We develop a technique for solving first-order linear differential equations.

### Linear First-Order Differential Equations

#### Identifying Linear First-Order Differential Equations

We say that a first-order equation is *linear* if it can be expressed in the form:

*linear*. Any other kind of first-order term is

*nonlinear*. So, for example:

- is not linear, because of the nonlinear term .
- is not linear, because of the nonlinear term .
- is not linear, because of the nonlinear term .

In the following sections we develop a general method for solving linear first-order equations.

#### A Special Kind of Linear Equation

We now want to devise a method to find the general solution of a linear first order differential equation. With the goal of developing intuition about the method, we start with an equation of very special form.

Let’s consider the following DE:

This equation is not in the form eq:linear-first-order-de, but it is actually easier to solve it in its given form. Notice the following “coincidence”: In words, we can express this remarkable coincidence as:

*The derivative of the function multiplying is equal to the function multiplying *.

(Repeat this to yourself until you fully understand what it means.)

This coincidence allows us to rewrite the left-hand side of the equation in a very convenient form. Using the product rule we have: Using this, eq:linear-de-example1 becomes: The nice thing is that this equation can be directly integrated: so that: where is an arbitrary constant. We can write the solution in explicit form by dividing by : Notice that this solution will only be defined on intervals that do not contain the value . Typically, the right interval is determined by an initial condition. For example, the solution of the differential equation that satisfies the initial condition is: We choose the interval because this interval contains the value , where the initial condition is specified.

It is not too difficult to identify the general case in which this method can be applied. The differential equation must be in the special form:

where is a function of the variable only. Then, the equation can be rewritten as: which can be directly integrated. Equation eq:linear-equation-special seems to be of a very special kind. In the next section we will see that there is a method to transform any equation of the form eq:linear-first-order-de into an equivalent equation of the form eq:linear-equation-special.#### General Method — An Example

We now want to move in the direction of finding a general method to solve equations like eq:linear-first-order-de. Let’s consider the DE: We would like to make this equation have the same pattern as eq:linear-equation-special. We can achieve this by multiplying the equation by a function to be determined:

Comparing eq:linear-equation-special and eq:integrating-factor-example, we see that we need: This is a first order separable equation, so we write it as: Integrating we get: We can simplify the work a bit by noticing that we only need*one*particular solution for , so we can choose : It’s now easy to see that this can be solved algebraically to yield: Let’s now go back to eq:integrating-factor-example and plug in the function we found above: As expected, this equation presents the same pattern observed in equation eq:linear-de-example1, and we can rewrite it as: Integrating this we get: and we get the following explicit general solution: In the next section we will see how we can simplify this solution method.

#### A Streamlined Method

The goal of this section is to present the *method of integrating factors*, a streamlined
procedure for the solution of a linear first order differential equation. The
method is simply a reorganization of the procedure illustrated in the previous
section.

To find the general solution of a first order linear differential equation such as eq:linear-first-order-de, we can proceed as follows:

- (a)
- Compute . It is safe to ignore the constant of integration here.
- (b)
- Let , which is called the
*integrating factor*. - (c)
- Multiply both sides of eq:linear-first-order-de, obtaining the equation:
- (d)
- Rewrite the left-hand side of this equation as a derivative, leaving the right-hand side unchanged:
- (e)
- Integrate: Notice that, here, you have to add an integration constant to the antiderivative.
- (f)
- Solve the equation explicitly for . Notice that this is always possible because for all , since is an exponential.

Let’s now consider some examples.

- (a)
- Compute . (No constant of integration needed here!)
- (b)
- Compute the integrating factor: .
- (c)
- Multiply both sides of the equation by the integrating factor :
- (d)
- Rewrite the left-hand side as :
- (e)
- Integrate:
- (f)
- Solve explicitly for :

We conclude that the general solution of the initial value problem is:

So, we have: and . We are now ready to find the general solution by the method of integrating factors:

- (a)
- Compute , where we used the assumption that .
- (b)
- Evaluate .
- (c)
- Multiply both sides of the equation by :
- (d)
- Rewrite the left-hand side as :
- (e)
- Integrate:
- (f)
- Solve for :

Once we have the general solution, we can plug in the initial condition: Solving for we get: Using this constant, we can now write the particular solution to the IVP:

We conclude this example by providing an interactive Sage cell which will display the particular solution to the IVP where . If you change the values of or , you will get a different solution curve, but notice that these curves are valid only for or .

### Text Source

Felipe L. Martins, Lecture Notes