First Order Linear Equations

In the first section under Linearity, you learned about the general form of a linear equation. A first order linear differential equation is this type and you can write it as...

...where both coefficients and the function g depend on the independent variable x alone and not y. If you devide both sides by the coefficient of the derivative, , the equation is placed into a standard form:

The two new functions P and Q merely replace the two functions of x that have been created by the division. Since they are both functions of x they can be replaced by new functions of x, which may or may not represent quotients in an actual problem.

To solve these equations an integrating factor is known and a standard method for finding it is available. Before the derivation of this integrating factor is presented, an example which is particularly insightful is shown.

Expl #17 A First 1st Order Linear Equation

Because of its simplicity, the following equation should help you understand why the integrating factor for 1st Order Linear equations works the way it does:

Input the following differential equation:

To place this equation into the form shown above, add y to each side of the equation.

It is now in standard form with P(x) = 1 and Q(x) = x.

Notice how the LHS looks suspiciously like the result of taking the derivative of the product of two functions.

To show this, take the derivative a function y times an unknown function u.

Now it really looks like the given equation. On the LHS of the derivative above, u and du/dx are multiplied onto the two same terms that make up the LHS of the equation you want to solve (dy/dx and y). If they can be equated, the LHS will be in the form of a derivative. Below is a graphic comparing the given equation to the derivative just obtained.

Since you are after a single factor you need to find a function that is equal to the derivative of itself?

Of course the answer to this question is !

Trying this out you find that the derivative of these two functions is the answer you are after. The factor is therefore equal to .

Using LiveMath, the following case theory demonstrates how integrating this function gives back the product. Make sure you input an Independence Declaration before the integration.

The integrating factor has now been determined to be , so you can solve the given problem by applying it to both sides of the equation. The case theory below demonstrates the method that you will use to find solutions to these equations. First define the integrating factor as u. Then apply u to both sides of the equation. After an Expand, the LHS is a derivative of a product and the RHS is the integrating factor times Q(x).

All that remains is to integrate both sides of the equation for the solution.

The RHS above is solved by using integration by parts. To do this you must commute the integrand as shown above. Select dx and integrate using Integration by Parts.

Because of the simplicity of this problem, the integrating factor was easily obtained. The question now becomes; what happens if the coefficient P(x) is something other than 1 ?

Expl #18 A Second 1st Order Linear Equation

This example increases the complexity of the first problem by making P(x) = 2. A pattern will emerge which will lead you into a general method for finding these integrating factors.

For reference the derivative of the product of y and u is shown again.

You have the problem this time of finding a factor where its derivative is itself times 2.

By the chain rule, a function is equal to times the derivative of v.

So you need to find a function v where the derivative is equal to 2. Below this equation is input and integrated giving you the power e must be raised to.

It looks like v = 2x, so the integrating factor is . Below this is verified using LiveMath.

You use e to some power as the basis of the integrating factor because it is the only function that is the derivative of itself. The argument (power) is achieved by taking the integral of P(x)!!

So the general equation giving the integrating factor can be given as:

For the first example you have:

And for this example:

The following case theory completes the solution to this example using this method of finding the factor.

Expand the LHS and apply the integral using the Palette. Again, Integration by Parts is used to solve the RHS:

As demonstrated in past examples, Auto Casing could have been turned on for the integration of the LHS and off for the RHS. This would have eliminated all those messy constants on the right and saved a couple of steps.

The method for finding the integrating factor of linear first order equations has been developed in the last two examples. To summarize, first place the equation in the standard form; dy/dx+P(x)y=Q(x). Then apply to both sides the integrating factor given by:

Next, solve the equation by integrating both sides and solving for y. The general solution can be stated as follows:

NOTE: A couple of things you should keep in mind when using LiveMath. First, when P(x) is a more complicated function, LiveMath may not be able to directly solve the integration of the LHS. To get around this, you may find it more efficient to just re-input it as u y. Secondly, notice that there is no constant of integration generated when solving for the integrating factor (u). This is because only one function is required for the factor, not a family of functions. Additionally, the assumption that P(x) and Q(x) are continuous over the interval in question, is a requirement.

By looking the general form of a first order linear equation, you can derive the general rule to find the integrating factor. First, the general form of the equation, as stated above, is:

First multiply both sides of the function above by the function u(x), so that the LHS is equal to the derivative of the product u y. With this requirement met, and looking only at the LHS, you have:

The integrating factor u(x) is not known at this point, but from the equation above it must take the form: (simplify the equation above)

Again, this is because you have made the requirement that u(x) make the LHS a derivative of a product.

You can manipulate the previous equation into the following:

Integrating both sides gives:

Exponentiating each side of this last equation gives you the integrating factor.

An interesting fact concerning first order linear equations is that the integrating factor, which is found by the method just discussed, turns this non exact equation into an exact one. To demonstrate this, use LiveMath to test the exactness of the same equation used in Example 18.

Below the equation is tested before the integrating factor is applied. Notice how the equation is first manipulated into the form used to study exact equations.

The next case theory demonstrates how, by applying the integrating factor, this non-exact equation is turned into an exact equation. The resulting equation could be solved using the methods studied in the section on Exact Equations.

This completes the study of analytical methods for 1st Order ODE's. Although these methods should help you solve many real world problems, there are many more that require other methods. The next section introduces some of these.