The H.L.D.E. with Constant Coefficient
Let’s consider the H.L.D.E. with constant coefficients
a
0
y
n
( )
+
a
1
y
n
!
1
(
)
+
.....
+
a
n
!
1
"
y
+
a
n
y
=
0
(1)
where a
0
, a
1
, …. , a
n
are constants.
Notice that the kth derivative of e
rx
is
d
k
dx
k
e
rx
!
"
#
$
=
r
k
e
rx
so any derivative of e
rx
is a multiple of e
rx
.
Assume that the function y = e
rx
is a solution of equation (1). Substituting y = e
rx
and its
derivatives in equation (1), we get
a
o
r
n
e
rx
+
a
1
r
n
!
1
e
rx
+
a
2
r
n
!
2
e
rx
+
…
+
a
n
!
1
re
rx
+
a
n
e
rx
=
0
factoring out e
rx
, and dividing both sides by e
rx
≠
0, we get
a
o
r
n
+
a
1
r
n
!
1
+
a
2
r
n
!
2
+
…
+
a
n
!
1
r
+
a
n
=
0
(2)
Then, the function y = e
rx
is a solution of (1), if r is a zero of the polynomial (2).
Equation (2) is called the
characteristic equation
of the given H.L.D.E. (1).
Therefore, y = e
rx
is a solution of (1), if r is a solution of the corresponding characteristic
equation.
Remark
: We obtain the corresponding characteristic equation of (1) by replacing the kth
derivative of y by r
k
; y itself is consider the derivative order 0, so r
0
= 1
According to the Fundamental Theorem of Algebra, every nth degree polynomial (such
as the characteristic equation) has n zeros, though not necessary distinct and not
necessary real numbers.
So, four different case arise, according to the nature of the zeros of the characteristic
equation:
Case 1: the zeros are real numbers and distinct
Case 2: the zeros are real numbers and repeated
Case 3: the zeros are complex numbers and distinct
Case 4: the zeros are complex numbers and repeated
Case 1
: Distinct real zeros
Suppose that the zeros of the characteristic polynomial are n distinct real numbers:
r
1
, r
2
, r
3
, … , r
n
Then the n functions
y
1
=
e
r
1
x
,y
2
=
e
r
2
x
,y
3
=
e
r
3
x
,
…
,y
n
=
e
r
n
x
are n distinct solutions of
equation (1). They are linearly independent and the general solution of equation (1) can
be expressed as:
y(x)
=
c
1
e
r
1
x
+
c
2
e
r
2
x
+
c
3
e
r
3
x
+
…
+
c
n
e
r
n
x
Example
:
Solve the equation
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View Full Document1) y’’ + 2y’ – 8y = 0
Characteristic equation: r
2
+ 2r – 8 = (r  2)(r + 4) = 0
The zeros are r
1
= 2, and r
2
= 4,
Then y
1
(x) = e
2x
and y
2
(x) = e
4x
are solutions or the set { e
2x
, e
4x
} is the fundamental set
of solutions.
The general solution is: y(x) = c
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 Fall '08
 STAFF
 Complex number, k2, characteristic equation, $ 4, erx, nth

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