Fall 2003 Math 308/501–502
4 Linear SecondOrder Equations
4.2 Homogeneous Linear Equations
Wed, 24/Sep
c 2003, Art Belmonte
Summary
In the Section 6.1 lecture handout, there is an overview of the
theory of
n
th order linear differential equations for
n
≥
1,
q.v.
Accordingly, the homogeneous linear secondorder constant
coefficient equations dealt with here are just a special case. Rather
than recapitulate what is covered in §6
.
1, we simply highlight how
to actually find solutions to these special types of equations.
Terminology and Solutions
Consider the ODE
ay
00
+
by
0
+
cy
=
0, where
a
,
b
,
c
are constants
(with
a
6=
0) and
t
is the independent variable (for specificity).
The ODE has an associated
auxiliary
or
characteristic equation
ar
2
+
br
+
c
=
0. This quadratic equation has roots
r
=
r
1
,
r
2
=

b
±
p
b
2

4
ac
2
a
, the discriminant of which is
b
2

4
ac
. In the following ,
c
1
and
c
2
are arbitrary constants.
•
When
b
2

4
ac
>
0, the roots are real and distinct. In this
case,
y
1
(
t
)
=
e
r
1
t
and
y
2
(
t
)
=
e
r
2
t
form a fundamental set of
solutions to the ODE. A general solution is
y
(
t
)
=
c
1
e
r
1
t
+
c
2
e
r
2
t
.
•
When
b
2

4
ac
=
0, there is a repeated or double root. In
this case
y
1
(
t
)
=
e
rt
and
y
2
(
t
)
=
te
rt
form a fundamental set
of solutions to the ODE. A general solution is
y
(
t
)
=
c
1
e
rt
+
c
2
te
rt
.
•
When
b
2

4
ac
<
0, the roots are complex conjugate
numbers. We shall deal with this last case in Section 4.3.
Hand Examples
Example A
Find a general solution of 2
y
00

y
0

y
=
0.
Solution
The characteristic equation is 0
=
2
r
2

r

1
=
(
2
r
+
1
)(
r

1
)
,
whence
r
= 
1
2
or
r
=
1. A general solution of the DE is
y
(
t
)
=
c
1
e

t
/
2
+
c
2
e
t
.
Example B
Find a general solution of 4
y
00
+
12
y
0
+
9
y
=
0.
Solution
The characteristic equation is 4
r
2
+
12
r
+
9
=
0. The quadratic
formula gives
r
=

12
±
√
144

144
8
= 
3
2
, a double root.
Accordingly, a general solution is
y
(
t
)
=
c
1
e

3
2
t
+
c
2
te

3
2
t
.
Example C
Explain why
y
1
(
t
)
=
e

3
t
and
y
2
(
t
)
=
te

3
t
are linearly
independent solutions of the DE
y
00
+
6
y
0
+
9
y
=
0 by using the
definition of linear independence. Also show why the Wronskian
of
y
1
and
y
2
implies this linear independence.
Solution
By computing derivatives and substituting into the differential
equation, we see that both
y
1
and
y
2
are solutions of the DE.
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 Linear Equations, Equations, Derivative, Elementary algebra, Characteristic polynomial, general solution

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