 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 use examples to explain this method.

## Integrating Factor: Topping-up My Coffee An enjoyable feature of restaurant breakfasts is the coffee service.

### Best services for writing your paper according to Trustpilot

From \$18.00 per page
4,8 / 5
4,80
Writers Experience
4,80
Delivery
4,90
Support
4,70
Price
Recommended Service
From \$13.90 per page
4,6 / 5
4,70
Writers Experience
4,70
Delivery
4,60
Support
4,60
Price
From \$20.00 per page
4,5 / 5
4,80
Writers Experience
4,50
Delivery
4,40
Support
4,10
Price
* All Partners were chosen among 50+ writing services by our Customer Satisfaction Team

To be more precise, it’s the regular coffee refills. I take my coffee with 2 sugars, but how do I obtain this concentration of sugar when my 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 (LHS) must equal its right-hand side (RHS).Say we’re given the following differential equation in standard form: 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. and Adding those two parts of the LHS and simplifying 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. 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 which simplifies to

[Image_Link]/cimages/multimages/16/eintfac9.

png” alt=”null” />

which is the result of our quick test when we differentiated yx

4.Multiplying both sides of the differential equation by the integrating factor Recognizing the LHS as the derivative of yx4 Integrating both sides

[Image_Link]/cimages/multimages/16/eintfac12.

png” alt=”null” />

On the LHS, the integration undoes the derivative. On the RHS, simplify by multiplying through by the x

4. Continuing with the integration gives Multiply through by x-4 to isolate y on the LHS Do you recognize this expression? It’s the solution we verified earlier.

## Solving a Mixing Problem

The mixing problem with coffee and sugar can be expressed as The number of teaspoons of sugar in the coffee at any time is y.

The volume of the coffee cup is 20 gulps. The cup is regularly refilled to the top, additional sugar is added and the coffee-sugar mix is stirred. The rate at which coffee is drunk is 2 gulps per 5 minutes. One teaspoon of sugar is added for every 5 gulps of coffee. Does this mean we have to count gulps? Yes, but this is an experiment, right? Our differential equation is in standard form making it easy to identify P = 1/50.

The integral of P with respect to t is Exponentiating we get the integrating factor A quick check gives us Multiplying both sides of the differential equation by the integrating factor The LHS is now which matches our quick check!Our differential equation becomes Integrating both sides isolates the y. This solution can be checked by substituting into the differential equation. At t = 0, y = 1.

Then 4 + C = 1. Which means C = -3. Our solution is now If we plot this we see at t = 20 minutes, y = 2 teaspoons of sugar in the coffee. Thus, if I stop adding coffee and sugar at 20 minutes, I can drink that last cup of coffee sweetened just the way I like it.

## Lesson Summary

First-order differential equations may be solved using the integrating factor method.

Once the differential equation is in standard form, we identify the P term. Integrating this P and then exponentiating produces the integrating factor. This is followed by multiplying both sides of the differential equation by the integrating factor, integrating and then isolating the y variable. 