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Spring 2003 Math 308/501–502
4 Linear SecondOrder Equations
4.5 Method of Undetermined Coeffs
Wed, 01/Oct
c
±
2003, Art Belmonte
Summary
The
method of undetermined coefﬁcients
is dealt with in full
generality in the Section 6.3 lecture,
q.v.
In this lecture handout,
we’ll restrict ourselves to examples of secondorder equations.
By hand or with MATLAB, you’re a winner!
The hand work involved in computing derivatives, substitution,
collecting terms, and solving linear systems can be tedious and
errorprone. It lends itself quite well, however, to machine power,
as you’ll observe in the
MATLAB Examples
.
You may have felt that the formulation for
y
p
given in Section 6.3
is overly complicated. Take solace in the fact that under that stated
formulation there is only
one
case (the general case). What’s
more, there is
no
guesswork as to the correct
form
of
y
p
. With
practice, you’ll get it right every time!
Note that in all examples (hand or MATLAB), we carry the work
through to a full general solution (or to the unique solution of an
initial value problem). Please do this in your homework problems
as well.
Hand Examples
Example A
Find a general solution of
y
00
+
3
y
0

18
y
=
18
e
2
t
.
Solution
1. The homogeneous equation
y
00
+
3
y
0

18
y
=
0has
characteristic equation 0
=
r
2
+
3
r

18
=
(
r

3
)(
r
+
6
)
,
with roots
r
=
3
,

6. Hence
y
h
=
c
1
e
3
t
+
c
2
e

6
t
.The
forcing function is
f
(
t
)
=
18
e
2
t
,whichis
not
a solution of
the homogeneous equation. (“There is no interference.”)
2. So let
y
p
=
ae
2
t
.Then
y
0
p
=
2
2
t
and
y
00
p
=
4
2
t
.
Substituting into the nonhomogeneous equation gives
(
4
2
t
)
+
3
(
2
2
t
)

18
(
2
t
)
=
18
e
2
t
(

8
a

18
)
e
2
t
=
0
=
0
e
2
t
3. Equating coefﬁcients of like entities, we have

8
a

18
=
0,
whence
a
=
9
4
. Thus
y
p
9
4
e
2
t
is our particular
solution.
4. A general solution (veriﬁed by
dsolve
)is
y
=
y
p
+
y
h
9
4
e
2
t
+
c
1
e
3
t
+
c
2
e

6
t
Example B
Find a general solution of
y
00
+
7
y
0
+
10
y
4sin3
t
.
Solution
1. The homogeneous equation
y
00
+
7
y
0
+
10
y
=
characteristic equation 0
=
r
2
+
7
r
+
10
=
(
r
+
2
)(
r
+
5
)
,
with roots
r
2
,

5. Hence
y
h
=
c
1
e

2
t
+
c
2
e

5
t
forcing function is
f
(
t
)
t
not
a solution
of the homogeneous equation.
2. Let
y
p
=
a
cos 3
t
+
b
sin 3
t
y
0
p
3
a
sin 3
t
+
3
b
cos 3
t
y
00
p
9
a
cos 3
t

9
b
sin 3
t
Thus
(

9
a
cos 3
t

9
b
sin 3
t
)
+
7
(

3
a
sin 3
t
+
3
b
cos 3
t
)
+
10
(
a
cos 3
t
+
b
sin 3
t
)
t
. Accordingly, we have
(
a
+
21
b
)
cos3
t
+
(
4
+
b

21
a
)
sin3
t
=
0
=
0cos3
t
+
0sin3
t
3. Equating coefﬁcients of like entities gives
a
+
21
b
=
0and
4

21
a
+
b
=
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