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Mathematical Expression Editor

We discuss the solution of an th order nonhomogeneous linear differential equation,
making use of the method of undetermined coefficients to find a particular
solution.

Undetermined Coefficients for Higher Order Equations

In this section we consider the constant coefficient equation

where and is a linear combination of functions of the form
or

From Theorem thmtype:9.1.5, the general solution of (eq:9.3.1) is , where is a particular solution of (eq:9.3.1)
and is the general solution of the complementary equation
In Trench 9.2 we learned how to find . Here we will learn how to find when the
forcing function has the form stated above. The procedure that we use is a
generalization of the method that we used in Trench 5.4 and 5.5, and is again called
method of undetermined coefficients. Since the underlying ideas are the same as
those in Trench 5.4 and 5.5, we’ll give an informal presentation based on
examples.

Forcing Functions of the Form

We first consider equations of the form

Find a particular solution of

Substituting

into (eq:9.3.2) and canceling yields
or
Since the unknown appears on the left, we can see that (eq:9.3.3) has a particular solution
of the form
Then

Substituting from the last four equations into the left side of (eq:9.3.3) yields

Comparing coefficients of like powers of on the right sides of this equation and (eq:9.3.3)
shows that satisfies (eq:9.3.3) if
Solving these equations successively yields , , , . Therefore
is a particular solution of (eq:9.3.3), so
is a particular solution of (eq:9.3.2).

or
Since neither nor appear on the left, we can see that (eq:9.3.5) has a particular solution of
the form
Then

Substituting , , and into the left side of (eq:9.3.5) yields

Comparing coefficients of like powers of on the right sides of this equation and (eq:9.3.5)
shows that satisfies (eq:9.3.5) if
Solving these equations successively yields , , . Substituting these into (eq:9.3.6) shows
that
is a particular solution of (eq:9.3.5), so
is a particular solution of (eq:9.3.4).

Forcing Functions of the Form

We now consider equations of the form
where and are polynomials.

Find a particular solution of

Substituting

into (eq:9.3.7) and canceling yields
or
Since and are not solutions of the complementary equation
a theorem analogous to Theorem thmtype:5.5.1 implies that (eq:9.3.8) has a particular solution of the
form
Then

so
Comparing the coefficients of , , , and here with the corresponding coefficients in (eq:9.3.8)
shows that is a solution of (eq:9.3.8) if
Solving the first two equations yields , . Substituting these into the last two
equations yields

so , . Substituting , , , into (eq:9.3.9) shows that
is a particular solution of (eq:9.3.8), so
is a particular solution of (eq:9.3.7).

Find a particular solution of

Substituting

into (eq:9.3.10) and canceling yields
or
Since and are solutions of the complementary equation
a theorem analogous to Theorem thmtype:5.5.1 implies that (eq:9.3.11) has a particular solution of the
form

Then

so
Comparing the coefficients of , , , and here with the corresponding coefficients in (eq:9.3.11)
shows that is a solution of (eq:9.3.11) if
Solving the first two equations yields , . Substituting these into the last two
equations yields

so and . Substituting , , , into (eq:9.3.12) shows that
is a particular solution of (eq:9.3.11), so
The figure below shows a particular solution of (eq:9.3.10).

Text Source

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