We explore the solution of nonhomogeneous linear equations in the case where the forcing function is the product of an exponential function and a polynomial.

The Method of Undetermined Coefficients I

In this section we consider the constant coefficient equation

where is a constant and is a polynomial.

From Theorem thmtype:5.3.2, the general solution of (eq:5.4.1) is , where is a particular solution of (eq:5.4.1) and is a fundamental set of solutions of the complementary equation In Trench 5.2 we showed how to find . In this section we’ll show how to find . The procedure that we’ll use is called the method of undetermined coefficients.

The preceding examples illustrate the following facts concerning the form of a particular solution of a constant coefficent equation where is a nonzero constant:

(a)
If isn’t a solution of the complementary equation then , where is a constant. (See Example example:5.4.1).
(b)
If is a solution of (eq:5.4.8) but is not, then , where is a constant. (See Example example:5.4.2.)
(c)
If both and are solutions of (eq:5.4.8), then , where is a constant. (See Example example:5.4.3.)

In all three cases you can just substitute the appropriate form for and its derivatives directly into and solve for the constant , as we did in Example example:5.4.1. However, if the equation is where is a polynomial of degree greater than zero, we recommend that you use the substitution as we did in Examples example:5.4.2 and example:5.4.3. The equation for will turn out to be

where is the characteristic polynomial of the complementary equation and however, you shouldn’t memorize this since it’s easy to derive the equation for in any particular case. Note, however, that if is a solution of the complementary equation then , so (eq:5.4.9) reduces to while if both and are solutions of the complementary equation then and , so and (eq:5.4.9) reduces to

Summary

The preceding examples illustrate the following facts concerning particular solutions of a constant coefficient equation of the form where is a polynomial

(a)
If isn’t a solution of the complementary equation then , where is a polynomial of the same degree as . (See Example example:5.4.4).
(b)
If is a solution of (eq:5.4.16) but is not, then , where is a polynomial of the same degree as . (See Example example:5.4.5.)
(c)
If both and are solutions of (eq:5.4.16), then , where is a polynomial of the same degree as . (See Example example:5.4.6.)

In all three cases, you can just substitute the appropriate form for and its derivatives directly into and solve for the coefficients of the polynomial . However, if you try this you will see that the computations are more tedious than those that you encounter by making the substitution and finding a particular solution of the resulting equation for . In Case item:partsolutionfactsa the equation for will be of the form with a particular solution of the form , a polynomial of the same degree as , whose coefficients can be found by the method used in Example example:5.4.4. In Case item:partsolutionfactsb the equation for will be of the form (no term on the left), with a particular solution of the form , where is a polynomial of the same degree as whose coefficents can be found by the method used in Example example:5.4.5. In Case item:partsolutionfactsc the equation for will be of the form with a particular solution of the form that can be obtained by integrating twice and taking the constants of integration to be zero, as in Example example:5.4.6.

Using the Principle of Superposition

The next example shows how to combine the method of undetermined coefficients and Theorem thmtype:5.3.3, the principle of superposition.

Text Source

Trench, William F., ”Elementary Differential Equations” (2013). Faculty Authored and Edited Books & CDs. 8. (CC-BY-NC-SA)

https://digitalcommons.trinity.edu/mono/8/