Download Second Order Linear Differential Equations A second order linear

Second Order Linear Differential
Equations
A second order linear differential equation is an equation which can be written in the form
y 00 + p(x)y 0 + q(x)y = f (x)
where
p, q,
and
f
(1)
are continuous
functions on some interval I.
The functions p and q are called the
coefficients of the equation.
1
The function f
is called the forcing
function or the nonhomogeneous term.
“Linear”
Set L[y] = y 00 + p(x)y 0 + q(x)y.
Then, for any two twice differentiable
functions y1(x) and y2(x),
L[y1(x) + y2(x)] = L[y1(x)] + L[y2(x)]
and, for any constant c,
L[cy(x)] = cL[y(x)].
That is,
L
is a linear differential
operator.
2
Existence and Uniqueness
THEOREM: Given the second order
linear equation (1).
Let
a
be any
point on the interval I, and let α and
β be any two real numbers. Then the
initial-value problem
y 00 + p(x) y 0 + q(x) y = f (x),
y(a) = α, y 0(a) = β
has a unique solution.
3
Homogeneous/Nonhomogeneous
Equations
The linear differential equation
y 00 + p(x)y 0 + q(x)y = f (x)
(1)
is homogeneous if the function f on
the right side is
0
for all
x ∈ I. In
this case, the equation becomes
y 00 + p(x) y 0 + q(x) y = 0.
(1) is nonhomogeneous if f
is not
the zero function on I.
4
Section 3.2.
Homogeneous Equa-
tions
y 00 + p(x) y 0 + q(x) y = 0
(H)
where p and q are continuous functions on some interval I.
The zero function,
y(x) = 0
for all
x ∈ I, ( y ≡ 0) is a solution of (H).
The zero solution is called the trivial
solution. Any other solution is a nontrivial solution.
5
Basic Theorems
THEOREM 1:
If
solution of (H) and if
y = y(x)
C
is a
is any real
number, then
u(x) = Cy(x)
is also a solution of (H).
Any constant multiple of a solution of
(H) is also a solution of (H).
6
THEOREM 2:
y = y2(x)
If
y = y1(x)
and
are any two solutions of
(H), then
u(x) = y1(x) + y2(x)
is also a solution of (H).
The sum of any two solutions of (H) is
also a solution of (H). (Some call this
property the superposition principle).
7
DEFINITION:
Let
f = f (x)
and
g = g(x) be functions defined on some
interval I, and let C1 and C2 be
real numbers. The expression
C1f (x) + C2g(x)
is called a linear combination of
and g.
8
f
Theorems 1 & 2 can be restated as:
THEOREM 3:
y = y2(x)
If
y = y1(x)
and
are any two solutions of
(H), and if C1 and C2 are any two
real numbers, then
y(x) = C1y1(x) + C2y2(x)
is also a solution of (H).
Any linear combination of solutions of
(H) is also a solution of (H).
9
NOTE:
y(x) = C1y1(x) + C2y2x
is
a two-parameter family which ”looks
like“ the general solution. Is it???
10
Some Examples from Chapter 1:
y = C1x + C2x3 is the general solution
of
x2y 00−3xy 0+3y
=0
y 00
3 0
3
− y + 2y =0
x
x
y = C1 cos 3x + C2 sin 3x general solution of
y 00 + 9y = 0
11
!
Example:
y 00
1 0 15
− y − 2y = 0
x
x
a. Solutions
y1(x) = x5, y2(x) = 3x5
Gen. solution: y = C1x5 + C2(3x5) ?
12
b. Solutions:
y1(x) = x5, y2(x) = x−3
General solution: y = C1x5 + C2x−3 ?
13
DEFINITION:
y = y2(x)
Let
y = y1(x)
and
be solutions of (H). The
function W defined by
W [y1, y2](x) = y1(x)y20 (x) − y2(x)y10 (x)
is called the Wronskian of y1, y2.
Determinant notation:
W (x) = y1(x)y20 (x) − y2(x)y10 (x)
=
y1(x) y2(x)
y10 (x) y20 (x)
14
THEOREM 4:
Let y = y1(x) and
y = y2(x) be solutions of equation (H),
and let W (x) be their Wronskian. Exactly one of the following holds:
(i) W (x) = 0 for all x ∈ I and y1 is
a constant multiple of y2 or v v
(ii) W (x) 6= 0 for all x ∈ I and
y = C1y1(x) + C2y2(x)
is the general solution of (H)
15
Fundamental Set; Solution basis
DEFINITION A pair of solutions
y = y1(x),
y = y2(x)
of equation (H) forms a fundamental
set of solutions (also called a solution
basis) if
W [y1, y2](x) 6= 0
for all x ∈ I.
16
Section 3.3.
Homogeneous Equa-
tions with Constant Coefficients
Fact:
In contrast to first order lin-
ear equations, there are no methods for
solving
y 00 + p(x)y 0 + q(x)y = 0.
(H)
But, there is a special case of (H) for
which there is a solution method, namely
17
y 00 + ay 0 + by = 0
(1)
where a and b are constants.
Solutions:
(1) has solutions of the
form
y = erx
18
y = erx is a solution of (1) if and only
if
r2 + ar + b = 0
(2)
Equation (2) is called the characteristic equation of equation (1)
19
Note the correspondence:
Diff. Eqn:
y 00 + ay 0 + by = 0
Char. Eqn:
r2 + ar + b = 0
20
The solutions of the differential equation (1) depend on the roots of its
characteristic equation (2). There are
three cases:
1.
(2) has two, distinct real roots,
r1 = α, r2 = β.
2.
(2) has only one real root, r = α.
3.
(2) has complex conjugate roots,
r1 = α + i β,
r2 = α − i β,
β 6= 0.
21
Case I: Two, distinct real roots.
r2 + ar + b = 0 has two distinct real
roots:
r1 = α,
r2 = β, α 6= β.
Then
y1(x) = eαx
and
y2(x) = eβx
are solutions of (1).
22
y1 and y2 are not constant multiples
of each other, {y1, y2} is a fundamental set,
W [y1, y2] =
y = C1 eαx + C2 eβx (gen. solution)
23
Example:
Find the general solution
of
y 00 − 2y 0 − 15y = 0.
Answer: y = C1e5x + C2e−3x.
24
Case II: Exactly one real root.
r = α; (α is a double root). Then
y1(x) = eαx
is one solution of (1).
We need a second solution which is independent of y1.
NOTE: In this case (1) has the form
y 00 − 2αy 0 + α2y = 0
25
y = Ceαx is a solution for any constant
C. Replace C by a function u which
is to be determined so that
y = u(x)eαx
is a solution of (1)
26
y1 and y2 are not constant multiples
of each other, {y1, y2} is a fundamental set,
W [y1, y2] =
y = C1 eαx + C2 xeαx gen. solution.
27
Example:
Find the general solution
of
y 00 + 6y 0 + 9y = 0.
Answer: y = C1e−3x + C2 xe−3x.
28
Popper #2
1.
Given
y 00 − 6y 0 + 8y = 0.
A
fundamental set of solutions is:
(a) {e2x, e6x}
(b) {e−2x, e−4x}
(c) {e2x, e4x}
(d) {e−x, e8x}
(e) None of the above.
29
2.
Given
y 00 + 6y 0 + 9y = 0.
A
fundamental set of solutions is:
(a) {e3x, e−3x}
(b) {e−3x, xe−3x}
(c) {e3x, xe3x}
(d) {e3x}
(e) None of the above.
30
3.
{e3x, e−4x} is a fundamental set
of solutions. The differential equation
is:
(a) y 00 − 7y 0 + 12y = 0
(b) y 00 − y 0 − 12y = 0
(c) y 00 + 7y 0 − 12y = 0
(d) y 00 − 6y 0 − 12y = 0
(e) None of the above
31
Case III: Complex conjugate roots.
r1 = α + i β,
r2 = α − i β,
β 6= 0
In this case
y1(x) = e(α+iβ)x
y2(x) = e(α−iβ)x
are linearly independent solutions of equation (1) and
y = C1 e(α+iβ)x + C2 e(α−iβ)x
is the general solution. BUT, these are
complex-valued functions!!
We want
real-valued solutions!!
32
Recall from Calculus II:
2
3
n
x
x
x
ex = 1 + x +
+
+ ··· +
+ ···
2!
3!
n!
x2
x4
x2n
cos x = 1 −
+
− ··· ±
+ ···
2!
4!
(2n)!
x5
x2n−1
x3
+
···±
+···
sin x = x −
3!
5!
(2n − 1)!
Euler’s Formula: eix = cos x+i sin x
33
y1 and y2 are not constant multiples
of each other, {y1, y2} is a fundamental set,
W [y1, y2] =
y = C1 eαx cos βx + C2 eαx sin βx
34
Example:
Find the general solution
of
y 00 − 4y 0 + 13y = 0.
Answer: y = C1e2x cos 3x+C2e2x sin 3x.
35
Examples:
1. Find the general solution of
y 00 + 6y 0 + 8y = 0.
36
2. Find a solution basis for
y 00 − 10y 0 + 25y = 0.
37
3. Find the solution of the initial-value
problem
y 00−4y 0+8y = 0,
y(0) = 1, y 0(0) = −2.
38
4.
Find the differential equation that
has
y = C1e2x + C2e−3x
as its general solution. (C.f. Chap 1.)
39
5.
y = 5xe−4x is a solution of a sec-
ond order homogeneous equation with
constant coefficients.
a. What is the equation?
b. What is the general solution?
40
From Exercises 1.3:
15. y = C1ex + C2e−2x.
18. y = C1e2x + C2xe2x
19. y = C1 cos 3x + C2 sin 3x.
22. y = C1e2x cos 3x + C2e2x sin 3x.
41
Popper #3
1.
Given y 00 − 2y 0 + 10y = 0. The
general solution is:
(a) y = C1e5x + C2e−2x
(b) y = C1e−x cos 3x + C2e−x sin 3x
(c) y = C1ex cos 3x + C2ex sin 3x
(d) y = C1e2x + C2e−5x
(e) None of the above.
42
2.
y = 7xe−2x
is a solution of
a 2nd order, linear homogeneous equation with constant coefficients.
The
general solution of the equation is:
(a) y = C1e2x + C2e−2x
(b) y = C1e−2x + C2xe−2x
(c) y = C1xe2x + C2xe−2x
(d) None of the above.
43
3.
y = 5e4x
is a solution of a
2nd order, linear homogeneous equation with constant coefficients.
The
equation could be:
(a) y 00 − 7y 0 + 12y = 0
(b) y 00 − 8y 0 + 16y = 0
(c) y 00 − 2y 0 − 8y = 0
(d) All of the above
44
Section 3.4. Second Order Nonhomogeneous Equations
y 00 + p(x)y 0 + q(x)y = f (x)
(N)
The corresponding homogeneous equation
y 00 + p(x)y 0 + q(x)y = 0
(H)
is called the reduced equation of (N).
45
General Results
THEOREM 1: If
z = z1(x)
and
z = z2(x)
are solutions of equation (N), then
y(x) = z1(x) − z2(x)
is a solution of equation (H).
46
THEOREM 2: Let
y = y1(x)
and
y = y2(x)
be linearly independent solutions of the
reduced equation (H) and let z = z(x)
be a particular solution of (N). Then
y(x) = C1y1(x) + C2y2(x) + z(x)
is the general solution of (N).
47
The general solution of (N) consists
of the general solution of the reduced
equation (H) plus a particular solution
of (N):
y=C
y (x) +
C2y2(x)} + z(x).
| 1 1
{z
| {z }
gen soln (H)
+
part soln (N)
48
To find the general solution of (N)
you need to find:
(i) a linearly independent pair of solutions y1, y2 of the reduced equation
(H), and
(ii) a particular solution z of (N).
49
THEOREM: (Superposition Principle)
If z = zf (x) and z = zg (x) are particular solutions of
y 00 + p(x)y 0 + q(x)y = f (x)
and
y 00 + p(x)y 0 + q(x)y = g(x)
respectively, then
z(x) = zf (x) + zg (x)
is a particular solution of
y 00 + p(x)y 0 + q(x)y = f (x) + g(x).
50
Variation of Parameters
Let
y = y1(x)
and
y = y2(x)
be
independent solutions of the reduced
equation (H) and let
W (x) = W (x) = y1y20 − y2y10
be their Wronskian.
Set
z(x) = y1(x)u(x) + y2(x)v(x)
where u and v are to be determined
so that z is a solution of (N).
51
u0(x)
u(x) =
−y2(x)f (x)
=
,
W (x)
Z
v 0(x)
v(x) =
−y2(x)f (x)
dx;
W (x)
y1(x)f (x)
=
W (x)
Z
y1(x)f (x)
dx
W (x)
and
z(x) =
y1(x)
Z
Z y (x)f (x)
−y2(x)f (x)
1
dx + y2(x)
dx
W (x)
W (x)
52
Examples:
1.
{y1(x) = x2, y2(x) = x4}
is a
fundamental set of solutions of the reduced equation of
y 00
5 0
8
− y + 2 y = 4x3
x
x
Find a particular solution
z
of the
equation.
Answer:
4 5
z(x) = x
3
53
{y1(x) = e2x, y2(x) = xe2x}
2.
is
a fundamental set of solutions of the
reduced equation of
2x
e
y 00 − 4y 0 + 4y =
x
Find the general solution of the equation.
Answer:
y(x) = C1 e2x + C2 xe2x + xe2x ln x
54
{y1(x) = e2x, y2(x) = e−3x}
3.
is
a fundamental set of solutions of the
reduced equation of
y 00 + y 0 − 6y = 3e2x
Find the general solution of the equation.
Answer:
y(x) = C1
e2x
+ C2
e−3x
3 2x
+ xe
5
55
Section 3.5. Undetermined Coefficients aka ”Guessing”
Find a particular solution of
y 00 + ay 0 + by = f (x)
NOTE: This method can be used
only when:
1.
constant coefficients
2.
f is an ”exponential” function
56
• If f (x) = cerx
set
z(x) = Aerx
Example:
y 00 − 5y 0 + 6y = 7e−4x.
Set
z = Ae−4x
where A is to be
determined.
57
−4x.
Answer: z = 1
e
6
The general solution of the differential
equation is:
y = C1
e2x
+ C2
e3x
1 −4x
.
+ e
6
58
• If f (x) = c cos βx,
or
set
d sin βx,
c cos βx + d sin βx,
z(x) = A cos βx + B sin βx
where A, B are to be determined
59
Example:
y 00 − 2y 0 + y = 3 cos 2x.
Set
z = A cos 2x + B sin 2x
60
9 cos 2x − 12 sin 2x.
Answer: z = − 25
25
The general solution of the differential
equation is:
y=
C1ex+C2xex−
12
9
cos 2x−
sin 2x.
25
25
61
Example:
Find a particular solution
of
y 00 − 2y 0 + 5y = 2 cos 3x − 4 sin 3x + e2x
Set
z = A cos 3x + B sin 3x + Ce2x
where A, B, C are to be determined.
Answer
8
1
1 2x
z=−
cos 3x +
sin 3x + e .
13
13
5
62
• If f (x) = ceαx cos βx,
or
set
deαx sin βx
ceαx cos βx + deαx sin βx
z(x) = Aeαx cos βx + Beαx sin βx
where A, B are to be determined.
63
Example:
y 00 + 9y = 4ex sin 2x.
Set z = Aex cos 2x + Bex sin 2x
4 ex cos 2x+ 6 ex sin 2x.
Answer: z = − 13
13
64
A BIG Difficulty:
The trial solution
z is a solution of the reduced equation.
65
Examples:
1. Find a particular solution z1 of
y 00 + y 0 − 6y = 3e2x.
66
2. Find a particular solution z2 of
y 00 − 6y 0 + 9y = 5e3x.
67
3. Find a particular solution z3 of
y 00 + 4y = 2 cos 2x.
68
4. Find a particular solution z4 of
y 00 − 5y 0 + 6y = 4e2x + 3
69
Answers:
3 2x
z1 = x e
5
5 2 3x
z2 = x e
2
1
z3 = x sin 2x
2
z4 =
−4xe2x
1
+
2
70
The Method of Undetermined Coefficients
A.
Applies only to equations of the
form
y 00 + ay 0 + by = f (x)
where a, b are constants and f is an
“exponential” function.
c.f Variation of Parameters
71
B. Basic Case:
• f (x) = aerx
If:
set z = Aerx.
• f (x) = c cos βx, d sin βx, or
c cos βx + d sin βx,
set z = A cos βx + B sin βx.
• f (x) = ceαx cos βx,
deαx sin βx or
ceαx cos βx + deαx sin βx,
set z = Aeαx cos βx + Beαx sin βx.
72
BUT:
•
If
z
satisfies the reduced
equation, try
xz;
• if xz also satisfies the reduced
equation, then x2z
will give a
particular solution.
73
C. General Case:
• If
f (x) = p(x)erx
where p is a polynomial of degree n,
then
set
z = P (x)erx
where P is a polynomial of degree n
with undetermined coefficients.
74
Example:
Find a particular solution
of
y 00 − y 0 − 6y = (2x2 − 1)e2x.
Set z = (Ax2 + Bx + C)e2x.
Answer: z =
1 x2
−2
−
3x − 9
4
16
e2x.
75
• If
f (x) = p(x) cos βx + q(x) sin βx
where p, q are polynomials of degree
n, then
set
z = P (x) cos βx + Q(x) sin βx
where P, Q are polynomials of degree
n with undetermined coefficients.
76
Example:
y 00 − 2y 0 − 3y = 3 cos x + (x − 2) sin x.
Set z = (Ax+B) cos x+(Cx+D) sin x.
Answer:
1
47
1
2
x−
cos x − x −
sin x.
z=
10
50
5
25
!
!
77
• If
f (x) = p(x)eαx cos βx + q(x)eαx sin βx
where p, q are polynomials of degree
n, then
set
z = P (x)eαx cos βx + Q(x)eαx sin βx
where P, Q are polynomials of degree
n with undetermined coefficients.
78
Example:
y 00 + 4y = 2x ex cos x.
Set
z = (Ax + B)ex cos x + (Cx + D)ex sin x
Answer:
1
1
x
z=
(10x−7)e cos x+ (5x−1)ex sin x.
25
25
79
BUT:
Warning!!!
• If any part of z satisfies the
reduced equation, try
xz;
• if any part of xz also satisfies
the reduced equation, then x2z
will give a particular solution.
80
Examples:
1.
Give the form of a particular
solution of
y 00 − 4y 0 − 5y = 2 cos 3x − 5e5x + 4.
Answer:
z = A cos 3x + B sin 3x + Cxe5x + D
81
2.
Give the form of a particular
solution of
y 00 + 8y 0 + 16y = 2x − 1 + 7e−4x.
Answer:
z = Ax + B + Cx2e−4x
82
3.
Give the form of a particular
solution of
y 00 + y = 4 sin x − cos 2x + 2e2x.
Answer:
z = Ax cos x+Bx sin x+C cos 2x+D sin 2x
+Ee2x
83
4.
Give the form of the general
solution of
y 00 + 9y = −4 cos 2x + 3 sin 2x
Answer:
z = A cos 2x + B sin 2x
84
5.
Give the form of the general
solution of
y 00 + 9y = −4 cos 3x + 3 sin 2x
z = Ax cos 3x+Bx sin 3x+D cos 2x+E sin 2x
85
6.
Give the form of the general
solution of
y 00 + 4y 0 + 4y = 4xe−2x + 3
Answer:
z = (Ax3 + Bx2)e−2x + C
86
7.
Give the form of the general
solution of
y 00 + 4y 0 + 4y = 4e−2x sin 2x + 3x
Answer:
z = Ae−2x cos 2x+Be−2x sin 2x+Cx+D
87
8.
Give the form of the general
solution of
y 00 + 4y 0 = 4 sin 2x + 3
Answer: z = A cos 2x + B sin 2x + Cx
88
9.
Give the form of the general
solution of
y 00 + 2y 0 + 10y = 2e3x sin x + 4e3x
Answer:
z = Ae3x cos x + Be3x sin x + Ce4x
89
10.
Give the form of the general
solution of
y 00 + 2y 0 + 10y = 2e−x sin 3x + 2e−x
Answer:
z = Axe−x cos 3x+Bxe−x sin 3x+Ce−x
90
11.
Give the form of a particular
solution of
y 00 −2y 0 −8y = 2 cos 3x−(3x+1)e−2x −4
Answer:
z = A cos 3x+B sin 3x+(Cx2+Dx)e−2x+E
91
12.
Give the form of a particular
solution of
y 00 − 2y 0 − 8y = 2 cos 3x − 3xe−2x − 3x
Answer:
z = A cos 3x+B sin 3x+(Cx2+Dx)e−2x
+Ex + F
92
13.
Find the general solution of
2x
e
y 00 − 4y 0 + 4y = −4e2x +
x
Answer:
y = C1e2x + C2xe2x − 2x2e2x + xe2x ln x
93
Summary:
1.
Variation of parameters:
• Can be applied to any linear nonhomogeneous equations, but
• requires a fundamental set of solutions of the reduced equation.
94
2.
Undetermined coefficients:
• Is limited to linear nonhomogeneous
equations with constant coefficients, and
•
f
must be an “exponential func-
tion,”
f (x) = aerx, f (x) = c cos βx + d sin βx,
f (x) = ceαx cos βx + deαx sin βx,
or p(x)f (x) p a polynomial.
95
In cases where both methods are
applicable, the method of undetermined coefficients is usually more
efficient and, hence, the preferable
method.
96
Section 3.6. Vibrating Mechanical
Systems
97
I. Undamped, Free Vibrations
Hooke’s Law: The restoring force of
a spring is proportional to the displacement:
F = −ky, k > 0.
Newton’s Second Law: Force equals
mass times acceleration:
d2 y
F = ma = m 2 .
dt
Mathematical model:
d2 y
m 2 = −ky
dt
98
which can be written
d2 y
2y = 0
+
ω
dt2
where ω =
q
k/m.
ω/2π is called the natural frequency
of the system.
The general solution of this equation
is:
y = C1 sin ωt + C2 cos ωt
which can be written
y = A sin (ωt + φ).
A is called the amplitude, φ is called
the phase shift.
99
II. Damped, Free Vibrations: A resistance force R (e.g., friction) proportional to the velocity
v = y0
and
acting in a direction opposite to the
motion:
R = −cy 0
with c > 0.
Force Equation:
F = −ky(t) − cy 0(t).
Newton’s Second Law:
F = ma = my 00
100
Mathematical Model:
my 00(t) = −ky(t) − cy 0(t)
or
y 00 +
c 0 k
y + y=0
m
m
(c, k, m constant)
Characteristic equation:
r2
c
k
+ r+
= 0.
m
m
Roots
q
−c ± c2 − 4km
.
r=
2m
There are three cases to consider:
c2 − 4km < 0,
c2 − 4km > 0,
c2 − 4km = 0.
101
Case 1: c2 −4km < 0. Complex roots:
(Underdamped)
c
r1 = −
+ iω,
2m
q
where ω =
c
r2 = −
− iω
2m
4km − c2
.
2m
General solution:
y = e(−c/2m)t (C1 cos ωt + C2 sin ωt)
102
or
y(t) = A e(−c/2m)t sin (ωt + φ0)
where
A
and
φ0
are constants,
NOTE: The motion is oscillatory AND
y(t) → 0
as
t → ∞.
103
Underdamped Case:
104
Case 2:
c2 − 4km > 0. Two distinct
real roots:
(Overdamped)
q
q
−c + c2 − 4km
−c − c2 − 4km
r1 =
, r2 =
.
2m
2m
General solution:
y(t) = y = C1er1t + C2er2t.
The motion is nonoscillatory.
105
NOTE: Since
q
c2 − 4km <
q
c2 = c,
r1 and r2 are both negative and
y(t) → 0
as
t → ∞.
106
Case 3: c2 − 4km = 0. One real root:
(Critically Damped)
−c
r1 =
,
2m
General solution:
y(t) = y = C1e−(c/2m) t + C2 t e−(c/2m) t.
The motion is nonoscillatory and
y(t) → 0 as t → ∞.
107
Overdamped and Critically Damped Cases:
108
III. Forced Free Vibrations
Apply an external force G to an undamped, freely vibrating system
Force Equation:
F = −ky + G.
Mathematical Model:
my 00
= −ky+G
or
G
y= ,
m
m
k
00
y +
a nonhomogeneous equation.
109
A periodic external force:
G = α cos γt,
α, γ > 0 const.
Force Equation:
F = −ky + α cos γt
Mathematical Model:
y 00
α
k
cos γt
+ y=
m
m
y 00
+ ω 2y
where ω =
α
cos γt
=
m
q
k/m.
110
ω/2π is called the natural frequency
of the system,
γ/2π is called the applied frequency.
111
Case 1: γ 6= ω.
y 00
+ ω 2y
α
=
cos γt
m
General solution, reduced equation:
y = A sin (ωt + φ0).
Form of particular solution (undetermined coefficients):
z = A cos γt + B sin γt.
A particular solution:
α/m
cos γt.
z= 2
2
ω −γ
112
General solution:
α/m
y = A sin (ωt + φ0) + 2
cos γt.
2
ω −γ
ω/γ rational: periodic motion
ω/γ irrational: not periodic
113
Case 2: γ = ω.
y 00
+ ω 2y
α
=
cos ωt
m
General solution, reduced equation:
y = A sin (ωt + φ0).
Form of particular solution (undetermined coefficients):
z = A t cos γt + B t sin γt.
A particular solution:
α
t sin ωt.
2ωm
114
General solution:
α
y = A sin (ωt + φ0) +
t sin ωt.
2ωm
Resonance: Unbounded oscillation!
Resonance:
α
y = A sin (ωt + φ0) +
t sin ωt.
2ωm
115
IV. Forced Damped Vibrations
Apply an external force G to a damped,
freely vibrating system
Force Equation:
F = −ky − cy 0 + G.
Mathematical Model:
my 00 = −ky − cy 0 + G
or
y 00
k
G
c 0
+ y + y= ,
m
m
m
a nonhomogeneous equation.
116
A periodic external force:
G = α cos γt,
α, γ > 0 const.
Force Equation:
F = −ky − cy 0 + α cos γt
Mathematical Model:
y 00
or
y 00
k
α
c 0
cos γt
+ y + y=
m
m
m
+ α y0
α
+βy =
cos γt
m
where α = c/m, β = k/m.
117
General solution, reduced equation:
Case I: distinct real roots:
Yc(t) = C1 er1t + C2 er2t,
r1 < 0, r2 < 0
Note:
lim Yc(t) = 0.
t→∞
118
Case II: real, equal roots:
Yc(t) = C1ert + C2 t ert,
r1 = r2 = r < 0
Note:
lim Yc(t) = 0.
t→∞
Case III: complex roots:
Yc(t) = C1 ert cos ωt + C2 ert sin ωt,
r1 = r + ωi, r2 = r − ωi < 0, r < 0.
Note:
lim Yc(t) = 0.
t→∞
119
Particular solution of
(N )
y 00
+ α y0
α
+βy =
cos γt
m
will have the form:
Z(t) = A cos γt + B sin γt.
General solution of (N):
y(t) = Yc(t) + Z(t)
Note:
lim y(t) = Z(t)
t→∞
120
Yc(t),
the general solution of the re-
duced equation, is called the transient
solution.
Z(t)
the particular solution of (N), is
called the steady state solution.
121
Examples:
1.
Find the general solution of:
y 000 + 3y 00 − 6y 0 − 8y = 0
r = 2 is a root of the char. poly.
122
2.
Find the general solution of:
y 000 + 5y 00 + 7y 0 + 3y = 0
r = −3 is a root of the char. poly.
123
3.
Find the general solution of:
y (4) + 2y 000 + 9y 00 − 2y 0 − 10y = 0
r = −1 + 3i is a root of the char. poly.
124
4.
y = C1e2x+C2e−2x+C3 cos 2x+C4 sin 2x
is the gen. soln. of a homogeneous
equation. What’s the equation?
125
5.
y = 2e−x − 3 sin 4x + 2x
is a solution of a homogeneous equation. What’s the equation?
126
6.
Give the form of a particular
solution of
y (4) + 4y 000 + 13y 00 + 36y 0 + 36y =
5e−2x + sin 2x + 6
127
7.
Give the form of the general
solution of
y (4) − 16y = 2 cos 2x − 3xe−2x − 3x
128