1
e

t
+
c
2
e
4
t

1
2
e
2
t
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Second Order Linear Equations
— (18/32)
Introduction
Theory for
2
nd
Order DEs
Linear Constant Coefficient DEs
Homogeneous Equations
Method of Undetermined Coefficients
Forced Vibrations
Method of Undetermined Coefficients
Method of Undetermined Coefficients  Example 2:
Consider
y
00

3
y
0

4
y
= 5 sin(
t
)
From before, the homogeneous solution is
y
c
(
t
) =
c
1
e

t
+
c
2
e
4
t
Neither solution matches the
forcing function
, so try
y
p
(
t
)
=
A
sin(
t
) +
B
cos(
t
)
so
y
0
p
(
t
)
=
A
cos(
t
)

B
sin(
t
)
and
y
00
p
(
t
) =

A
sin(
t
)

B
cos(
t
)
It follows that
(

A
+ 3
B

4
A
) sin(
t
) + (

B

3
A

4
B
) cos(
t
) = 5 sin(
t
)
or 3
A
+ 5
B
= 0 and 3
B

5
A
= 5 or
A
=

25
34
and
B
=
15
34
The solution combines these to obtain
y
(
t
) =
c
1
e

t
+
c
2
e
4
t
+
15
34
cos(
t
)

25
34
sin(
t
)
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Second Order Linear Equations
— (19/32)
Introduction
Theory for
2
nd
Order DEs
Linear Constant Coefficient DEs
Homogeneous Equations
Method of Undetermined Coefficients
Forced Vibrations
Method of Undetermined Coefficients
Method of Undetermined Coefficients  Example 3:
Consider
y
00

3
y
0

4
y
= 2
t
2

7
From before, the homogeneous solution is
y
c
(
t
) =
c
1
e

t
+
c
2
e
4
t
Neither solution matches the
forcing function
, so try
y
p
(
t
) =
At
2
+
Bt
+
C
It follows that
2
A

3(2
At
+
B
)

4(
At
2
+
Bt
+
C
) = 2
t
2

7
,
so matching coefficients gives

4
A
= 2,

6
A

4
B
= 0, and
2
A

3
B

4
C
=

7, which yields
A
=

1
2
,
B
=
3
4
and
C
=
15
16
The solution combines these to obtain
y
(
t
) =
c
1
e

t
+
c
2
e
4
t

t
2
2
+
3
t
4
+
15
16
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Second Order Linear Equations
— (20/32)
Subscribe to view the full document.
Introduction
Theory for
2
nd
Order DEs
Linear Constant Coefficient DEs
Homogeneous Equations
Method of Undetermined Coefficients
Forced Vibrations
Method of Undetermined Coefficients
Superposition Principle:
Suppose that
g
(
t
) =
g
1
(
t
) +
g
2
(
t
). Also,
assume that
y
1
p
(
t
) and
y
2
p
(
t
) are
particular solutions
of
ay
00
+
by
0
+
cy
=
g
1
(
t
)
ay
00
+
by
0
+
cy
=
g
2
(
t
)
,
respectively.
Then
y
1
p
(
t
) +
y
2
p
(
t
) is a solution of
ay
00
+
by
0
+
cy
=
g
(
t
)
From our previous examples, the solution of
y
00

3
y
0

4
y
= 3
e
2
t
+ 5 sin(
t
) + 2
t
2

7
satisfies
y
(
t
) =
c
1
e

t
+
c
2
e
4
t

1
2
e
2
t
+
15
34
cos(
t
)

25
34
sin(
t
)

t
2
2
+
3
t
4
+
15
16
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Second Order Linear Equations
— (21/32)
Introduction
Theory for
2
nd
Order DEs
Linear Constant Coefficient DEs
Homogeneous Equations
Method of Undetermined Coefficients
Forced Vibrations
Method of Undetermined Coefficients
Method of Undetermined Coefficients  Example 4:
Consider
y
00

3
y
0

4
y
= 5
e

t
From before, the homogeneous solution is
y
c
(
t
) =
c
1
e

t
+
c
2
e
4
t
Since the
forcing function
matches one of the solutions in
y
c
(
t
), we
attempt a particular solution of the form
y
p
(
t
) =
Ate

t
,
so
y
0
p
(
t
) =
A
(1

t
)
e

t
and
y
00
p
(
t
) =
A
(
t

2)
e

t
It follows that
(
A
(
t

2)

3
A
(1

t
)

4
At
)
e

t
=

5
Ae

t
= 5
e

t
,
Thus,
A
=

1
The solution combines these to obtain
y
(
t
) =
c
1
e

t
+
c
2
e
4
t

te

t
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Second Order Linear Equations
— (22/32)
Introduction
Theory for
2
nd
Order DEs
Linear Constant Coefficient DEs
Homogeneous Equations
Method of Undetermined Coefficients
 Fall '08
 staff