CHAPTER 2: HIGHER ORDER LINEAR ODES
MINGFENG ZHAO
October 18, 2014
• Fundamental Set of Solutions:
Theorem 1. Let p(x) and q(x) be continuous functions, y1 and y2 are two linearly independent solutions to a
homogeneous equation y 00 + p(x)y 0 + q(x)y = 0. Then the general solution to y 00 + p(x)y 0 + q(x)y = 0 is:
y = C1 y1 (x) + C2 y2 (x).
In this case, {y1 (x), y2 (x)} is called a fundamental set of solutions to y 00 + p(x)y 0 + q(x)y = 0.
Example 1. Verify y1 (x) = x−1 and y2 (x) = x2 form a fundamental set of solutions to x2 y 00 − 2y = 0 for x > 0.
In fact, we have
y10 (x)
= −x−2
y100 (x)
=
2x−3
y20 (x)
=
2x
y200 (x)
=
2.
So we have
x2 y100 − 2y1 = x2 · 2x−3x − 2 · x−1 = 0,
and x2 y200 − 2y2 = x2 · 2 − 2 · x2 = 0.
It’s easy to see that y1 = x−1 and y2 (x) = x2 are linearly independent, then
y1 (x) = x−1 and y2 (x) = x2 form a fundamental set of solutions to x2 y 00 − 2y = 0 .
• Constant coefficient second order linear ODEs:
To find the general solution to a constant coefficient second order liner differential equation ay 00 + by 0 + cy = 0:
1) Write the characteristic equation of ay 00 + by 0 + cy = 0:
ar2 + br + c = 0.
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2
MINGFENG ZHAO
2) Find the solutions to the characteristic equation ar2 + br + c = 0.
∗ Compute b2 − 4ac.
∗ Solve ar2 + br + c = 0:
· If ∆ = b2 − 4ac > 0, then r1 , r2 are two different real numbers, and
√
√
−b
b2 − 4ac
b2 − 4ac
−b
r1 =
+
, and r2 =
−
.
2a
2a
2a
2a
· If b2 − 4ac = 0, then we have the same real root, and
r1 = r2 = −
b
.
2a
· If b2 − 4ac < 0, then r1 and r2 are two different complex numbers, and
√
√
4ac − b2
b
4ac − b2
b
, and r2 = − − i
r1 = − + i
2a
2a
2a
2a
3) Write the general solution to ay 00 + by 0 + cy = 0:
∗ If b2 − 4ac > 0, then
y = C1 er1 x + C2 er2 x .
∗ If b2 − 4ac = 0, then
−b
−b
y = C1 e 2a ·x + C2 xe 2a ·x .
∗ If b2 − 4ac < 0, then
√
y = C1 e
b
·x
− 2a
cos
4ac − b2
x
2a
√
!
b
·x
− 2a
+ C2 e
sin
!
4ac − b2
x .
2a
Example 2. Find the solution to 2y 00 + y 0 − 2y = 0, y(0) = a, y 0 (0) = b.
The characteristic equation is
2r2 + r − 3 = 0.
Solve 2r2 + r − 3 = 0, then
r1 = 1,
3
and r2 = − .
2
So, the general solution to 2y 00 + y 0 − 2y = 0 is
3
y(x) = Aex + Be− 2 x .
Then
3
y 0 (x) = Aex − Be−x .
2
CHAPTER 2: HIGHER ORDER LINEAR ODES
3
The initial conditions give
3
b = y 0 (0) = A − B.
2
a = y(0) = A + B,
Solve A and B, we have
A=
3a + 2b
,
5
and B =
2(a − b)
.
5
Therefore, the solution to 2y 00 + y 0 − 3y = 0, y(0) = a, y 0 (0) = b is:
y(x) =
2(a − b) − 3x
3a + 2b x
e 2 +
e .
5
5
Example 3. Find the general solution to y 00 − 4y 0 + 4y = 0.
The characteristic equation of y 00 − 4y 0 + 4y = 0 is:
r2 − 4r + 4 = 0.
Solve r2 − 4r + 4 = 0, then
r1 = r2 = 2.
Then the general solution to y 00 − 4y 0 + 4y = 0 is:
y(x) = C1 e2x + C2 xe2x .
Example 4. Find the general solution to 2y 00 + y 0 + 2y = 0.
The characteristic equation of 2y 00 + y 0 + 2y = 0 is:
2r2 + r + 2 = 0.
Solve 2r2 + r + 2 = 0, then
√
√
15
−1 + 12 − 4 · 2 · 2
1
r1 =
=− +
i,
4
4
4
1
and r2 = − −
4
Then the general solution to 2y 00 + y 0 + 2y = 0 is:
!
√
x
15
−x
y(x) = C1 e 4 cos
x + C2 e− 4 sin
4
√
15
x
4
!
.
√
15
i.
4
4
MINGFENG ZHAO
• Mechanical vibration:
I. Mass-Spring System:
Figure 1. Mass-Spring System
Let x(t) be the displacement of the mass, then
mx00 + cx0 + kx = F (t) .
II. RLC Circuit System:
Figure 2. RLC Circuit System
Let I(t) be the current in the circuit, then
LI 00 (t) + RI 0 (t) +
1
I(t) = E 0 (t) .
C
CHAPTER 2: HIGHER ORDER LINEAR ODES
III. Pendulum Problem:
Figure 3. Pendulum Problem
Let θ(t) be the angle between the vertical line and the pendulum at time t (in seconds), then
θ00 +
g
θ=0.
L
• Unforced Mass-Spring System:
Figure 4. Mass-Spring System
Let x(t) be the displacement of the mass, then
mx00 + cx0 + kx = F (t) .
For free motion (that is, F (t) = 0), rewrite the equation, we have
x00 + 2px0 + ω02 x = 0,
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6
MINGFENG ZHAO
where
c
p=
,
2m
r
and ω0 =
k
.
m
The characteristic equation of x00 + 2px0 + ω02 x = 0 is:
r2 + 2pr + ω02 = 0.
Solve r2 + 2pr + ω02 = 0, we get
r1,2 = −p ±
q
p2 − ω02 .
Then
I. Free undamped motion: c = 0 (that is, p = 0)
The general solution to x00 + ω02 x = 0 is:
x(t) = A cos(ω0 t) + B sin(ω0 t) .
II. Overdamping: c2 − 4km > 0 (that is, p2 − ω02 > 0)
The general solution to x00 + 2px0 + ω02 x = 0 is:
y(t) = Aer1 t + Ber2 t ,
where
r1 = −p −
q
p2 − ω02 < 0,
and r2 = −p +
q
p2 − ω02 < 0.
III. Critical damping: c2 − 4km = 0 (that is, p2 − ω02 = 0)
The general solution to x00 + 2px0 + ω02 x = 0 is:
y(t) = Ae−pt + Bte−pt .
IV. Underdamping: c2 − 4km < 0 (that is, p2 − ω02 < 0)
The general solution to x00 + 2px0 + ω02 x = 0 is:
y(t) = Ae−pt cos
q
q
ω02 − p2 t + Be−pt sin
ω02 − p2 t ,
Example 5. Suppose you want to use a spring to weigh items. You place the mass on the spring and put it in
motion. You have two reference weight 1 kg and 2 kg to calibrate your setup. You put each in motion on your
spring an measure the frequency. For the 1 kg weight you measured 1.1 Hz, for the 2 kg weight you measured
0.8 Hz. a) Find the spring constant k and the damping constant c. b) Find a formula for the mass in terms of
CHAPTER 2: HIGHER ORDER LINEAR ODES
7
the frequency in Hz. c) For an unknown object you measure 0.2 Hz, what is the mass of the object? Suppose
that you know that the mass of the unknown object is more than a kilogram.
The differential equation for the mass spring system is:
mx00 + cx0 + kx = 0.
Rewrite the equation, we have
x00 + 2px0 + ω02 x = 0,
where
c
p=
,
2m
r
and ω0 =
k
.
m
a) By the assumption, we know that the motion is undamped, and the frequency is
r
p
q
4ω02 − 4p2
1
1
k
c2
1
2
2
·
=
· ω0 − p =
·
−
.
2π
2
2π
2π
m 4m2
By the assumption, we know that
r
k
c2
1
·
−
= 1.1,
2π
1 4 · 12
and
1
·
2π
r
k
c2
−
= 0.8.
2 4 · 22
Then we have
k = 4π 2 · 1.35,
and c = 2π ·
√
0.56 .
b) Let f be the frequency corresponding to the weight m, then we have
r
4π 2 · 1.35 4π 2 · 0.56
1
·
−
= f.
2π
m
4m2
That is, we have
f2 =
1.35 0.14
− 2 .
m
m
c) If f = 0.2, then
1.35 0.14
− 2 = 0.22
m
m
That is, we have
0.04m2 − 1.35m + 0.14 = 0.
Then
m=
1.35 +
√
1.352 − 4 · 0.04 · 0.14
≈ 33.645 .
2 · 0.04
8
MINGFENG ZHAO
• Undetermined Coefficients:
Let a, b and c be constants, consider the equation:
ay 00 + by 0 + cy = f (x).
Let pn (x) and p̃n (x) be polynomials with degree n, a particular solution yp (x) to ay 00 + by 0 + cy = f (x) can be
taken as:
f (x)
yp (x)
pn (x)emx cos(kx) + p̃n (x)emx sin(kx)
xα [qn (x)emx cos(kx) + q̃n (x)emx sin(kx)]
where
– α is one of 0, 1 and 2 (α is the multiplicity of m + ki as the solutions to ar2 + br + c = 0):
∗ If m + ki is not a root of ar2 + br + c = 0, then α = 0.
∗ If m + ki is a root of ar2 + br + c = 0 and b2 − 4ac 6= 0, then α = 1.
∗ If m + ki is a root of ar2 + br + c = 0 and b2 − 4ac = 0, then α = 2.
– qn (x) and q̃n (x) are undetermined polynomials with degree n.
Example 6. Solve y 00 + 2y 0 + y = x2 , y(0) = 1, y 0 (0) = 2.
The characteristic equation of y 00 + 2y 0 + y = 0 is:
r2 + 2r + 1 = 0.
Solve r2 + 2r + 1 = 0, then
r1 = r2 = −1.
Then the general solution to y 00 + 2y 0 + y = 0 is:
y(x) = C1 e−x + C2 xe−x .
Let y(x) = Ax2 + Bx + C be a particular solution to y 00 + 2y 0 + y = x2 , then
y 0 (x) = 2Ax + B,
and y 00 (x) = 2A.
Then
2A + 4Ax + 2B + Ax2 + Bx + C = x2 .
That is,
(A − 1)x2 + (2A + B)x + 2A + 2B + C = 0.
CHAPTER 2: HIGHER ORDER LINEAR ODES
9
So we get
A − 1 = 0,
4A + B = 0,
and
2A + 2B + C = 0.
Hence we have
B = −4,
A = 1,
and C = 6.
So a particular solution to y 00 + 2y 0 + y = x2 is:
y(x) = x2 − 4x + 6
Hence the general solution to y 00 + 2y 0 + y = x2 is:
y(x) = C1 e−x + C2 xe−x + x2 − 4x + 6.
Then
y 0 (x) = −C1 e−x + C2 e−x − C2 xe−x + 2x − 4.
Since y(0) = 1 and y 0 (0) = 2, then
C1 + 6 = 1,
and
− C1 + C2 − 4 = 2.
So
C1 = −5,
and C2 = 1.
So the solution to y 00 + 2y 0 + y = x2 , y(0) = 1, y 0 (0) = 2 is:
y(x) = −5e−x + xe−x + x2 − 4x + 6 .
Example 7. Solve the initial value problem y 00 + 9y = cos(3x) + sin(3x) for y(0) = 2, y 0 (0) = 1.
The characteristic equation is
r2 + 9 = 0,
which has the imaginary roots r = ±3i. Therefore the solutions to the homogeneous equation are cos(3x) and
sin(3x). Notice that the inhomogeneous terms are of the same form as the homogeneous solutions, so look for
a particular solution of the form
yp = αx cos(3x) + βx sin(3x).
Then
yp00 = −6α sin(3x) − 9αx cos(3x) + 6β cos(3x) − 9βx sin(3x),
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MINGFENG ZHAO
so the differential equation becomes
−6α sin(3x) + 6β cos(3x) = cos(3x) + sin(3x).
Therefore α = −1/6 and β = 1/6, so the particular solution is
1
1
yp (x) = − x cos(3x) + x sin(3x).
6
6
Then the general solution becomes
1
1
y = A − x cos(3x) + B + x sin(3x).
6
6
Using the initial conditions we get
2 = y(0) = A,
1
1 = y 0 (0) = − + 3B,
6
which gives A = 2 and B = 7/18. So the solution to y 00 + 9y = cos(3x) + sin(3x) for y(0) = 2, y 0 (0) = 1 is:
y(x) =
1
1
(12 − x) cos(3x) + (7 + 3x) sin(3x) .
6
18
Example 8. For an arbitrary constant c find a particular solution to y 00 − y = ecx . (Hint: Make sure to handle
every possible real c)
Observe that any derivative of ecx will be proportional to ecx . Hence we look for a particular solution of the
form h(x)ecx where h(x) is an unknown function to be determined. We have y 00 = h00 ecx + 2ch0 ecx + c2 hecx so
the differential equation becomes
h00 ecx + 2ch0 ecx + c2 hecx − hecx = ecx =⇒ h00 + 2ch0 + (c2 − 1)h = 1.
We now have to look at the particular solutions to this differential equation in h. If c 6= 1 then the particular
solution is h = 1/(c2 − 1) as is easily verified by substituting it into the differential equation. On the other hand
when c = 1 the term (c2 − 1)h disappears and we are left only with h00 + 2h0 = 1. Hence a particular solution
which is a first order polynomial will work. In fact the particular solution is h = x/2. Similarly, when c = −1
we get the particular solution h = −x/2. Putting these together along with y(x) = h(x)ecx we obtain that the
particular solution to y 00 − y = ecx is:

xe−x


y(x) = −
, if c = −1


2

xex
y(x) =
,
if c = 1

2 cx


e


y(x) = 2
, if c =
6 ±1
c −1
.
CHAPTER 2: HIGHER ORDER LINEAR ODES
11
• Variation of Parameters:
To find a particular solution to the nonhomogeneous equation y 00 + p(x)y 0 + q(x)y = f (x):
I. Find a fundamental set of solutions y1 (x) and y2 (x) to the homogeneous equation y 00 + p(x)y 0 + q(x)y = 0.
II. Let yp (x) = u1 (x)y1 (x) + u2 (x)y2 (x) be a particular solution to y 00 + p(x)y 0 + q(x)y = f (x)
III. Compute yp0 (x), we get
yp0 (x)
= u01 (x)y1 (x) + u02 (x)y2 (x)
+u1 (x)y10 (x) + u2 (x)y20 (x)
Take
u01 (x)y1 (x) + u2 (x)y2 (x) = 0.
Then
yp0 (x)
= u1 (x)y10 (x) + u2 (x)y20 (x)
yp00 (x)
= u01 (x)y10 (x) + u1 (x)y100 (x) + u02 (x)y20 (x) + u2 (x)y200 (x).
IV. Plug yp (x), yp0 (x) and yp00 (x) into y 00 + p(x)y 0 + q(x)y = f (x), we get
u01 (x)y10 (x) + u02 (x)y20 (x) = f (x).
V. Solve u01 (x) and u02 (x) from the system:


 u01 (x)y1 (x) + u02 (x)y2 (x) = 0

 u0 (x)y 0 (x) + u0 (x)y 0 (x) = f (x).
1
1
2
2
Then
−y2 (x)f (x)
,
W (y1 , y2 )
and u02 (x) =
−y2 (x)f (x)
dx,
W (y1 , y2 )
and u2 (x) =
u01 (x) =
y1 (x)f (x)
.
W (y1 , y2 )
VI. Solve u1 (x) and u2 (x), then
Z
u1 (x) =
Z
y1 (x)f (x)
dx.
W (y1 , y2 )
VII. Write down the solution:
Z
yp (x) = −y1 (x)
y2 (x)f (x)
dx + y2 (x)
W (y1 , y2 )
Z
y1 (x)f (x)
dx .
W (y1 , y2 )
12
MINGFENG ZHAO
where
y1 (x) y2 (x)
W (y1 , y2 ) = y10 (x) y20 (x)
= y1 (x)y20 (x) − y10 (x)y2 (x).
Example 9. Let α be a real constant.
a) Find a fundamental set of solutions to y 00 − 4y 0 + α(4 − α)y = 0. (Hint: You need to consider two cases
which depend on α)
The characteristic equation of y 00 − 4y 0 + α(4 − α) = 0 is:
r2 − 4t + α(4 − α) = 0.
Solve r2 − 4t + α(4 − α) = 0, we get
r1 = α,
and r2 = 4 − α.
We have the following two cases:
I. If r1 = r2 , that is, α = 4 − α, then α = 2. Then y1 (x) = e2x and y2 (x) = xe2x form a fundamental
set of solutions to y 00 − 4y 0 + 4y = 0.
II. If r1 6= r2 , that is, α 6= 4 − α, then α 6= 2. Then y1 (x) = eαx and y2 (x) = e(4−α)x form a
fundamental set of solutions to y 00 − 4y 0 + α(4 − α) = 0.
b) Compute the Wronskian of those two solutions. (Hint: You need to consider two cases which depend on α)
I. When α = 2, since y1 (x) = e2x and y2 (x) = xe2x , then
2x
e
xe2x
W (y1 , y2 ) = 2e2x e2x + 2xe2x = e2x · e2x + 2xe2x − 2e2x · xe2x
=
=
e4x + 2xe4x − 2xe4x
e4x .
II. When α 6= 2, since y1 (x) = eαx and y2 (x) = e(4−α)x , then
αx
e
e(4−α)x
W (y1 , y2 ) = αeαx (4 − α)e(4−α)x = eαx · (4 − α)e(4−α)x − αeαx · e(4−α)x
=
(4 − α)e4x − αe4x
CHAPTER 2: HIGHER ORDER LINEAR ODES
13
(4 − 2α)e4x .
=
Example 10. Find the general solution to y 00 − 2y 0 + y =
ex
.
1 + x2
The characteristic equation of y 00 − 2y 0 + y = 0 is:
r2 − 2r + 1 = 0.
Solve r2 − 2r + 1 = 0, then
r1 = r2 = 1.
Then the general solution to y 00 − 2y 0 + y = 0 is:
y(x) = C1 ex + C2 xex .
Then y1 (x) = ex and y20 (x) = xex form a fundamental set of solutions to y 00 − 2y 0 + y = 0. Let yp (x) =
ex
u1 (x)y1 (x) + u2 (x)y2 (x) = u1 (x)ex + u2 (x)xex be a particular solution to y 00 − 2y 0 + y =
such that
1 + x2
u01 (x)y1 (x) + u02 (x)y2 (x) = u01 (x)ex + u02 (x)xex = 0.
Then
u01 (x)y10 (x) + u02 (x)y20 (x) = u01 (x) · (ex ) + u02 (x) · (ex + xex ) =
ex
.
1 + x2
So we get


 u01 (x) + xu02 (x) = 0

 u01 (x) + (1 + x)u02 (x) =
1
.
1 + x2
Solve u01 (x) and u02 (x), we get
u01 (x) = −
x
,
x2 + 1
and u02 (x) =
1
.
x2 + 1
Then
Z
u1 (x) = −
x
1
dx = ln(x2 + 1),
x2 + 1
2
and u02 (x) =
Z
So
yp (x) =
ex
ln(x2 + 1) + xex arctan(x).
2
1
dx = arctan(x).
x2 + 1
14
MINGFENG ZHAO
Therefore, the general solution to y 00 − 2y 0 + y =
y(x) = C1 ex + C2 xex +
ex
is:
1 + x2
ex
ln(x2 + 1) + xex arctan(x) .
2
• Forced Motion and Resonance:
Mass-Spring System:
Figure 5. Mass-Spring System
Let x(t) be the displacement of the mass, then
mx00 + cx0 + kx = F (t) .
We are interested in periodic forcing, that is, F (t) = F0 cos(ωt). So we have
mx00 + cx0 + kx = F0 cos(ωt) .
Rewrite the equation, we have
x00 + 2px0 + ω02 x =
F0
cos(ωt) ,
m
where
c
p=
≥ 0,
2m
r
and ω0 =
k
> 0.
m
CHAPTER 2: HIGHER ORDER LINEAR ODES
15
I. Undamped (c = 0):
The differential equation for the undamped forced motion (c = 0) is:
F0
cos(ωt).
m
x00 + ω02 x =
The general solution to x00 + ω02 x =
x(t) =
F0
cos(ω0 t) is:
m



 A cos(ω0 t) + B sin(ω0 t) +
F0
cos(ωt), if ω0 6= ω,
m[ω02 − ω 2 ]


 A cos(ω0 t) + B sin(ω0 t) +
F0
t sin(ω0 t),
2mω0
.
if ω0 = ω
The case of ω0 = ω is called the pure resonance.
II. Dameped (c > 0):
The differential equation for the damped forced motion (c > 0) is:
x00 + 2px0 + ω02 x =
F0
cos(ωt).
m
The transient solution to x00 + 2px0 + ω02 x = 0 is:



Aer1 t + Ber2 t ,
overdamping, that is, p2 − ω02 > 0



Ae−pt + Bte−pt ,
critical damping, that is, p2 − ω02 = 0 .
xtr (t) =
q
q




 Ae−pt cos
ω02 − p2 · t + Be−pt sin
ω02 − p2 · t , underdamping, that is, p2 − ω02 < 0
The steady periodic solution to x00 + 2px0 + ω02 x =
xsp (t) =
F0
cos(ωt) is:
m
2ωpF0
F0 [ω02 − ω 2 ]
· cos(ωt) +
sin(ωt) .
m [(2ωp)2 + (ω02 − ω 2 )2 ]
m [(2ωp)2 + (ω02 − ω 2 )2 ]
The amplitude of xsp is:
C(ω) =
F0
1
·p
.
m
(2ωp)2 + (ω02 − ω 2 )2
Then
∗ If ω02 − 2p2 ≤ 0, then C(ω) has the maximum value at ω = 0. But we assume that ω > 0, so C(ω)
can not attain its maximum, that is,
C(ω) < C(0) =
F0
,
mω02
∀ω > 0.
16
MINGFENG ZHAO
∗ If ω02 − 2p2 > 0, then C(ω) has the maximum value at ω =
p
ω02 − 2p2 , that is,
q
2F0
1
F0
=p
max C(ω) = C( ω02 − 2p2 ) =
·p
2
2
2
2
m
c [4k 2 − c2 ]
4p [ω0 − p ]
p
F0
In this case, ω =
ω02 − 2p2 is called the practical resonance frequency, max C(ω) =
·
m
1
p
is called the practical resonance amplitude.
4p2 [ω02 − p2 ]
Example 11. A mass of 4 kg on a spring with k = 4 and a damping constant c = 1. Suppose F0 = 2.
Using forcing function F0 cos(ωt), find the ω that causes practical resonance and find the amplitude.
By the assumption, m = 4, c = 1, k = 4, and F0 = 2, we have
4x00 + x0 + 4x = 2 cos(ωt).
Let the stead periodic solution be:
xsp (t) = A cos(ωt) + B sin(ωt).
That is, xsp is a particular solution to 4x00 + x0 + 4x = 2 cos(ωt). Then
x0sp (t)
= −Aω sin(ωt) + Bω cos(ωt)
x00sp (t)
= −Aω 2 cos(ωt) − Bω 2 sin(ωt).
Then
4x00sp + x0sp + 4xsp
=
4 −Aω 2 cos(ωt) − Bω 2 sin(ωt) + [−Aω sin(ωt) + Bω cos(ωt)] + 4 [A cos(ωt) + B sin(ωt)]
=
=
2 cos(ωt).
−4Aω 2 + Bω + 4A cos(ωt) + −4Bω 2 − Aω + 4B sin(ωt)
Then
−4Aω 2 + Bω + 4A = 2,
and
− 4Bω 2 − Aω + 4B = 0.
Solve A and B, we get
A=
8 − 8ω 2
,
ω 2 + (4 − ω 2 )2
and B =
ω2
2ω
.
+ (4 − ω 2 )2
Then
xsp (t) = A cos(ωt) + B sin(ωt) =
8 − 8ω 2
2ω
cos(ωt) + 2
sin(ωt).
ω 2 + (4 − 4ω 2 )2
ω + (4 − 4ω 2 )2
CHAPTER 2: HIGHER ORDER LINEAR ODES
17
So the amplitude is:
C(ω)
=
=
=
=
=
=
=
=
p
A2 + B 2
s
2 2
2ω
8 − 8ω 2
+
ω 2 + (4 − 4ω 2 )2
ω 2 + (4 − 4ω 2 )2
p
(8 − 8ω 2 )2 + 4ω 2
ω 2 + (4 − 4ω 2 )2
p
2 (4 − 4ω 2 )2 + ω 2
ω 2 + (4 − 4ω 2 )2
2
p
2
ω + (4 − 4ω 2 )2
2
ω 2 + 16 − 32ω 2 + 16ω 4
2
√
16ω 4 − 31ω 2 + 16
2
q
2
2
31
16 ω 2 − 32 − 16 · 31
+ 16
32
√
2
q
.
31 2
63
2
16 ω − 32 + 64
r
31
, that is, the practical resonance frequency is:
So C(ω) will achieve its maximum at ω =
32
r
31
≈ 0.984 .
ωpr =
32
=
The practical resonance amplitude is:
r
max C(ω) = C(ωpr ) = C
ω>0
31
32
!
2
=q
63
64
16
= √ ≈ 2.016 .
63
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca