The H.L.D.E. with Constant Coefficient
Let’s consider the H.L.D.E. with constant coefficients
a 0 y( n ) + a1y( n!1) + ..... + a n!1y" + a n y = 0 (1)
where a0, a1, …. , an are constants.
Notice that the k-th derivative of erx is
d k rx
! e # = r k e rx
dx k " $
so any derivative of erx is a multiple of erx.
Assume that the function y = erx is a solution of equation (1). Substituting y = erx and its
derivatives in equation (1), we get
a o r n e rx + a1r n!1e rx + a 2 r n!2 e rx + … + a n!1re rx + a n e rx = 0
factoring out erx, and dividing both sides by erx ≠ 0, we get
a o r n + a1r n!1 + a 2 r n!2 + … + a n!1r + a n = 0 (2)
Then, the function y = erx 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 = erx 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 k-th
derivative of y by rk; y itself is consider the derivative order 0, so r0 = 1
According to the Fundamental Theorem of Algebra, every n-th 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:
r1, r2, r3, … , rn
r1 x
r2 x
Then the n functions y1 = e , y 2 = e , y 3 = e r3 x ,…, y n = e rn 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) = c1e r1x + c 2 e r2 x + c 3e r3 x + … + c n e rn x
Example:
Solve the equation
1) y’’ + 2y’ – 8y = 0
Characteristic equation: r2 + 2r – 8 = (r - 2)(r + 4) = 0
The zeros are r1 = 2, and r2 = -4,
Then y1(x) = e2x and y2(x) = e-4x are solutions or the set { e2x, e-4x} is the fundamental set
of solutions.
The general solution is: y(x) = c1e2x + c2e-4x.
2) y’’’ – 4y’’ + y’ + 6y = 0
Characteristic equation: r3 – 4r2 + r + 6 = 0.
Using the fact that a rational number p/q is a zero if p is a factor of 6 and q is a factor of 1
So p = ±1, ±2, ±3, ±6 and q = ±1, therefore the possible rational zeros are: ±1, ±2, ±3,
±6. Substituting, we see that -1 is a zero. Now using synthetic division
1!!!!!!4!!!!!!!!1!!!!!!6
!1 "!!!!!!1!!!!!!!!5!!!!6
#!1! !!!#!1! !!#!1!
!!! 1!
!5
6 !!0
so r3 – 4r2 + r + 6 = (r +1)(r2 – 5r + 6) = (r +1)(r -2)(r -3) = 0
the zeros are: r1 = -1, r2 = 2,and r3 = 3.
Fundamental set of solutions: {e-x, e2x, e3x}
General solution: y(x) = c1e-x + c2e2x + c3e3x.
3) Solve the I.V.P.
y’’ – 4y = 0
y(0) = 0
y’(0) = 3
Characteristic equation: r2 – 4 = (r – 2)(r + 2) = 0
The zeros are r1 = 2, and r2 = -2,
Then y1(x) = e2x and y2(x) = e-2x are solutions or the set { e2x, e-2x} is the fundamental set
of solutions.
The general solution is: y(x) = c1e2x + c2e-2x.
Then, y’(x) = 2c1e2x - 2c2e-2x.
The initial conditions yield
y(0) = c1+ c2 = 0, then c1= -c2 and y’(0) = 2c1- 2c2 = 3, then c1 = ¾ c2 = -¾
Solution I.V.P.: y(x) = ¾e2x - ¾e-2x = 3/2 sinh(2x).
Case 2: Repeated real zeros
Consider the equation y’’ – 6y’ + 9y = 0.
The characteristic equation is: r2 – 6r + 9 = (r – 3)2, so r1 = 3, and r2 = 3 are repeated root,
repeated twice.
Corresponding to r1 = 3, we have the solution y1(x) = e3x and corresponding to r2 = 3 we
have the solution y2(x) e3x, they are not linearly independent.
Since y1(x) = e3x is a solution of the equation, we can apply the method of reduction of
order.
Let y = e3xv a solution,
dy
dv
= e 3x
+ 3e 3x v
dx
dx
2
2
d y
dv
3x d v
=
e
+ 6e 3x
+ 9e 3x v
2
2
dx
dx
dx
Substituting,
! 3x d 2 v
! 3x dv
3x dv
3x $
3x $
3x
# e dx 2 + 6e dx + 9e v& ' 6 #" e dx + 3e v&% + 9e v = 0
"
%
or
d2v
=0
dx 2
Since e3x ≠ 0, divide both side by e3x and call w = dv/dx,
Then dw/dx = 0 or w = c.
Choosing c = 1, w = dv/dx = 1 or v(x) = x.
Therefore, y2(x) = e3xv = xe3x is the second independent solution of the equation.
The general solution is: y(x) = c1e3x + c2xe3x= (c1 + c2x)e3x.
e 3x
Theorem: Consider the n-th order H.L.D.E with constant coefficients (1).
If the corresponding characteristic equation has the real zero r repeating k-times, then
corresponding to this zero we can find k linearly independent solutions
erx, xerx, x2erx, ….. , x(k-1)erx
and as part of the general solution of the equation, we have
y(x) = (c0 + c1x + c2x2 + …… + ck-1 x(k-1))erx + …….
Example: Solve the equation
1) y(4) – 7 y’’’ + 18y’’ – 20y’ + 8y = 0
Characteristic equation: r4 – 7r3 + 18r2 – 20r + 8 = 0
Using the fact that a rational number p/q is a zero if p is a factor of 8 and q is a factor of 1
So p = ±1, ±2, ±4, ±8 and q = ±1, therefore the possible rational zeros are: ±1, ±2, ±4,
±8. Substituting, we see that 1 is a zero. Now using synthetic division
1!!!!!!!7!!!!18!!!20!!!!!8
1 "!!!!!!!1!!!!!!6!!!!12!!!!8
#1! #1! !!#1! #1!
!!!1!
!6 ! 12 !! !8 !!!!!0
so r4 – 7r3 + 18r2 – 20r + 8 = (r - 1)(r3 – 6r2 + 12r – 8) = (r -1)(r – 2)3
The zeros are: 1, 2, 2, 2.
Fundamental set of solutions: {ex, e2x, xe2x, x2e2x}
General solution: y(x) = c1ex + c2e2x + c3xe2x + c4x2e2x = c1ex + (c2 + c3x + c4x2) e2x.
2) y(4) + 2y’’’ + y’’ = 0
Characteristic equation: r4 + 3r3 + r2 = r2(r2 + 2r + 1) = r2(r + 1)2
The zeros are: 0, 0, -1, -1.
Fundamental solution set {e0x, xe0x, e-x, xe-x}={1, x , e-x, xe-x}
General solution: y(x) = c1 + c2x + c3e-x + c4e-x.
Case 3: Distinct conjugate complex zeros
The characteristic equation has non-repeated conjugate complex zeros:
a + bi and a – bi
The corresponding part of the general solution to these two zeros is:
k1e( a+bi)x + k 2 e( a!bi) = k1e ax e ibx + k 2 e ax e !ibx = e ax k1e ibx + k 2 e !ibx , k1 and k2 constants
Now, using the Euler’s Formula:
(
)
(
)
k1e( a+bi)x + k 2 e( a!bi) = k1e ax e ibx + k 2 e ax e !ibx = e ax k1e ibx + k 2 e !ibx =
e ax "# k1 ( cos bx + isin bx ) + k 2 ( cos bx ! isin bx ) $% = e ax "#( k1 + k 2 ) cos bx + i ( k1 ! k 2 ) sin bx $% =
e ax [ c1 cos bx + c 2 sin bx ]
where c1 = k1 + k2 and c2 = i(k1 - k2) are new arbitrary constants.
Theorem: Consider the n-th order H.L.D.E with constant coefficients (1).
If the corresponding characteristic equation has non-repeated conjugate complex zeros
a + bi and a - bi, then corresponding to these zeros we can find two linearly independent
solutions
eax cos bx, and eaxsin bx
and as part of the general solution of the equation, we have
y(x) = ……+ (c0 cos bx + c1 sin bx)eax+ …….
Example: Solve the equation
1) y’’’ – 3y’’ + 9y’ + 13y = 0
Characteristic equation: r3 – 3r2 + 9r + 13 = 0.
Using the fact that a rational number p/q is a zero if p is a factor of 6 and q is a factor of 1
So p = ±1, ±13 and q = ±1, therefore the possible rational zeros are: ±1, ±13.
Substituting, we see that -1 is a zero. Now using synthetic division
1!!!!!!3!!!!!!!!9!!!!!!13
!1 "!!!!!!1!!!!!!!!4!!!!!13
#!1! !!!#!1! !!#!1!
!!! 1!
!4 13 !!0
3
2
2
so r – 3r + 9r + 13 = (r +1)(r – 4r + 13) = 0
let’s use quadratic formula:
4 ± 4 2 ! 4 (13)
4 ± !36
= 2 ± 3i
2
2
the zeros are: r1 = -1, r2 = 2+3i, and r3 = 2 – 3i.
Fundamental set of solutions: {e-x, e2x cos(3x), e2x sin(3x)}
General solution: y(x) = c1e-x + c2e2x cos 3x+ c3e2x sin 3x.
2) y(4) + y = 0
=
Characteristic equation: r4 + 1 = 0 or r4 = -1.
In order to solve the equation we must find the 4-th root of -1,
we represent -1 as a complex number in polar representation:
!1 = cos ( " + 2"k ) + i!sin ( " + 2"k ) ,!k!any!int eger
4
!1 . To accomplish that
then
4
# " + 2"k &
# " + 2"k &
!1 = cos %
+ i!sin %
(
( ,!k = 0,1, 2, 3
$
$
4 '
4 '
2
2
# "&
# "&
k = 0,!cos % ( + i!sin % ( =
+i
$ 4'
$ 4'
2
2
2
2
# " "&
# " "&
# 3" &
# 3" &
k = 1,!cos % + ( + i!sin % + ( = cos % ( + i!sin % ( = !
+i
$ 4 2'
$ 4 2'
$ 4'
$ 4'
2
2
2
2
#"
&
#"
&
# 5" &
# 5" &
k = 2,!cos % + " ( + i!sin % + " ( = cos % ( + i!sin % ( = !
!i
$4
'
$4
'
$ 4 '
$ 4 '
2
2
2
2
# " 3" &
# " 3" &
# 7" &
# 7" &
k = 3,!cos % + ( + i!sin % + ( = cos % ( + i!sin % ( =
!i
$4 2 '
$4 2 '
$ 4 '
$ 4 '
2
2
The general solution is:
y(x) = c1e
2
x
2
! 2 $
cos #
x& + c2e
" 2 %
2
x
!
2 sin
'
2 $
x
+
c
e
3
# 2 &
"
%
2
x
2
'
! 2 $
cos #
x& + c 4 e
" 2 %
2
x
!
2 sin
2 $
# 2 x&
"
%
Case 4: Repeated conjugate complex zeros
Theorem: Consider the n-th order H.L.D.E with constant coefficients (1).
If the corresponding characteristic equation has the conjugate complex zeros a + bi and
a - bi repeating k-times, then corresponding to these zero we can find k linearly
independent solutions
eax cos bx, xeax cos bx, ….. , x(k-1)eax cos bx, eax sin bx, xeax sin bx, ….. , x(k-1)eax sin bx
and as part of the general solution of the equation, we have
y(x) = (c0 + c1x + c2x2 + …… + ck-1 x(k-1))eax cos bx
+ (d0 + d1x + d2x2 + …… + dk-1 x(k-1))eax sin bx+ …..
Example: Solve the equation
1) y(4) + 8y’’ + 16y = 0
Characteristic equation: r4 + 8r2 + 16 = (r2 + 4)2 = 0
The solutions of r2 + 4 = 0 are ±2i, so each of them is a double zero of the characteristic
equation, so the zeros are: 2i, 2i, -2i, -2i.
General solution: y(x) = (c1 + c2x) cos 2x + (d1 + d2x) sin 2x.
3) y(7) + 14 y(5) + 49 y’’’ = 0
Characteristic equation: r7 + 14 r5 + 49r3 = r3(r4 + 14 r2 + 49) = r3(r2 + 7)2 = 0
Then, r = 0 is a triple zero and r = ± 7i is a double zero
{
}
Fundamental set of solutions: 1, x, x 2 , cos 7x, x cos 7x,sin 7x, x sin 7x .
General solution:
y(x) = c1 + c 2 x + c 3x 2 + c 4 cos 7x + c 5 x cos 7x + c 6 sin 7x + c 7 x sin 7x