This document discusses the method of undetermined coefficients for solving nonhomogeneous linear differential equations. It begins by introducing the method and when it can be used. The method involves finding a particular solution of the form of the nonhomogeneous term. Several examples are then worked through step-by-step to demonstrate applying the method to equations with polynomial, exponential, sine, and cosine nonhomogeneous terms. The examples show determining the complementary function first before applying the method of undetermined coefficients to find the particular solution.

1. Ch 4.3: Nonhomogeneous Equations:
Method of Undetermined Coefficients
The method of undetermined coefficients can be used to
find a particular solution Y of an nth order linear, constant
coefficient, nonhomogeneous ODE
provided g is of an appropriate form.
As with 2nd
order equations, the method of undetermined
coefficients is typically used when g is a sum or product of
polynomial, exponential, and sine or cosine functions.
Section 4.4 discusses the more general variation of
parameters method.
[ ] ),(1
)1(
1
)(
0 tgyayayayayL nn
nn
=+′+++= −
−


2. Example 1
Consider the differential equation
For the homogeneous case,
Thus the general solution of homogeneous equation is
For nonhomogeneous case, keep in mind the form of
homogeneous solution. Thus begin with
As in Chapter 3, it can be shown that
t
eyyyy 433 =−′+′′−′′′
t
eAttY 23
)( =
( ) 010133)(
323
=−⇔=−+−⇒= rrrrety rt
ttt
C etctececty 2
321)( ++=
ttttt
etetctecectyettY 232
321
23
3
2
)(
3
2
)( +++=⇒=

3. Example 2
Consider the equation
For the homogeneous case,
Thus the general solution of homogeneous equation is
For the nonhomogeneous case, begin with
As in Chapter 3, it can be shown that
tBtAtY cossin)( +=
tttY cos
3
1
sin
9
2
)( −=
ttyyy cos3sin2168)4(
−=+′′+
( )( ) 0440168)( 224
=++⇔=++⇒= rrrrety rt
( ) ( )ttcttctctcty 2sin2cos2sin2cos)( 4321 +++=

4. Example 3
Consider the equation
As in Example 2, the general solution of homogeneous
equation is
For the nonhomogeneous case, begin with
As in Chapter 3, it can be shown that
tBttAttY 2cos2sin)( 22
+=
tttttY 2cos
32
3
2sin
16
1
)( 22
+−=
ttyyy 2cos32sin2168)4(
−=+′′+
( ) ( )ttcttctctcty 2sin2cos2sin2cos)( 4321 +++=

5. Example 4
Consider the equation
For the homogeneous case,
Thus the general solution of homogeneous equation is
For nonhomogeneous case, keep in mind form of
homogeneous solution. Thus we have two subcases:
As in Chapter 3, can be shown that
t
etyy 3
9 +=′−′′′
( ) ,)(,)( 3
21
t
CtetYtBtAtY =+=
( ) ( )( ) 033909)( 23
=+−⇔−⇔=−⇒= rrrrrrrety rt
tt
C ececcty 3
3
3
21)( −
++=
t
ettYttY 3
2
2
1
18
1
)(,
18
1
)( =−=

6. Example 4
Consider the equation
For the homogeneous case,
Thus the general solution of homogeneous equation is
For nonhomogeneous case, keep in mind form of
homogeneous solution. Thus we have two subcases:
As in Chapter 3, can be shown that
t
etyy 3
9 +=′−′′′
( ) ,)(,)( 3
21
t
CtetYtBtAtY =+=
( ) ( )( ) 033909)( 23
=+−⇔−⇔=−⇒= rrrrrrrety rt
tt
C ececcty 3
3
3
21)( −
++=
t
ettYttY 3
2
2
1
18
1
)(,
18
1
)( =−=