative of y with respect to the other variable. In this case, the other variable is x. If it were a differential equation for y as a function of time t, then the derivative of y would be with respect to time.


Plugging in our previous equation, we get:


Adding those two parts of the LHS and simplifying, we get the following:


The LHS equals the RHS. Okay, this y is the solution for this differential equation.

Getting Into the Details

We know how to verify a solution, but how would we find it in the first place? This is where the integrating factor method comes in. Finding the integrating factor involves two steps: integrating P from the standard form equation and exponentiating the result. Combining those two steps into one statement gives a complicated-looking but easy to evaluate expression for the integrating factor:


We then multiply both sides of the equation by the integrating factor and integrate to solve for y. Let's test this process using the same differential equation as earlier.

First, we take the P term and integrate it with respect to x but don't add a constant C:


We then take the result and exponentiate it:


Our integrating factor is x4.

Ultimately, we're solving for y, so we hope one side of our equation yields y multiplied by our integration factor after integration. We can do a quick check by differentiating yx4 to see how it relates to parts of the earlier differential equation.

Differentiate yx4 using the product rule to get:


Here's the fascinating part. When we multiply both sides of the differential equation by the integrating factor, the LHS becomes


This simplifies to


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