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Mathematical Expression Editor

We explore the solution of nonhomogeneous linear equations with other forcing
functions.

The Method of Undetermined Coefficients II

In this section we consider the constant coefficient equation

where and are real numbers, , and and are polynomials. We want to find a
particular solution of (eq:5.5.1). As in the previous module, the procedure that we will use is
called the method of undetermined coefficients.

Forcing Functions Without Exponential Factors

We begin with the case where in (eq:5.5.1); thus, we we want to find a particular solution of

where and are polynomials.

Differentiating and yields
and
This implies that if
where and are polynomials, then
where and are polynomials with coefficients that can be expressed in terms of the
coefficients of and . This suggests that we try to choose and so that and ,
respectively. Then will be a particular solution of (eq:5.5.2). The next theorem tells us how
to choose the proper form for .

Suppose is a positive number and and are polynomials. Let be the larger of the
degrees of and . Then the equation
has a particular solution
where
provided that and are not solutions of the complementary equation. The solutions
of
(for which and are solutions of the complementary equation) are of the form (eq:5.5.3),
where

Find a particular solution of

In (eq:5.5.4) the coefficients of and are both zero degree polynomials (constants). Therefore
Theorem thmtype:5.5.1 implies that (eq:5.5.4) has a particular solution
Since
replacing by in (eq:5.5.4) yields

Equating the coefficients of and here with the corresponding coefficients on the
right side of (eq:5.5.4) shows that is a solution of (eq:5.5.4) if

Solving these equations yields , . Therefore
is a particular solution of (eq:5.5.4).

Find a particular solution of

The procedure used in Example example:5.5.1 doesn’t work here; substituting for in (eq:5.5.5)
yields
for any choice of and , since and are both solutions of the complementary equation
for (eq:5.5.5). We’re dealing with the second case mentioned in Theorem thmtype:5.5.1, and should
therefore try a particular solution of the form
Then

so
Therefore is a solution of (eq:5.5.5) if
which holds if and . Therefore
is a particular solution of (eq:5.5.5).

Find a particular solution of

The coefficients of and in (eq:5.5.7) are polynomials of degree one and zero, respectively.
Therefore Theorem thmtype:5.5.1 tells us to look for a particular solution of (eq:5.5.7) of the form
Then
and
so
Comparing the coefficients of , , , and here with the corresponding coefficients in (eq:5.5.7)
shows that is a solution of (eq:5.5.7) if
Solving the first two equations yields , . Substituting these into the last two
equations yields

Solving these equations yields , . Substituting , , , into (eq:5.5.8) shows that
is a particular solution of (eq:5.5.7).

A Useful Observation

In (eq:5.5.9), (eq:5.5.10), and (eq:5.5.11) the polynomials multiplying can be obtained by replacing , and by ,
, , and , respectively, in the polynomials multiplying .

If
where and are polynomials with coefficients …, and , …, then the polynomials
multiplying in
can be obtained by replacing , … by … and … by … in the corresponding
polynomials multiplying .

We won’t use this theorem in our examples, but we recommend that you use it to
check your manipulations when you work on the exercises.

Find a particular solution of

According to Theorem thmtype:5.5.1, we should look for a particular solution of the form
since and are solutions of the complementary equation. However, let’s try
first, so you can see why it doesn’t work. From (eq:5.5.10),
which together with (eq:5.5.14) implies that
Since the right side of this equation does not contain or , (eq:5.5.14) can’t satisfy (eq:5.5.12) no
matter how we choose , , , and .

so
Comparing the coefficients of and here with the corresponding coefficients in (eq:5.5.12)
shows that is a solution of (eq:5.5.12) if
The solution of this system is , , , . Therefore
is a particular solution of (eq:5.5.12).

Forcing Functions with Exponential Factors

To find a particular solution of

when , we recall from the previous module that substituting into (eq:5.5.15) will produce a
constant coefficient equation for with the forcing function . We can find a particular
solution of this equation by the procedure that we used in Examples example:5.5.1–example:5.5.4. Then is a
particular solution of (eq:5.5.15).

Find a particular solution of

Let . Then

if
Since and aren’t solutions of the complementary equation
Theorem thmtype:5.5.1 tells us to look for a particular solution of (eq:5.5.17) of the form
Then

so

Comparing the coefficients of , , , and here with the corresponding coefficients on
the right side of (eq:5.5.17) shows that is a solution of (eq:5.5.17) if
Solving the first two equations yields , . Substituting these values into the last two
equations of (eq:5.5.19) yields

Solving this system yields , . Substituting , , , and into (eq:5.5.18) shows that
is a particular solution of (eq:5.5.17). Therefore
is a particular solution of (eq:5.5.16).

Find a particular solution of

Let . Then

if
Since and are solutions of the complementary equation
Theorem thmtype:5.5.1 tells us to look for a particular solution of (eq:5.5.21) of the form
Then

so
Equating the coefficients of , , , and here with the corresponding coefficients on the
right side of (eq:5.5.21) shows that is a solution of (eq:5.5.21) if
The solution of this system is , , , . Therefore
is a particular solution of (eq:5.5.21), and
is a particular solution of (eq:5.5.20).

You can also find a particular solution of (eq:5.5.20) by substituting
for in (eq:5.5.20) and equating the coefficients of , , , and in the resulting expression
for
with the corresponding coefficients on the right side of (eq:5.5.20). This leads to the same
system (eq:5.5.22) of equations for , , , and that we obtained in Example example:5.5.6. However, if you
try this approach you’ll see that deriving (eq:5.5.22) this way is much more tedious than the
way we did it in Example example:5.5.6.

Text Source

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