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GRE Math: Study Guide & Test Prep27 chapters | 182 lessons | 16 flashcard sets

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Lesson Transcript

Instructor:
*Christopher Haines*

In this lesson you'll learn how to solve a first-order linear differential equation. We first define what such an equation is, and then we give the algorithm for solving one of that form. Specific examples follow the more general description of the method.

A **first-order linear differential equation** is an equation of the form

This is where

are all functions of the independent variable *x*. We can alternatively write (1) as

where

It's important to note what the equation here is showing you: that D has the property of linearity.

Therefore, D is a linear operator.

We're solving the equation for *y* with *A* and *f* as fixed functions. Examples of a first-order linear differential equation include:

and

Consider the differential equation:

First, observe that (2) takes the form of (1). We would like to write the left side of (2) in the form:

This is where *g* is to be determined. This identity is known as the product rule for differentiation. This way, when we integrate both sides, the left side will be the integral of a derivative, which in turn is the original function *yg*. We do not know the function *y*, because it's what we are solving for. If this can be accomplished, the only integration we will need to really think about is that on the right side of (2).

However, the left side of (2) doesn't have the desired form of a derivative of a product. Thus, we introduce an **integrating factor**, which is a function that, upon multiplication of both sides of (2), forces the left side to be the derivative of a product of two functions. That is, we want the integrating factor *h* to have the following property:

for some function *g*.

In the next section, we're going to learn that a formula for the integrating factor can be derived. For now, let's use the method of trial and error and suppose the integrating factor is *h(x) = exp(x).* Multiplying both sides of (2) by this factor produces the equation:

Now we can rewrite the left side of (3) by applying the product rule for differentiation, so it becomes:

Next, recognize that if we integrate the left side of (4), we obtain the original function. The right side of (4) becomes the integral of the respective function on the right side. A constant of integration is also introduced in the process. A **constant of integration** indicates that a function has infinitely many anti-derivatives, but all of them with the exception of a constant.

For example, an anti-derivative of the function *f = 1* is *F* = *x*, while another anti-derivative is *F* = *x* + 2. *C*, in the equation here, is assumed here, since we haven't imposed any initial conditions.

Before continuing, let's pause to reference an integration method we're going to frequently need in the examples to come: integration by parts. For any two functions *u* and *v*:

Now, using integration by parts, we can evaluate the integral on the right side of the equation, as you can see here:

Then, by substituting equation (6) back into equation (5), and then multiplying both sides by *exp(-x)*, we obtain the general solution to (2):

This is where *K* is a real number.

We can then check to see if (7) is the general solution by direct substitution, which, as you can see, it is:

Let's take a look at an in-depth example to help illustrate all this.

Question: Find the function which passes through the point (0, 5) and satisfies the given differential equation of:

Solution: Notice that the differential equation is not in the desired form given in (1). However, by some basic algebra, we can rewrite our previous example as:

By comparing this last equation with our original equation, we see that the integrating factor is present:

Now, applying multiplication of the factor given in our integration equation to the previous equation, and then integrating, we obtain:

We can simplify the right side of this equation by first applying integration by parts:

The integral on the right side of this equation can be evaluated by the add and subtract trick:

Now combining the last three steps and solving for *y* in the usual algebraic fashion, we arrive at the general solution:

This is our general solution, but with the additional information that the curve passes through (0, 5), we can plug in *x* = 0 into (17) and solve for the constant *C*:

Thus, the unique solution is:

That was a lot, wasn't it? So, let's take a few moments to review the important points that we've learned about first-order linear differential equations. A **first-order linear differential equation** is an equation of the form:

To solve such a differential equation, we can use an **integrating factor**, which is a function that forces the left side to be the derivative of a product of two functions. The purpose of the integrating factor is to express one side of the differential equation in the form of:

This effectively simplifies the integration on one side of the equation, and really leaves all the integration work for the other side of the equation. From there, it comes down to integrating the other side of the equation using the learning methods of integration, and then solving algebraically for *y*.

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GRE Math: Study Guide & Test Prep27 chapters | 182 lessons | 16 flashcard sets

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