chapter 3 slides
TRANSCRIPT
Second Order Linear Differential
Equations
A second order linear differential equa-
tion is an equation which can be writ-
ten in the form
y′′ + p(x)y′ + q(x)y = f(x) (1)
where p, q, and f 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′′ + p(x)y′ + 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′′ + p(x) y′ + q(x) y = f(x),
y(a) = α, y′(a) = β
has a unique solution.
3
Homogeneous/Nonhomogeneous
Equations
The linear differential equation
y′′ + p(x)y′ + 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′′ + p(x) y′ + q(x) y = 0.
(1) is nonhomogeneous if f is not
the zero function on I.4
Homogeneous Equations
y′′ + p(x) y′ + q(x) y = 0 (H)
where p and q are continuous func-
tions on some interval I.
Trivial Solution The zero function,
y(x) = 0 for all x ∈ I, ( y ≡ 0) is
a solution of (H). (y ≡ 0 implies y′ ≡
0 and y′′ ≡ 0). The zero solution is
called the trivial solution. Any other
solution is a nontrivial solution.5
Basic Theorems
THEOREM 1 If y = y(x) is a solu-
tion of (H) and if C is any real num-
ber, 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 If y = y1(x) and
y = y2(x) 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 (Linear Combinations)
Let f = f(x) and g = g(x) be func-
tions 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 f
and g.
8
THEOREM 3 If y = y1(x) and
y = y2(x) 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???
Example: y′′ −1
xy′ −
15
x2y = 0
a. y1(x) = x5, y2(x) = 3x5
y = C1x5 + C2(3x5)
b. y1(x) = x5, y2(x) = x−3
y = C1x5 + C2x−3
10
DEFINITION: Wronskian
Let y = y1(x) and y = y2(x) be solu-
tions of (H). The function W defined
by
W [y1, y2](x) = y1(x)y′2(x) − y2(x)y
′1(x)
is called the Wronskian of y1, y2.
Determinant notation:
W (x) = y1(x)y′2(x) − y2(x)y
′1(x)
=
∣∣∣∣∣∣y1(x) y2(x)y′1(x) y′2(x)
∣∣∣∣∣∣
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THEOREM 4 Let y = y1(x) and
y = y2(x) be solutions of equation (H),
and let W (x) be their Wronskian. Ex-
actly one of the following holds:
(i) W (x) = 0 for all x ∈ I and y1 is
a constant multiple of y2 or vv.
(ii) W (x) 6= 0 for all x ∈ I and
y = C1y1(x) + C2y2(x)
is the general solution of (H)
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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.
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Homogeneous Equations with Con-
stant Coefficients
y′′ + ay′ + by = 0 (1)
where a and b are constants.
Solutions: y = erx
is a solution if and only if
r2 + ar + b = 0 (2)
Equation (2) is called the character-
istic equation of equation (1)
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The solutions of the differential equa-
tion (1) depend on the roots of its
characteristic equation (2).
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.
15
Case I: (2) has two, distinct real roots,
r1 = α, r2 = β. Then
y1(x) = eαx and y2(x) = eβx
are solutions of (1). α 6= β, y1 and
y2 are not constant multiples of each
other, {y1, y2} is a fundamental set,
W [y1, y2] = (β − α)e(α+β)x 6= 0,
and
y = C1 eαx + C2 eβx
is the general solution.
16
Example: Find the general solution
of
y′′ − 2y′ − 15y = 0.
Answer: y = C1e5x + C2e−3x.17
Case II: The characteristic equation
has only one real root, r = α; (α is
a double root). Then
y1(x) = eαx and y2(x) = x eαx
are linearly independent solutions of equa-
tion (1) and
y = C1 eαx + C2 x eαx
is the general solution.
18
Example: Find the general solution
of
y′′ + 6y′ + 9y = 0.
Answer: y = C1e−3x + C2 xe−3x.19
Case III: The characteristic equation
has complex conjugate roots:
r1 = α + i β, r2 = α + i β, β 6= 0
In this case
y1(x) = eαx cos βx, y2(x) = eαx sin βx
are linearly independent solutions of equa-
tion (1) and
y = C1 eαx cos βx + C2 eαx sin βx
is the general solution.
20
Example: Find the general solution
of
y′′ − 4y′ + 13y = 0.
Answer: y = C1e2x cos 3x+C2e2x sin 3x.
21
Examples:
1. Find the general solution of
y′′ + 6y′ + 8y = 0.
2. Find the general solution of
y′′ − 10y′ + 25y = 0.
3. Find the solution of the initial-value
problem
y′′−4y′+8y = 0, y(0) = 1, y′(0) = −2.
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4. Find the differential equation that
has
y = C1e2x + C2e−3x
as its general solution.
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?
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6. y = 4e−2x sin 3x is a solution of
a second order homogeneous equation
with constant coefficients.
a. What is the equation?
b. What is the general solution?
24
Second Order Nonhomogeneous
Equations
y′′ + p(x)y′ + q(x)y = f(x) (N)
The corresponding homogeneous equation
y′′ + p(x)y′ + q(x)y = 0 (H)
is called the reduced equation of (N).
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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).
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).
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The general solution of (N) consists of the gen-
eral solution of the reduced equation (H) plus
a particular solution of (N):
y = C1y1(x) + C2y2(x)︸ ︷︷ ︸general solution of (H)
+ z(x).︸ ︷︷ ︸part. soln. of (N)
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).
27
THEOREM (Superposition Principle) If z =
zf(x) and z = zg(x) are particular solutions
of
y′′ + p(x)y′ + q(x)y = f(x)
and
y′′ + p(x)y′ + q(x)y = g(x)
respectively, then
z(x) = zf(x) + zg(x)
is a particular solution of
y′′ + p(x)y′ + q(x)y = f(x) + g(x).
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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) = y1y′2 − y2y′1
be their Wronskian.
Set z(x) = y1(x)u(x) + y2(x)v(x)
where
u′(x) = −y2(x)f(x)
W (x)dx, v′(x) =
y1(x)f(x)
W (x)dx,
Then, z is a particular solution of the nonho-
mogeneous equation (N).
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That is,
u(x) =∫ −y2(x)f(x)
W (x)dx,
v(x) =∫
y1(x)f(x)
W (x)dx,
and
z(x) =
y1(x)∫ −y2(x)f(x)
W (x)dx + y2(x)
∫y1(x)f(x)
W (x)dx
is a particular solution of (N).
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Examples:
1. {y1(x) = x2, y2(x) = x4} is a fundamental
set of solutions of the reduced equation of
y′′ −5
xy′ +
8
x2y = 4x3
Find a particular solution z of the equation.
Answer: z(x) =4
3x5
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2. {y1(x) = e2x, y2(x) = e−3x} is a funda-
mental set of solutions of the reduced equation
of
y′′ + y′ − 6y = 3e2x
Find the general solution of the equation.
Answer: y(x) = C1 e2x + C2 e−3x +3
5x e2x
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Undetermined Coefficients
A particular solution of y′′ + ay′ + by = f(x) :
• If f(x) = cerx set z(x) = Aerx
Example: Find a particular solution of
y′′ − 5y′ + 6y = 7e−4x.
Set
z = Ae−4x
where A is to be determined
Answer: z = 16 e−4x.
The general solution of the differential equa-
tion is:
y = C1e2x + C2e3x +1
6e−4x.
33
• If f(x) = c cos βx, d sin βx,
or c cos βx + d sin βx
set z(x) = A cos βx + B sin βx
Example: Find a particular solution of
y′′ − 2y′ + y = 3cos 2x.
Set
z = A cos 2x + B sin 2x
where A, B are to be determined
Answer: z = − 925 cos 2x − 12
25 sin 2x.
The general solution of the differential equa-
tion is:
y = C1ex + C2xex −9
25cos 2x −
12
25sin 2x.
34
Example: Find a particular solution of
y′′ − 2y′ + 5y = 2cos 3x − 4 sin 3x + 7
Set
z = A cos 3x + B sin 3x + C
where A, B, C are to be determined.
Answer: z = − 813 cos 3x + 1
13 sin 3x + 75.
35
• If f(x) = ceαx cos βx, deαx sin βx
or ceαx cos βx + deαx sin βx
set z(x) = Aeαx cos βx + Beαx sin βx
Example: Find a particular solution of
y′′ + 9y = 4ex sin 2x.
Set
z = Aex cos 2x + Bex sin 2x
where A, B are to be determined.
Answer: z = − 213 ex cos 2x + 6
13 ex sin 2x.
36
A Difficulty: The trial solution z is a solution
of the reduced equation.
Examples:
1. Find a particular solution of
y′′ − 6y′ + 8y = 3e2x.
2. Find a particular solution of
y′′ − 6y′ + 9y = 5e3x.
3. Find a particular solution of
y′′ + 4y = 4cos 2x.
Answers:
z1 = −3
2x e2x
z2 =5
2x2 e3x
z3 = x sin 2x +1
4cos 2x
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The Method of Undetermined
Coefficients
A. Applies only to equations of the form
y′′ + ay′ + by = f(x)
where a, b are constants and f is an “ex-
ponential” function.
38
B. Basic Case: If:
• f(x) = aerx 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.
Note: If z satisfies the reduced equation, try
xz; if xz also satisfies the reduced equation,
then x2z will give a particular solution.
39
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.
Example: Find a particular solution of
y′′ − y′ − 6y = (2x2 − 1)e2x.
Set z = (Ax2 + Bx + C)e2x.
Answer: z =(−1
2x2 − 34x − 9
16
)e2x.
40
• 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.
Example: Find a particular solution of
y′′ − 2y′ − 3y = x cos 2x + (3x − 1) sin 2x.
Set z = (Ax + B) cos 2x + (Cx + D) sin 2x.
Answer:
z =1
169(13x−46) cos 2x−
1
169(65x−9) sin 2x.
41
• 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.
Example: Find a particular solution of y′′ +
4y = 2x ex cos x.
Answer:
z =1
25(10x − 7)ex cos x +
1
25(5x − 1)ex sin x.
42
Note: 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.
43
Examples
Give the form of a particular solution of the
differential equation:
1. y′′ − 4y′ − 5y = 2cos 3x − 5e5x + 4.
2. y′′ + 8y′ + 16y = 2x − 1 + 7e−4x.
3. y′′ + y = 4sin x − cos 2x + 2e2x.
Answers:
z1 = A cos 3x + B sin 3x + Cxe5x + D.
z2 = Ax + B + Cx2e−4x.
z3 = Ax cos x+Bx sin x+C cos 2x+D sin 2x+Ee2x.
44
Summary:
1. Variation of parameters:
Can be applied to any linear nonhomogeneous
equations, but
requires a fundamental set of solutions of the
reduced equation.
2. Undetermined coefficients:
Is limited to linear nonhomogeneous equations
with constant coefficients, and
f must be an “exponential function,” i.e.,
f(x) = aerx, f(x) = c cos βx + d sin βx,
f(x) = ceαx cos βx + deαx sin βx
45
In cases where both methods are applicable,
the method of undetermined coefficients is usu-
ally more efficient and, hence, the preferable
method.
Vibrating Mechanical Systems
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:
F = ma = md2y
dt2.
Mathematical model:
md2y
dt2= −ky
which can be written
d2y
dt2+ ω2y = 0
where ω =√
k/m.
46
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.
47
Damped, Free Vibrations A resistance force
R proportional to the velocity v = y′ and
acting in a direction opposite to the motion:
R = −cy′ with c > 0.
Force Equation:
F = −ky(t) − cy′(t).
Newton’s Second Law: F = ma = my′′
Mathematical Model:
my′′(t) = −ky(t) − cy′(t)
or
y′′ +c
my′ +
k
my = 0 (c, k, m constant)
48
Characteristic equation:
r2 +c
mr +
k
m= 0.
Roots
r =−c ±
√c2 − 4km
2m.
There are three cases to consider:
c2 − 4km < 0,
c2 − 4km > 0,
c2 − 4km = 0.
49
Case 1: c2 − 4km < 0. Complex roots:
(Underdamped)
r1 = −c
2m+ iω, r2 = −
c
2m− iω
where ω =
√4km − c2
2m.
General solution:
y = e(−c/2m)t (C1 cos ωt + C2 sin ωt)
or
y(t) = A e(−c/2m)t sin (ωt + φ0)
where A and φ0 are constants, A > 0, φ0 ∈
[0,2π). The motion is oscillatory.
50
Underdamped Case:
51
Case 2: c2 − 4km > 0. Two distinct real
roots:
(Overdamped)
r1 =−c +
√c2 − 4km
2m, r2 =
−c −√
c2 − 4km
2m.
General solution:
y(t) = y = C1er1t + C2er2t.
The motion is nonoscillatory.
Since√
c2 − 4km <
√c2 = c,
r1 and r2 are both negative and y(t) → 0
as t → ∞.
52
Case 3: c2 − 4km = 0. One real root:
(Critically Damped)
r1 =−c
2m,
General solution:
y(t) = y = C1e−(c/2m) t + C2t e−(c/2m) t.
The motion is nonoscillatory and y(t) → 0 as
t → ∞.
53
Overdamped and Critically Damped Cases:
54
Forced Vibrations
Apply an external force G to an undamped,
freely vibrating system
Force Equation:
F = −ky + G.
Mathematical Model:
my′′ = −ky + G or y′′ +k
my =
G
m,
a nonhomogeneous equation.
55
A periodic external force:
G = F0 cos γt,
F0 and γ positive constants.
Force Equation:
F = −kx + F0 cos γt
Mathematical Model:
y′′ +k
my =
F0
mcos γt
or
y′′ + ω2y =F0
mcos γt
where ω =√
k/m.
ω/2π is called the natural frequency of the
system, γ/2π is called the applied frequency.
56
Case 1: γ 6= ω.
y′′ + ω2y =F0
mcos γt
General solution, reduced equation:
y = A sin (ωt + φ0).
Form of particular solution (undetermined co-
efficients):
z = A cos γt + B sin γt.
A particular solution:
z =F0/m
ω2 − γ2cos γt.
General solution:
y = A sin (ωt + φ0) +F0/m
ω2 − γ2cos γt.
57
Case 2: γ = ω.
y′′ + ω2y =F0
mcos ωt
General solution, reduced equation:
y = A sin (ωt + φ0).
Form of particular solution (undetermined co-
efficients):
z = A t cos γt + B t sin γt.
A particular solution:
F0
2ωmt sin ωt.
General solution:
y = A sin (ωt + φ0) +F0
2ωmt sin ωt.
Resonance: Unbounded oscillation!
58
Resonance
y = A sin (ωt + φ0) +F0
2ωmt sin ωt.
59