工程數學(I)
Advanced Engineering Mathematics (I)
機械與自動化工程系
教師 王曉剛
中華民國 98 年 9 月
1
Advanced Engineering Mathematics (I)
Course Outlines:
(A). Ordinary differential equations
— Mathematic modeling of physical systems
— Separable ODE and exact equation
— 1st order ODE’s
— Linear ODE’s
— Systems of ODE’s
(B). Laplace transform
— Definition of integral transform
— Transform and inverse transform
— Properties of Laplace transform
— Solution of ODE using Laplace transform
— Systems of ODE
2
Chap. 1
Introduction of Differential Equation &
First Order Differential Equations
Ex: Free fall of a body (with no air friction)
x
g
x  xt 
dx
v
dt
dv
 a  g
dt
This is an ordinary differential equation.
2
d  dx  d x
 g
 
dt  dt  dt 2
Question: What are the “dependent variable 相依變數” and the
“independent variable 獨立變數” of this differential
equation?
Answer:
3
This equation can be solved simply by integrating this equation
twice:
 d 2x 
  dt 2 dt    g dt
dx
v
  gt  c1
dt
 dx 
x    dt    gt  c1 dt
 dt 
1
  gt 2  c1t  c2
2
Question: What are the constants c1 and c2?
How many
unknown constants are there in an equation?
How to solve
these constants?
Answer:
4
Ex: Free fall of a pin-pon ball (with air friction)
Balance of forces:
ma  kv  mg
dv
 kv  mg
dt
vt   ?
m
Question: Can this equation be solved simply by integration?
Answer:
Ex: Spring-mass system (Hooke’s law 虎克定律)
X=0
d 2x
m 2  kx
dt
-kx
5
Question: What about if the spring is located vertically?
Answer:
Ex: Swing of a pendulum
ma  mg sin   mg
θ
d 2S
S
m
 mg  mg
dt
L
d 2
S  L , L 2   g
dt
6
Ex: Parachute jumping
m
dv
 kv  mg
dt
Ex: Electrical circuit
(a)
7
Electrical circuit:
Question: What are the Kirchihoff’s laws(柯企賀夫定律)?
Answer:
We can use electrical charge q as the dependent variable as the
following:
di
1
  Ri   idt  E (t )  This is not a differential equation
dt
C
dq
i
dt
d 2q
dq 1
 L 2 R
 q  E (t )  This is a differential equation
dt
dt C
L
Or, we can use current i as the dependent variable as the
following:
8
q   idt
di
1
  Ri   idt  E (t )  This is not a differential equation
dt
C
Differentiate again :
L
d 2i
di 1
dE (t )
 L 2 R  i
dt
dt C
dt
 This is a differential equation
Question: Can you write the differential equation for a
spring-mass system with friction and external force? Is there
any “analogy 類此” between electrical circuit and spring-mass
systems?
Answer:
9
Differential Equations 微分方程式:
An equation containing the derivatives of one or more
dependent variables, with respect to one or more independent
variables, is said to be a differential equation.
Type : Ordinary Differential Equations (ODE)常微分方程式
＆ Partial Differential Equations (PDE)偏微分方程
式
ODE:
If an equation contains only ordinary derivatives of one or more
dependent variables with respect to a single independent
variables, it is said to be ODE.
variables : Dependent
Independent
≧
＝
1
==>ODE
1
Ex:
dy
 5 y  ex
dx
d 2 y dy

 6y  0
dx 2
dx
dx dy

 2x  y
dt
dt
10
PDE:
If there are two or more independent variables the derivatives
are partial derivatives and the equations are called Partial DE.
variables : Dependent
≧
Independent
>
1
==>PDE
1
Ex:
z  z ( x, y )
z
z
 zx
x
y
Order 微分次方: The order of a DE is the order of the highest
derivative in the equations.
3
d 2 y  dy 
 5   4 y  e x ……………………2nd order ODE
2
dx
 dx 
2nd order 1st order
11
Ex :
dy
 x  5            1st order
dx
d2y
dy
 3  2 y  0        2nd
2
dx
dx
dy
x  y  3            1st
dx
2
 d 2 y   dy 
 2      3 y  x 2    2nd
 dx   dx 
d4y
 k 4 y  0          4th
4
dx
3
First order DE can be written in a differential from
M(x,y)dy + N(x,y)dx = 0
Linear 線性 ＆ non-linear 非線性:
nth order DE is linear if it can be written as
a0 ( x) y ( n ) ( x)  a1 ( x) y ( n1) ( x)  ....  a n ( x) y( x)  f ( x)
where a0 ( x),..., an ( x) are constants or function of independent
variable x alone.
Otherwise, it is non-linear.
12
Two Characteristics of a linear ODE:
1. The dependent variable and all it’s derivatives are of the first
degree — the power of each team involving y is “1”. 所有
微分項只以下列形式出現:
(d(n)y/dxn)1, (dy(n-1)/dxn-1)1, …(dy/dx)1, (d(0)y/dx0)1(what is
this?).
2. Each coefficient depends at most on the independent
variables “x”. 所有係數 ao(x),…an(x) 不可為相依變數
(y) 之函數, 例如: a(y).
Question: Are the following equations linear or non-linear?
d 2x
m
 F (t )
dt 2
dx
m
 F0 t  A
dt
d 2x
m
 kx  F (t )
dt 2
d4y
EI
 w( x )
dx 4
dx
m
 k  x (t ) dt   F (t ) dt  A
dt
2
 dy  d y 
  y  F ( x)


2 
dx
dx



d2y
 dy 
 C 1 

2
dx
 dx 
2
13
Ex : Linear
( y  x)dx  4 xdy  0
y  2 y  y  0
d3y
dy

3
 5y  ex
3
dx
dx
ydx  xdy  4 xdy  0
3xdy  ydx  0
3xy  y  0
Non-linear :
(1  y ) y   2 y  e x
d2y
 sin y  0
dx 2
d4y
 y2  0
4
dx
(Why are they non-linear?)
Homogeneous 齊次 and non-homogenous 非齊次:
a0 ( x) y ( n) ( x)  a1 ( x) y ( n1) ( x)  ....  an ( x) y( x)  f ( x)
Homogeneous
f x   0
＆ non-homogeneous
f x   0
14
※ In general
Linear DEs are easier to solve and they have comprehensive
solutions.
Non-linear DEs are more difficult to solve and they can only be
solved qualitatively and numerically.
Solution of an ODE 微分方程式之解
A function of y = f(x) defined on some interval I in which it
has at least n-derivatives and satisfies the ODE:
F x, f x , f " x ,      , f
n
x   0
y  f x  is the solution of the ODE.
Initial-Value Problems (IVP) 起始值問題 and
Boundary-Value Problems (BVP)邊界值問題:
Differential equations involving “time” as the independent
variable are called IVP, and we need “initial conditions” (IC) to
solve the problems.
15
Differential equations involving “space” as the independent
variable are called BVP, and we need “boundary conditions”
(BC) to solve the problems.
Initial-Value Problem:
All solutions satisfy the
ODE (general solutions
通解)
Ex:
dy
 f ( x, y )
dx
IC : y ( x0 )  y0
( x0 , y 0 )
Only solution satisfies
the initial condition
(particular solution 特
Ex:
殊解)
d2y
 f ( x, y, y)
dx 2
IC #1 : y ( x0 )  y0
IC #2 : y( x0 )  y1
slope = y1
(xo, yo)
All curves satisfy IC#1, but only one curve satisfies both IC#1
and IC#2, which one is it?
16
General and Particular Solutions 通解與特殊解
/
( n)
For an n-th order ODE F ( x, y, y ,... y )  0 , a function
defined by y  f ( x, c1 , c 2, ,..., c n ) which satisfying the above
ODE is called the General Solution of the ODE (usually to a
linear ODE).

The function y  f ( x, c1 , c 2, ,..., c n ) is usually an n-parameter
family of curves.

A given DE usually has an “infinite” number of general
solutions.
Sometime, a general solution y  f ( x, c1 , c2, ,..., cn ) with chosen
values of c i that the curve of it will pass a predetermined point
(or points), then this obtained function y is called a Particular
Solution of the specific DE.
Nth order IVP differential equation with N initial conditions
Solve :
dny
 f ( x, y, y ,.., y ( n 1) )
n
dx
Subject to:
y( x0 )  y0 ,
y( x0 )  y1 ,... , y ( n1) ( x0 )  y( n1)
(initial conditions)
17
Boundary Value Problem → independent variable is often a
space variable
F ( x, y, y / , y // ,... y ( n ) )  g ( x)
y ( xn )  y n
y( x0 )  y0
y( x0 )  y01
y ( xn )  yn1
y ( x0 )  y02
y ( xn )  y n 2

Ex :

ky////  w( x)
y (0)  0
y / (0)  0
(0  x  L )
,
y // ( L)  0
,
,
y /// ( L)  0
(boundary conditions)
18
How to put physical problems into Mathematic Model and
Mathematic differential equations?
1. Newton’s law of motion, Kirchihoff’s laws
2. “Rate of Change”
3. Conservation laws (mass, momentum, and energy), “OIS”
Mathematical models:
1. Newton’s law of motion, Kirchihoff’s laws:
Ex: Electrical Circuits:
Kirchihoff’s law :
d 2q
dq 1
L 2 R
 q  E (t )
dt C
dt
Ex: Falling Body:
s0
d 2s
m
  mg
dt
s ( 0)  s 0
s (0)  v 0
19
If with air resistance
dv
 mg  kv.............1 IC
dt
d 2s
ds
m 2  mg  k ........2 IC ' s
dt
dt
m
Use velocity v as the dependent variable: m
dv
 kv  mg ,
dt
Its solutions are
v(t ) 
 kt
mg
 Ce m
k
for any values of C. These are general solutions.
However, there is only one solution (particular solution)
v(t ) 
mg mg  kt m 

e


k
k 
which satisfies the initial condition
v(0)  0
Conclusion: nth order DE can be expressed as a function
involving n arbitrary constants.
General solution:
 kt
mg
 Ce m
k
y ( x)  C1e kx  C2 e kx  C3 sin kx  C4 cos kx
v(t ) 
These are called general solutions, nth order DE has n
unknown constants.
20
If there is no constant in the solution, it’s a particular solution.
v (t )
C
v0
C
v(t ) 
mg
 v0
k
 kt
mg
 Ce m
k
v(0)  0 , v(t ) 
mg
k
t
general solution
mg mg kt m

e
k
k
particular solution .
Ex :
y(x) = C1cos2x + C2sin2x is a general solution of
d2y
 4y  0
dx 2
y = sin 2x is a particular solution ( C1 = 0, C2 = 1) which
satisfies the boundary condition:
y(0) = 0
;
y(
π
)=0
4
21
Ex: Deflection of a beam 長棒彎曲
d4y
k4y  0
4
dx
boundary conditions:
BC#1: y(0) = 0
BC#2: y(L) = 0
BC#3: y’’(0) = 0
BC#4: y’’(L) = 0
Solution:
y( x)  C1e kx  C2e  kx  C3 sin kx  C4 cos kx …….4 constants
Solve for the constants:
0  C1  C 2  C 4



kL
 kL
0  C1e  C 2 e  C 3 sin kL  C 4 coakL


0

C

C

C
1
2
4


0  C e kL  C e  kL  C sin kL  C coakL
1
2
3
4


22
2. “Rate of Change”:
Initial Value Problems use the concept of “rate of change 時間
變率” – derivatives, to derive their differential equations.
Ex: Population Dynamics 人口變率: Malthusian model
dp
 p
dt
dp
  kp (why “＋”?)
dt
p (t ) : population at time t . IC: p(0)  p0
Ex: Radioactive decay 放射性物質衰變:
dA
 A
dt
dA
  kA (Why “-”?)
dt
A(t ) : number of nuclei at time t . IC: A(0)  A0
Ex: Newton’s law of cooling 牛頓冷卻定律:
dT
 (T  T )
dt
dT
 k (T  T )
dt
ambient temperature
T (t ) : temperature at time t . IC : T (0)  T0
23
Ex: Spread of a Disease 疾病傳播:
dx
 kxy
dt
x(t ) : number of people who have contracted the disease .
y (t ) : number of people who have not yet been exposed .
1 (one infected person)
x+y=n+1
n people
dx
 kx(n  1  x)
dt
IC: x(0)  1
3. Conservation laws (mass, momentum, and energy), “OIS”:
Ex: Mixture of fluids:
Inflow: 3 gal/min fluid
(2 pound salts/gal)
Density: 300 gallon of
water + A pounds of
salts
Outflow: 3 gal/min fluid
24
Conservation of mass:
O  I  S  0
Outflow rate  Inflow rate
 Storage rate 

 
 

of
mass
(
kg
/
s
)

of
mass
(
kg
/
s
)

of
mass
(
kg
/
s
)

 
 
0
leaving CV  entering CV  in the CV


 
 

dA Inflow rate of  Outflow rate of 



dt salt entering CV  salt leaving CV 
= (3 gal/min)(2 lb/gal)－(3 gal/min)(
dA
A
 6
dt
100
A
lb/gal)
300
IC: A(0) = Ao
Ex: Draining a tank:
gh 
Aw
h(t)
Ah
v
v
1
v 2  v 
2
2 gh
dV
 vAh   Ah 2 gh
dt
V (t )  Aw h
dh
A
 h
dt
Aw
2 gh
25
First-Order Differential Equations
1st order Separable DE:
(1) Direct integration
dy
 g ( x) → y 
dx
 g ( x)dx  c
(2) Separable equation
y   f ( x, y )
dy
M ( x, y )
 f ( x, y ) 
dx
N ( x, y )
M ( x, y ) dx  N ( x, y ) dy
If
M ( x) dx  N ( y ) dy
Separable DE
 M ( x)dx   N ( y )dy  C
26
27
EX:
dy
3x2
=
dx
1-y
 (1  y)dy   3x dx
2
 y
1 2
y  x 3  C → general solution
2
dy
 y2  4
Ex :
dx
1 
 1
dy
4
 dx  
 4  dy  dx
2
y 4
 y  2 y  2


1
1
ln y  2  ln y  2  x  C
4
4
y2
ln
 4 x  C2
y2
y2
 e 4 x  C2
y2
1  Ce 4 x
y2
1  Ce 4 x
2y
Ex : cos x(e  y )
dy
 e y sin 2 x,
dx
e2y  y
sin 2 x
dy 
dx
y
cos x
e
 (e
y
y (0)  0
, sin 2x  2 sin x cos x
 ye  y )dy  2 sin xdx
e y  ye  y  e  y  2 cos x  C
Question: What is the particular solution which satisfies the
initial condition?
28
Answer:
29
Ex:
y(0)  yo
y   y 2 ,
dy
  y2
dx
dy
 y 2    dx  C
1
  x  C
y
1
 y ( x) 
xC
1
1
x  0, y  y0 
,C 
C
y0
y ( x) 
Ex :
y 
y0
1  y0 x
4x
1  2e y
 (1  2e
y
y ( 0)  1
)dy   4 xdx  C
y  2e y  2 x 2  C
y (0)  1  1  2e  C
same
Note :
Definite integral:

y
1
(1  2e y ) dy 

y
x
y  2e y
1
 2x2
x
0
4 xdx
0
y  2e y  (1  2e)  2 x 2
Initial condition is similar to the lower-bound limit of the
integration.
30
Ex :
y 
y ( y  2)
x( y  1)
1 
1
y 1
dx

2
2 dy  dx  C
dy


C


1
 y ( y  2)
x
 y y  2  x 1


1
1
ln y  ln y  2  ln x  C1
2
2
y ( y  2)
y ( y  2)
ln

C

 e C 2  C3
2
2
2
x
x
y ( x)  1  1  Ax 2
31
32
Ex: Radiocarbon dating 利用放射性碳之化石年齡估算:
Ordinary carbon C612
Radioactive carbon C614
T = 5730 yrs（half-life 半衰期）
If a fossilized bone contains 25﹪of the original amount of
C614 , what is it’s age ? Amount of C614 / C612 in atmosphere is a
33
constant. When an animal dies, the absorption of C614 by
breathing and eating terminates.
y(t) = concentration of C614
dy
= -ky
dt
dy
= ky , population growth )
dt
(
dy
= -kdt , ln(y) = -kt + C1
y
IC : y(0) = y0
Half-life T:
∴ e-kT =
∴C = y0
y=
y0
2
1
2
,
∴
,
,
y = Ce-kt
y = y0e-kt
y0
= y0e-kT
2
k = 0.000121
If 25﹪ C614 remains in the fossilized bone
y0
= y0e-0.000121t ∴ t = 11460 yrs
4
Ex : Flow of water through a hole
h(t): water level at time t
v(t): velocity of water at
outflow
34
v(t )  0.6 2 gh(t )
( Torricelli’s Law )
IC: h(0) = H (initially full)
Water flowing out of the tank in time △t is
 V   avt
Ah   V   avt
h
a
  (0.6) 2 gh
t
A
t  0

dh
a
 C h , C  (0.6) 2  9.8
dt
A
dh
 Cdt
h
2h
1
2
 Ct  C1
C 
C
h(t )   1  t 
2 
 2
2
2
C 
h(0)   1   H
 2 
Question:
solve for C1
When does the water level drop to
H
?
2
When is it empty?
Answer:
35
36
Separable by substitution:
Ex :
2 xyy  y 2  x 2  0
Not separable
2
y
 y
2 y     1  0
x
 x
2

dy 1 x  y 

   1  Separable ?
dx 2 y  x 

Reduce to separable by substitution
y
dy
du
 u  y  ux,   u  x
x
dx
dx
2u (u  xu)  u 2  1  0
Let
2 xuu  u 2  1  0
2udu
dx


1  u2
x
ln( 1  u 2 )   ln x  C1
1  u2 
C
,
x
x 2  y 2  Cx
2
C
C2

2
x   y 
2
4

Conclusion :
y
 y 
y  f  ..... u  
x
x 
u  ux  f (u )
du
f (u )  u
u 

dx
x
Separable !
37
38
39
40
x y
x y
Ex : y  
1 y
x  f  y ........u  y
x
x
1 y
x
1 u
u  u x 
1 u
1 u
1 u  u  u2
ux 
v 
1 u
1 u
1 u
du 1

1  2u  u 2 dx x
y 
u 1
u 1
1
du

du

ln u 2  2u  1
2
 u  2u  1
 (u  1)  2
2
2
 ln u 2  2u  1
1
2
 ln Cx
(u 2  2u  1) x 2  C1
y 2  2 xy  x 2  C1
Ex :
( x 2  y 2 )dx  xydy  0
2
1   y 
2
2
x y
 x
y 

y
xy
x
1  u2
1
ux  u 
, ux 
u
u
1
1 2
udu  dx,
u  ln Cx
x
2
y 2  2 x 2 ln Cx
41
y 
Ex :
x  y 1
X Y
 y 
x  y 1
X Y
X  x  h, Y  y  k 
dY
dY dx dy


dX
dx dX dx
dY
X  h  Y  k 1

dX
X  h  Y  k 1
h  k  1  0

 h  0, k  1
h

k

1

0

dY
X Y

dX
X Y
2
Y  2 XY  X 2  C
( y  1) 2  2 x( y  1)  x 2  C
Solution by substitution:
dy
 f ( x, y ),
dx
dy 
substitute : y  g ( x, u ), u  u ( x)
g
u
dx 
du
x u
x x

dy
du
 g x  gu
dx
dx
f ( x, y )  f ( x, g ( x, u ))  g x ( x, u )  g u ( x, u )

du
dx
du
 F ( x, u )
dx
solution : u   ( x)
sub.back : y  g ( x,  ( x))
42
Ex :
dy
 (2 x  y ) 2  7
dx
,
y (0)  0
du
dy
 2 
dx
dx
du
du
 2  u2  7 
 u 2  9......separable
dx
dx
du

 dx
(u  3)(u  3) 
u  2 x  y
,
43
44
Exact Differential Equation 正合方程式: (If an ODE is not
separable; try to see if it is an exact ODE.)
45
Ex :
( y cos x  sin y )dx  (sin x  x cos y )dy  0
M
N
try a function
 ( x, y )  y sin x  x sin y  C

 y cos x  sin y
x

 sin x  x cos y
y


 d ( x , y ) 
dx 
dy
x
y
 ( y cos x  sin y ) dx  (sin x  x cos y )  0
So  ( x, y )  y sin x  x sin y  C is the solution.
But how to find  ( x, y ) ?
46
If
 ( x, y )  C
d 
d
d
dx 
dy  0
dx y  const
dy x  const


dx 
dy  0
x
y
Mdx  Ndy  0
or
dy
M

dx
N
is called an exact DE and the solution is μ(x,y) = C
Test for exactness:
M
2
2
    N





y
yx xy x  y  x
 Mdx  Ndy  0
is exact if and only if
M
N

y
x
How to find μ(x,y) ?

 M ( x, y )
x

 ( x, y )  
dx
y c x

y c
Also;
y
 f ( x)
x
y
dy
dx   f ( x)dx  h( y )
dx
M ( x, y )dx h( y )

 N ( x, y )
y

 ( x, y )  
dy   N ( x, y )dy k ( x)
x c y
x c
Compare these two equations and find hand k.
47
48
Ex:
(e 2 y  y cos xy)dx  (2 xe2 y  x cos xy  2 y )dy  0
M

 2e 2 y  xy sin xy  cos xy
y

exact
N
 2e 2 y  cos xy  xy sin xy 

x



M 
,
N
x
y
49
 ( x, y )   e 2 y dx   y cos xydx
 xe2 y  sin xy  h( x)
 ( x, y )  2 x  e 2 y dy  x  cos xydy  2  ydy
 xe2 y  sin xy  y 2  k ( x)
 xe2 y  sin xy  y 2  C
dy
2x 3  3 y

dx
3x  y  1
Ex:
(2 x3  3 y)dx  (3x  y  1)dy  0 = dμ
M
N
M
N
3
 3  equation is
y
x
exact!
∴Solution is μ(x,y) = C
 ( x, y )  
(2 x 3  3 y )dx  h( y ) 
 ( x, y )  
(3x  y  1)dy  3xy 
y C
x C

1 4
x  3 xy  h( y )
2
1 2
y  y  k ( x)
2
1 4 1 2
x  y  y  3xy  C
2
2
What if the DE is not exact?
50
(Why “x” is an integration factor?)
51
Ex :
dx  (3x  e 2 y )dy  0
M
0
y
Try
,
N
3
x
Not
exact!
e3 y  ( dx  (3 x  e 2 y )dy )  0
e3 y dx  (3 xe3 y  e y )dy  0
M
N
 3e3 y 
 3e3 y
y
x
exact!
 ( x, y )   e3 y dx  xe3 y  h( y )
 ( x, y )   (3 xe3 y  e y )dy  xe3 y  e y  k ( x)
xe3 y  e y  C
How to find e3y ?
52
How to find Integration factor 積分因子 I ? (if not exact)
53
54
2
2
Ex : 2 sin( y )dx  xy cos( y )dy  0
P  2 sin( y 2 )
,
Q  xy cos( y 2 )
1
3
R
4 y cos( y 2 )  y cos( y 2 ) 
2
xy cos( y )
x
3
 I ( x)  exp  dx  x 3
x
2 x 3 sin( y 2 )dy  x 4 y cos( y 2 )dy  0

M

N
x 4 sin( y 2 )  C
55
56
Integrating Factors:
If M(x,y)dx+N(x,y)dy = 0 is not exact , the following rules can
be followed to find the integrating factor.
a).
If
M N

y
x
 f ( x) then
N
f ( x ) dx
e
is an int egrating
factor
b).
If
M N

y
x
  g ( y ) then
M
g ( x ) dx
I ( y)  e 
is an int egration
factor
57
c). If Mdx  Ndy  0 is homogenous and Mdx  Ndy  0
then
1
is an integration factors.
Mx  Ny
Homogenous equations:
A function is called homogenous of
n
degree n if f (x, y)   f ( x, y)
— f(x,y) = x4 + x3y is homogenous of degree 4 since
f(λx,λy) = (λx)4-(λx)3(λy) = λ4(x4-x3y) = λ4f(x,y)
— f(x,y) = e(y/x) + tan
y
is homogenous of degree 0 since
x
f(λx,λy) = e(λy/λx) + tan
λy
y
= e(y/x) + tan = λ0f(x,y)
λx
x
— f(x,y) = x2 +sin x cos y is not homogenous .
The D.E. M(x,y)dx + N(x,y)dy = 0 is called homogenous if
M(x,y) and N(x,y) both are homogenous and of the same
degree .
y
y2
y
( x ln )dx + (
sin(-1) )dy = 0 ………Homo
x
x
x
(x2+y2)dx – (xy2-y3)dy = 0 ………is not Homo
Ex : (x4+y4)dx – xy3dy = 0
Exact……(not) ;
Homo……(yes, degree 4)
58

1
1
1
 5
 5 … is an Integration
4
4
Mx  Ny x  xy  xy
x
factor
 1 y4 
y3
  5 dx  4 dy  0  d (  ( x, y))
x
x x 
y3
1 y4
 ( x, y )    4 dy  
  ( x)
x C
x
4 x4
 ( x, y ) y 4
1 y6
 5   ( x)   5
x
x
x x
1
 ( x) 
,
  ln x
x
1 y4
 ln x 
C
4 x4
d). If Mdx  Ndy  0 can be written in the form
y f(xy)dx + x g(xy)dy = 0
where f(xy)≠g(xy)
1
1

xy f ( xy)  g ( xy) Mx  Ny
Ex :
then
is an Integration factor.
y( x 2 y 2  2)dx  x(2  2 x 2 y 2 )dy  0
M = y f(xy)
N = x g(xy)
59

1
1
1


is an int egration factor
Mx  Ny xy x 2 y 2  2  2  2 x 2 y 2
3x 3 y 3


x2 y 2  2
2  2x2 y 2
dx 
dy  0
3x 3 y 2
3x 2 y 3
 ( x, y )  
y C
is exact
 1
x2 y2  2
2 

dx


  3x 3x 3 y 2 dx
3x 3 y 2
1
1
 ln x  2 2  h( y )
3
3x y
1
1
2
 ln x  2 2   ln y  C
3
3x y
3
e). Miscellaneous:
Group of terms
I.F.
Exact Differential
xdy – ydx
1
x2
xdy-ydx
y
= d(
)
2
x
x
xdy – ydx
1
y2
ydx-xdy
x
= d( - )
2
y
y
xdy – ydx
1
xy
dy dx
y
= d( ln )
y
x
x
xdy – ydx
1
x +y2
xdy-ydx
x2
xdy-ydx
y
-1
=
)
2
2
2 = d( tan
x +y
1+(y/x)
x
xdy + ydx
1
xy
xdy+ydx
= d(ln(xy))
xy
xdx + ydy
1
x +y2
xdx+ydy
1
= d｛ ln(x2+y2)｝
2
2
x +y
2
2
2
60
Ex : xdx + ydy + 4y3(x2 + y2)dy = 0
I(x,y) =
1
x2+y2
xdx+ydy
+ 4y3dy = 0
x2+y2
1
ln(x2+y2) + y4 = C
2
f). Other techniques:
— If M ( x, y )dx  N ( x, y )dy  0 can be written in the form
f1(x)．g2(y)dx + f2(x)．g1(y)dy =0
I.F :
1
f1(x)
g1(y)
→
dx +
dy = 0
f2(x)
g2(y)
f2(x)．g2(y)
Ex : (x-1)2ydx + x2(y+1)dy = 0
I.F :
1
(x-1)2
(y+1)
→
dx +
2
2
x
y
x ．y
dy = 0
— If Mdx  Ndy  0 is homogenous, then y = νx
will
reduce any homogenous of a separable equation .
P(x,ν)dx + Q(x,ν)dν = 0
61
Ex : (x3+y3)dx – 3xy2dy = 0
Exact……(not) ;
y = νx
,
Homo……(yes)
dy = νdx + xdν
x3｛(1 + ν3)dx - 3ν2(νdx + xdν)｝= 0
(1 - 2ν3)dx - 3ν2xdν = 0 ……separable
1
Integration factor I(x,v) =
,
x(1-2ν3)
㏑x+
dx 3ν2dν
=0
x
1-2ν3
1
㏑(1-2ν3) = C
2
y3
x (1- 2 3 ) = C(y/x)
x
2
Substitution:
if
dy
y
 f( )
dx
x
let
v
y
x
y  vx y  v  vx
v  vx  f (v)
v
1
dv
1

,
 dx
f (v )  v x
f (v )  v x
function of v
function of x
62
Equation of first order:
dy
+ P(x)y = Q(x)……linear
dx
(Why is it linear?)
dy 2 2
(
) , y ,…… not linear .
dx
Homogeneous and Non-homogeneous:
If Q(x) = 0, it is homogeneous.
If Q(x) not zero, it is non-homogeneous.
63
64
65
First order linear homogeneous equation:
dy
+ P(x)y = 0, IC: y(0) = yo
dx
Solution:
dy
  P( x)dx ( separable )
y
ln y    P( x)dx  C1
 P ( x ) dx
 P ( x ) dx
y ( x)  Ce 
 yo e 
Ex :
(Why ?)
dy
 2 xy  4 x
dx
I ( x)  e 
P ( x ) dx
 e
2 xdx
 ex
2
ye x   4 xex dx  2e x  C
2
2
 y  2  Ce x
Ex :
x
2
2
dy
 y  x 3  3x 2  2 x
dx
66
dy 1
 y  x 2  3x  2
dx x
 P ( x ) dx 
 P ( x ) dx dx  C 
ye 
Q
(
x
)
e
 

1
1

  dx 
 x dx   2
x
e
  ( x  3 x  2)e
dx  C 

 

2


 x   ( x  3  )dx  C 
x


 1


 x  x 2  3 x  2 ln x   C 

 2

Ex:
y' y  e t ,
y(0)  0
dt
 dt
y (t )  e  [  e   e t dt  C ]
1
  e t  Cet
2
1
y (0)  0,  C 
2
67
Bernoulli’s Equation:
dy
 P( x) y  Q( x) y n
dx
dy
y n
 P ( x) y n1  Q( x)
dx
y n1  v  (1  n) y n
Let
dy dv

dx dx
dv
 (1  n) P ( x)v  (1  n)Q( x)
dx
 (1 n ) P ( x ) dx 
 (1n ) P ( x ) dx dx  C 
ve 
(
1

n
)
Q
(
x
)
e
 


Ex :
dy
 y  xy5
dx
y 5
dy
 y4  x
dx
dy
1 dv

dx
4 dx
1 dv
dv

v  x
 4v  4 x
4 dx
dx
 P ( x ) dx 
 P ( x ) dx dx  C 
ve 
Q
(
x
)
e



Let
y  4  v  y 5
 e 4 x
  4 xe
4x
dx  C

1


 e  4 x  xe4 x  e 4 x  C 
4


1
1
  x   Ce 4 x
4
y
4


 u  x, dv  e 4 x dx 



1 4x 
 du  dx, v  e

4


  xe4 x dx



 1 4x 1 4x 
  4 xe  4  e dx 


  1 xe4 x  1 e 4 x 


16
 4

68
Physical Applications:
69
Question: Why i(t) = Eo/R when t is large?
70
71
72
Ex : L-R circuit with discontinuous emf
I 1 ( 0)  0
0  t 1
1
R t
1  e L 

R
1
R 
I1 (1)  1  e L 

R
I1 (t ) 
73
t  1, t '  t  1
dI
L 2  RI 2  0
dt '
I 2 (t ' )  C1e
 Rt '
I 2 (t  1)  C1e
 I1 (0)  C1
L
 R ( t 1)
L

R
1
 I 2 (0)  I1 (1)  1  e L   C1

R
1
1  e R / L
R

 Rt
1 

L
1

e
,
0

t

1


 R 


 I (t )  

R
 ( t 1)
1
 (1  e  R / L )e L , t  1
 R

74
Chap. 2
Linear Differential Equations of Higher Order
Linear nth order D.E:
dny
d n 1 y
dy
a0 ( x) n  a1 ( x) n 1  ...  a n 1 ( x)  a n ( x) y  F ( x)
dx
dx
dx
If
F(x) = 0 → homogenous
If
F(x)≠0 → non- homogenous
dy 2
dy
d 2y
What is linear : (
) ,(
)(
) , ……
dx
dx
dx2
dy
 a1 ( x) y  F ( x) linear 1st
dx
dy a1 ( x)
F ( x)

y
dx a0 ( x)
a0 ( x )
n  1 a0
order
D.E
dy
 P( x) y  Q( x)
dx
 P ( x ) dx 
 P ( x ) dx dx  C 
y ( x)  e 
Q
(
x
)
e
 

d
d2
2
Differential operator D =
,D =
……
dx
dx2
a ( x) D
0
n

 a1 ( x) D n1  ...  an1 ( x) D  an ( x) y  F ( x)
2
Ex: xy  y  xy  0 , ( xD  D  x) y  0
75
Linearity
Properties of linear operator L :
(1)
L(y1+y2) = Ly1+Ly2
(2)
L(Cy1) = C(Ly1)
Question: Are sin x, cos x, ln x, d/dx… linear or non-linear
operators?
Answer:
Linear D.E :
d2y
a ). 2  y  0  ( D 2  1) y  0
dx
y  y1  y2
 d 2 y1
  d 2 y2

d2



  0  0  0
(
y

y
)

(
y

y
)


y


y
1
2
1
2
1
2
 dx 2
  dx 2
dx 2

 

 ( y1  y2 )   ( y1 )   ( y2 )......linear
b).
d2y
 y2  0
2
dx
d
d 2 y1 d 2 y 2
2
( y1  y2 )  ( y1  y2 ) 

 y12  2 y1 y2  y22  2 y1 y2  0
2
2
2
dx
dx
dx
 not linear
  ( y1 )   ( y2 )
76
Linear Dependent 線性相依 / Linear Independent 線性獨立:
A set of functions f1(x) , f2(x) , … ,fn(x) is said to be LD if there
exist constants C1 , C2 , … , Cn , not all zero , such that
C1 f1(x) + C2 f2(x) +…+ Cn fn(x) = 0
for every x in the interval.
If the set of functions is not LD on the interval, it is said to be LI.

A set of functions is LI if the only constants for which
C1 f1(x) + C2 f2(x) +…+ Cn fn(x) = 0
for every x are
C 1 = C2 = … = C n = 0
Ex : C1 f1(x) + C2 f2(x) = 0
Assume
, f1 ＆ f2 LD
C1≠0 ∴ f1(x) = -(
C2
) f (x)
C1 2
 If two functions are LD, then one is simply a constant
multiple of the other.
f1(x) = sin2x = 2sinx cosx , f2(x) = sinx cosx , f1(x) - 2f2(x) = 0
∴ f1(x) = 2f2(x)
Ex : f1(x) = ex , f2(x) = e-x
C1 f1(x) + C2 f2(x) = 0  C1ex + C2e-x = 0
77
ex
ex = (-
C1 -x
)e
C2
(True?)
∴ C1 = C 2 = 0
e-x
Test for LI or LD:
The functions f1(x) , f2(x) , … ,fn(x) possesses at least n-1
derivatives, the determinant :
W ( f1 , f 2 ,..., f n )  det
f1
f1
f2
f 2


n 1
n 1
f1
f2


fn
f n


fn
n 1
is called the Wronskian of the functions .
The set of functions f1 , f2 , … ,fn are LI if and only if
W( f1 , f2 , … ,fn )≠0
3x
Ex : y1  e
W  det
,
for every x .
y 2  e 3 x
e3 x
e 3 x
3e3 x
 3e 3 x
 6  0......LI
78
Ex : f1  x ,
x
W  det 1
0
f2  x 1 ,
f3  x  3
x 1 x  3
1
1 0
0
0
C1 x  C2 ( x  1)  C3 ( x  3)  0
(C1  C2  C3 ) x(C2  3C3 )  0
C3  1 C2  3 C1  4
Not
all
zero
 LD
Superposition principle 重疊定律:
If f1(x) , f2(x) , … ,fn(x) are solutions of a homogenous linear
D.E
dn
d n1
 y  0   {an n  an1 n1  ......a1 d  a0 }
dx
dx
dx
Then so too is any combination thereof
C1 f1(x) + C2 f2(x) +…+ Cm fm(x)
For arbitrary constant C1,……,Cm ,
 y1  0 ,  y2  0 , ....  ym  0
 (C1 y1  C2 y2  ...  Cm ym )   (C1 y1 )   (C2 y2 )  ...   (Cm ym )
 0  0  ...  0  0
Ex:
79
y  y  12 y  0  Linear
y1 ( x)  e 4 x , y2 ( x)  e 3 x  particular
y ( x)  C1e 4 x  C2e 3 x  general
solution
solution
16C1e 4 x  9C2e 3 x  4C1e 4 x  3C2e 3 x  12 y  0
Theorem:
If f1(x) , f2(x) , … ,fn(x) are n linearly independent solutions
of an nth-order homogenous linear D.E , then
y(x) = C1 f1(x) + C2 f2(x) +…+ Cn fn(x) is a general
solutions of the D.E .
Linearly independent－a solutions cannot be written as a
linear combinations of other solutions .
Ex:
y  3 y  3 y  y  0
y1  e x , y2  xex , y3  2e x  3xex
y3
is
a linear
combination of
y1
and
y2
so y3 is a solution
80
Ex : y   3 y   3 y   y  0
y1  e x , y2  xex , y3  2e x  3xex  2 y1  3 y2 ... not
L.I
y  C1e x  C2 xex  C3 (2e x  3xex )
... is
not
a
general
solution
Ex : y   9 y  0
y1  cos 3x ,
y2  sin 3x are LI
solutions
y  C1 cos 3x  C2 sin 3x is the general
solution
81
Homogenous linear D.E with constant coefficient:
a0 D n y  a1 D n1 y  a2 D n2 y  ...  an1 Dy  an y  0
d
D
dx
d2
, D  2
dx
2
...
82
83
(Euler formula, 唸為 “oiler” 歐拉)
complex number e ix
x 2 x3
e  1 x 

 ...
2! 3!
i 2 x 2 i3 x3
ix
e  1  ix 

 ...
2!
3!
x 2 ix 3
 1  ix 

 ...
2! 3!

 

x2 x4
x3
 1 

 ...  i x 
 ...
2! 4!
3!

 

x2 x4
cos x  1 

 ...
2! 4!
x3
sin x  x 
 ...
3!
 e ix  cos x  i sin x
x
84
Ex : y   2 y   y  0
m 2  2m  1  0  (m  1) 2  0
y1  e  x
m  1,1 
that
we need
solution
y2
try
is
a
another linear
solution ,
independent
y  v ( x )e  x
y   v( x)e  x  e  x v( x)
y   v( x)e  x  e  x v( x)  e  x v( x)  e  x v( x)
ve  x  2e  x v  e  x v  2ve  x  2e  x v  ve x  0
 ve  x  0 , e  x  0 v  0
v  C , v  Cx
 y 2  Cx e  x
 a general solution y  C1 y1  C2 y2  C1e x  C2 xe x
85
86
87
88
89
N 次線性常係數微分方程式 Summary:
( i ). If  (m)  0 has single roots m1 , m2 ,…, mk , the general
solution of the D.E is
C1e m1x  C2 e m2 x  ...  Ck e mk x
( ii ). If  (m)  0 has a real root m of multiplicity k, then a
solution of the D.E is
(C1  C2 x  ...  Ck x k 1 )e mx
( iii ). If  (m)  0 has a pair of complex conjugate roots a ± bi,
each of multiplicity of k then a solution is

e ax (C1  C 2 x  ...  C k x k 1 ) cos bx  ( D1  D2 x  ...  Dk x k 1 ) sin bx
A general solution of (ii)＆(iii) is obtained by superposing
all solutions in (ii)＆(iii) .
90

Ex : y   3 y   4 y   6 y  0
 (m)  m3  3m 2  4m  6  0
(m  1)( m 2  2m  6)  0
m  1 , m  1 7
 y  C1e x  C2e(1
7)x
 C3e(1
7)x
Ex : y   y  0
 ( m)  m 3  1  0
(m  1)( m 2  m  1)  0
 1  3i
2
3
3
x 
 y  C1e x  e 2  C2 cos
x  C3 sin
2
2

m 1 ,
m

x 

Question:
 (m)  0, m  3,3,3,2i,2,1  3 ,4  i,4  i
What is the general solution of the DE?
Answer:
91
92
或
m2  k  0 ,   i k
x(t )  C cos t  D sin t
also
m
,  k
natural frequence
m
x(t )  E sin( t   )  E sin  cos t  E cos  sin t
 C  E sin 
,
D  E cos 
E  C 2  D2
,   tan 1
E sin t
C
D
E : amplitude
-φ
w
E sin(t+φ)
93
94
95
96
Non-homogeneous linear D.E with constant coefficient:
dny
d n1 y
dy
a0 n  a1 n1    an1
 an y  F ( x )  0
dx
dx
dx
(a0 , a1  an are const )
 ( D) y  F ( x)
 ( D)  a0 D n  a1 D n1   an1 D  an
 ( D) y  0 is called the hom ogeneous equation
associates with the non  hom o equation
 ( D) y  F ( x)
97
Theorem : a general solution of  ( D) y  F ( x) is
y ( x)  y h ( x)  y p ( x)
where y h (x) is a general solution
of  ( D) y  0 (homo) and
y p (x )
is a particular
solution of the given solution .
 ( D)( yh  y p )   ( D) yh   ( D) y p
 0  F ( x)
 F ( x)
∴ y ( x)  y h ( x)  y p ( x) is a general solution .
This is also true for variable coefficient.
How to determines
y p (x ) ?
1. Undetermined Coefficient method 未定係數法
2. Variation of Parameters method 參數變動法
98
Undetermined coefficient method:
99
Ex : y   y   6 y  F ( x)
F ( x)  6 x 2  2 x  3
(1).
y p  Ax 2  Bx  C
2 A  (2 Ax  B )  6( Ax 2  Bx  C )  6 x 2  2 x  3
 6A  6

2
17

2 A  6B  2
  A  1, B   , C  
3
18
2 A  B  6C  3

 y  C1e3 x  C2 e 2 x  x 2 
( 2).
2
17
x
3
18
F ( x)  2 sin 2 x
y p  A sin 2 x  B cos 2 x
(Why ?)
( 10 A  2 B ) sin 2 x  ( 2 A  10 B ) cos 2 x  2 sin 2 x
 10 A  2 B  2
 A   5 26 , B   1 26
2 A  10 B  0 
5
1
sin 2 x 
cos 2 x
26
26
F ( x)  xe x  e  x
 y  yh 
(3).
y p  Axe x  Be  x
( 6 A) xe x  (  A  6 B )e  x  xe x  e  x
 6A  1

1
7
A   , B 
 A  6 B  1
6
36
 y  yh 
1 x 7 x
xe 
e
6
36
100
3 x
Ex : y   y  x e
y h : m 3  1  0  m  1,
1
3

i
2 2

3
3
C
cos
x

C
sin
 2
3
2
2

y p  Ax 4 e x  Bx 3 e x  Cx 2 e x  Dxe x
y h  C1e  e
x

1
2

x

12 A  1


36 A  9 B  0
1
1
2
2

A  , B   ,C  , D  
24 A  18B  6C  0
12
3
3
3

6 B  6C  3D

2
2 x
Ex :  ( D) y  x  2 sin x  xe
 ( D) y  0  i,2,2,2,4,4
y h  C1 c o xs  C 2 s i nx  (C3  C 4 x  C 5 x 2 )e  2 x  (C 6  C 7 x)e 4 x
x 2 : Ax 2 , Bx , C
s i nx : D s i nx, E c o xs  Dx s i nx  Ex c o xs
xe 2 x : Fxe 2 x , Ge  2 x  Fx 4 e  2 x , Gx 3 e  2 x
F ( x)  x n f ( x)
101
2
Ex : y ' y   3x  sin 2 x
m4  m2  0
m 2 (m 2  1)  0
,
m  0,0,1,1
yh  C1  C2 x  (C3  C4 x)e x
f ( x)  3 x 2  sin 2 x


3 x 2 : 3 x 2 ,6 x,6,0 x 2 , x,1
 sin 2 x :  sin 2 x,2 cos 2 x,4 sin 2 x sin 2 x, cos 2 x
y p  Ax 2  Bx  C  D sin 2 x  E cos 2 x
( No good )
y p  Ax 4  Bx 3  Cx 2  D sin 2 x  E cos 2 x
102
Ex :
y' y  3 x 2
y p1: 24 A  12 Ax 2  6 Bx  2C  3 x 2
 12 A  3

1

 6B  0
  A   , B  0, C  3
4
24 A  2C  0

1 4
x  3x3
4
y' y  D sin 2 x  E cos 2 x
20 D sin 2 x  20 E cos 2 x   sin 2 x
1
D  
E0
20
1
 y p2  
sin 2 x
20
1
1
y ( x)  C1  C2 x  C3e x  C4 e  x  x 4  3 x 3 
sin 2 x
4
20
y p1  
Variation of parameter method:
Example :
y   y   2 y  tan x
undetermined coefficient method :
try y p  A tan x
yp  yp  2 y p  2 A sec 2 x tan x  A sec 2 x  2 A tan x
?
 tan x
2
2
try y p  A tan x  B sec x  C sec x tan?
sec 4 x ….
103
“undetermined coefficient ” method will work only for an
function that produces finitely many linearly indep. functions
when repeated differentiated.
x k e ax cos bx ， x k e ax sin bx
1
2
1
2
2
What if y  y  2 y  cos x ,   cos 2 x
e x  ex
sinh x , 
2
2 x  e ln 2  x
What if “undetermined coefficient ” method fails ?
How to determine y p  x 
Method of Variation of Parameters
y   a1 y   a0 y  f x 
If y1 & y 2 are two linearly indep. solutions to
y  a1 y  a0 y  0
Then assume y p  uy1  vy2
 Luy1  vy2   0

uy1  vy2  0
2 unknowns , 2 eqs.
for simplicity
y  uy1  vy2  uy1  vy2
0
104
y  u y1  vy2  uy1  vy2
( uy1  vy2  uy1  vy2 )+…..
(uy1  vy2 )  u( y1  a1 y1  a0 y1 )+v( y2  a1 y2  a0 y2 ) = f(x)
 u y1  vy 2  f ( x)
 u 
For
n th
 y 2 f x 
 y y2 

d et 1


 y1 y 2 
order
，
v 
y1 f ( x)
 y y2 

det 1


 y1 y 2 
ODE
y n   a n 1 y n 1  ....  a1 y   a 0 y  f x 
yc  c1 y1  c2 y 2  ....  cn y n
y p  u1 y1  u2 y2  ....  un yn
where u1 ,...., u n satisfy the eqs.：
u1 y1  u2 y2  ....  un yn  0
u1 y1  u2 y2  ....  un yn  0
…
…
u1 y n11  u2 y n12  ....  un y n1n  f ( x)
Ex:
y   y  sec x
105
y h  c1 cos x  c2 sin x
y p  u cos x  vsin x
y1
y2
uy1  vy2  0

uy1  vy2  sec x
 u 

 y2 sec x
 y y2 

det  1


y
y
2
 1
v 
y1 f ( x)
1
 y1 y2 

det 


 y1 y2 
 sin x sec x
 cos x sin x 

det 
  sin x cos x 
= －tan x
 u  
sin x
dx  ln cosx+ c3
cos x
v = x+ c4
 y  c1 cos x  c2 sin x  ln cos xcos x  x sin x
Ex:
y   y  
1
1 ex
y h  c1  c 2 e x  c3 e  x
y p  u1  u2e x  u3e x
106

x
x
u1  u 2 e  u 3 e  0

x
x
 u 2 e  u 3 e  0

1
x
x
u 2 e  u 3 e 
1 ex

1 ex

det  0 e x
0 ex

ex 

x
e   2
e  x 

 0
1
u1  d et 0
2  1

1  ex
ex
ex
ex

0
1
1
u2  d et0
0

2
1
0
1 ex


x
1 e
1
u3  d et0 e x

2
 0 ex


ex 
1
 ex  
 1  ex
x
e 


ex  1 ex
 ex   2 x
 1 e
x
e 


0  1 ex
0  2 x
1  1 e

1  ex 
dx
e  x dx
x
u1   



ln
e
 1  c4
 ex  1
1 ex


1 x
e
1  x
1 
2
u2   
dx

e


dx
1 ex
2 
1 ex 


1
1
  e  x  ln e  x  1  c5
2
2
107


1 e x dx
1
u3  

ln 1  e x  c6
x
2 1 e
2
y p  u1  u 2 e x  u3 e  x  ....
Forced Osculation 強迫振盪 (Resonance 共振):
mx’’ + cx’ + kx = F(t)≠0
F(t): driving fore, or input
(e.g.: external voltage, emf)
x(t) = output or response
If F(t) = Fo sin Ωt
xp = a cosΩt + b sinΩt
xp’ = -aΩsinΩt + bΩcosΩt
xp’’ = -aΩ 2 cosΩt – bΩ 2 sinΩt
[(k-mΩ 2 )a +Ωcb]cosΩt + [-Ωca + (k - mΩ 2 )b]sinΩt
= Fo sinΩt
  (k  m 2 )a  cb  0

2
 ca  (k  m )b  F0
∴ a=
c
 F 0 (k  m 2 ) 2   2 c
,
b=
(k  m 2 )
F 0 (k  m 2 ) 2   2 c
If no damping, C = 0
a = 0, b =
F
0
1
 2F 0 2
2
(k  m )   
108
k
is called “natural frequency 自然頻率”
m
ω=
x = x h + x p = C 1 cos ωt + C 2 sin ωt +
F
0
  2
2
sinΩt
if x(o) = 0, x’(0) = 0
∴ C 1 = 0, C 2 = - ΩFo/ω(  2   2 )
∴ x(t) =
what if
F
0
 (   2 )
2
Ω
(-Ωsinωt +ωsinΩt) ,
Ω≠ω
ω
xp

L’ H o pital’s Rule
x(t) = lim F 0
 
 sin t  t cos t
 2 .
F0
F0
sin

t

t cos t
=
2 2
2
This is called “resonance”.
OR
x’’ + ω 2 x = F 0 sin
t
,
x(o) = 0 ,
x’(o) = 0
x h  C1 cos t  C 2 sin t
x p  at cos t  bt sin t
Solve for xp.
109
LRC circuit:
mx’’+cx’+kx=f(t)
L
Lq”+Rq’+ 1 q=E(t)
c
Lm 2 +Rm+ 1 =0
E
c
m
R
 R  R 2  4L / C
L
R 2  4L / C
> 0 over damped
R 2  4L / C
=0 critically damped
R 2  4 L / C <0
under damped
Lq”+Rq’+ 1 q = Eo sinΩt
c
q p = A sinΩt + BsinΩt
1
)
C
,
2L
1
2
2
2
- (L  
R )
C C22
Eo( L -
A=
B=
= EoX/(-ΩZ 2 )
X = LΩ -
1
C
= EoR/(-ΩZ 2 )
X 2  R2
, Z=
Reactance
q = qh + q p = e
EoR
2L
1
- (L2  2  2 2  R2)
C C 
impedance
R
t
L
(
) + (  EoX3 )sinΩt Z
transient solution
→ 0 as t →∞
ZoR
Z 3
cos Ωt
steady – state
solutions
110
Linear model – Boundary Value Problems (side conditions)
deflection of a beam.
M(x) = bending moment
y(x)
w(x) = load per unit length
d 2M
 w( x)
dX 2
M(x) = E I κ
curvature
Young’s modulus

moment of inertia
y' '
 y ' ' for small deflection y’ ~ 0
[1  ( y ' ) 2 ]3 / 2
∴M(x) = E I y”
d4y
EI 4  w( x)
dx
Boundary conditions
(a)
Embedded at both ends
x =0
x=L
y(0) = 0
no deflection at 0
y(L) = 0
no deflection at L
y’(0) = 0
y’(L)=0
deflection curve is tangent to x-axis
(slope of deflection is zero )
111
If w(x) = wo
d4y
EI 4  wo
dx
w
y 1 (x) = C1  C 2 x  C3 x 2  C 4 x 3  o x 4
24EI
BC# 1
BC# 3
C =0
1
C
2
0
L2  C 4 L3 
BC# 2
C
BC# 4
2 C3 L  C 4 L2 
3
wo L2
C3 = 24EI
y(x) =
=
(b)
wo L2
24 EI
x2 
wo 3
L 0
6EI
wL
C =  12EI
,
wo L
12 EI
wo 4
L 0
24EI
o
4
x3+
wo 4
x
24EI
wo 2
x ( x  L) 2
24 EI
Free and at x = L
y(0) = 0
y’(0) = 0
y’’(L) = 0
y”’(L) = 0
x=0
x=L
bend moment is zero
shear force is zero
dM
d3y
 EI 3 )
(shear force F(x)=
dx
dx
112
(c)
simply supported (pin supported, hinged )
x=0
x=L
y(0)=0
y(L)=0
y”(0)=0
bending moment is zero
y”(L)=0
Eigenvalues 特徵值 and Eigenfunctions 特徵函數
Many 2-point boundary value problems involving a Linear DE
that contains a parameter λ. We need to seek the values ofλ
for which the boundary value problem has nontrivial solutions.
Ex: y” +λy = 0 , y(0) = 0 , y(L) = 0
Case I , λ= 0
y” = 0 , y =
C1 x  C2
c c
1
2
0
y = 0 → a trivial solution.
113
y' '  y  0
Case II , λ< 0
y(x) = C1 exp(   x)  C2 exp(  x)
c c
1
2
0
y = 0 → a trivial solution.
λ>0
Case III.
y”+λy=0
y ( x)  C1 cos  x  C 2 sin  x
y(0) = 0
C1 = 0
y(L) = 0
C2 sin
 L =0
∴sin
 L  0 , C2  0 (Why ?)
∴
L =
nπor

n 2 2
L2
, n = 1,2,3,…
Since the DE is Linear
∴sin 
L
x
, sin 2
L
x
, sin 3 x ….
L
are all nontrivial solutions of the original problem. The numbers
λn =
n 2 2
L2
, n = 1,2,3,…for which the boundary value problem
has a nontrivial solution are known as characteristic values or
eigenvalues. The corresponding solution y n  C sin nx are
L
called characteristic functions or eigenfuntions.
114
Ex. Buckling of a thin Vertical column.
d2y
EI 2  py  0
dx
x
y(x)
BC’s
y(0) = 0,
Leonard Euler
y(L) = 0
y = 0 →trivial solution
L
For what value of P does the boundary value problem gives
nontrivial solution (or for what value of P does the beam
deflect?)
y” + λy = 0 , λ= p/EI
λn =
pn
n 2 2

, n  1,2,3,...
EI
L2
The smallest load which causes beam deflation is
P1 
2
EI
L2
y  c sin(
1
(Euler load )
x
1
y  c sin(
2
n=1
2
L
)
2x
)
L
first buckling mode
2nd buckling mode
n=2
n=3
115
Laplace Transform 拉普拉氏轉換
Chap. 4
General Transformation
O (y) = Y
Transform
unknown
Linear operation
O-1(Y) = y
Inverse transform
Y can be solved algebraically. Then the original unknown is
solved by taking
O 1 (Y)
= y using some transformation skills.
Steps:
1. Transform a “hard” problem into a simple equation
(subsidiary equation)
2. The subsidiary equation is solved by purely algebraic
manipulations.
116
3. The solution of the subsidiary equation is transformed back
to obtain the solution of the original problem.
Laplace transform is particularly useful in problem where the
mechanical or electrical driving force has discontinuities.
E(t)
E(t)
t
Some Transformations:
d
dt
D {f} =
f(t)
t
I {f}= 0 f(t) dt
Definition of the
Laplace
Transform
Let f (t ) be defined for t  0 , then the integral £{f(t)}


0
b
e  st f (t )dt  lim  e  st f (t )dt  F ( s)
b 
0
117
is the Laplace Transform of f(t) if the limit exists.
Laplace transform:
L{f(t)} =
Or

F (s)

0

e -st f(t) dt

 e f(t) dt,
0
 st
a kind of integral transform
L-1{F(s)} = f(t)
118
Ex.
L {t}
F(t)
f(t) = t

L {f (t)} = 0 e  st t dt
t
improper integration.
119

u = t , dv = e  st dt
=
du = dt , v = - 1 e  st
=- 1 t e  st
s
B
lim 
B
0
 sB
s
1
s2
=
-e
B
s
 sB
s
t dt
+ 1 0 e  st dt
B
0
s
= - Be
e  st
2

1
s2
, s>0
Question: Why s > 0 in the last equation?
Answer:
1
s2
L {t} =
Ex:
L -1{
,
1
s2
}=t
f(t) = 1

F(s) = 0 e  st dt =
e _ st
B   s
lim

B
0
1
s
, s>0
L-1{ 1 } = 1
s
Ex: f(t) = sin at

F(s) =  sin at e  st dt
0
=
e  st a
e  st
sin
at

cos
at
[(
s s
s
=(0-0) + (0-
a
 s2
)–
a2
s2
)

0
a2
 2
s
F(s) ,

B
0
sin ate  st dt ]
s>0
120
∴F(s) =
L-1{
a
s  a2
a
s  a2
2
2
,
s>0
} = sin at
Piecewise continuous on [ 0 , ∞ ]
Existence of LT
Exponential order for t > T
Exponential order
A function is said to be of Exponential order c if there exist constants c,
M, and T, all positive, such that | f(t) | ≦ Me ct for all t > T.
If f is an increasing function, then the condition | f(t) | ≦ Me ct , t > T,
simply states that the graph of f on the interval (T, ) does not grow
faster than the graph of the exponential function Me ct , where c is a
positive constant
Question is, can we find M, c, T (all positive)?
Question: Are the following functions exponential order?
121
t
Question: Is the function f (t )  e exponential order?
2
Answer:
A positive integer power of
t is always of exponential order for c > 0.
122
| t n | ≦ Me ct
tn
e ct
or |
t=1 |
1n
ec
t=2 |
2n
e 2c
| ≦ M for t > T
1
ec
| ≦
(c>0)
2n
e 2c
| ≦
2 n ≦e 2 c
n ln2 ≦ 2c

tn
nt n 1
nt n 2
n(n  1)  (2)(1)
t   lim
|
|| ct || 2 ct |  |
|
t  e ct
ce
c e
c n e ct
n!
So, L{t n } exists. In fact, L{t n }= n 1
s
is finite ≦ M.
Piecewise continuous on [0, ]
0
a
t
t
1
t
2
3
in any interval 0 ≦ a ≦ t ≦ b
there are finite number of
discontinuities.
b
If f(t) is piecewise continuous on interval [0 , ∞] and of exponential
order c for t > T ,then L{f(t)} exists for s>c
Pf:
L{f(t)} =
T

0
T

0
e  st f(t) dt +


e  st f(t) dt =
T
I I
1
2
e -st f(t) dt : exists

finite


T


e  st f(t) dt ≦ | T e  st f(t) | dt ≦ M T e  st e ct f(t) dt
e  ( s c ) t
dt = -M
= M T e
sc



-(s -c)t
Since s -c  0

t
e  ( s  c )T
=M
sc
exists for s>c
∴ converges and exists
L{e at } =


0
e at e  st dt =


0
e ( s a )t dt
123
= B lim  L{e at } =
1
,
sa
e  ( s a )t B
1
|0 =
,
sa
sa
1
L-1{
} = e at
sa
s>0
Existence of Laplace transform:


e  st f(t) dt
T
exists if e  st f(t) goes to zero
as t
∞
f(t) can be piecewise continuous
f(t)
t
Existence theorem of L.T.
f(t) is a function that is piecewise continuous on every interval in the
range t ≧ 0 and satisfies
| f(t) | ≦ Me ct
for all t ≧ 0
and for some constant M and c , then the Laplace transform of f(t)
exists for all s > c.
Pf :
f(t) is piecewise continuous , ∴ e  st f(t) is integerable over any
finite interval on t
Assuming s > c then
L{f(t)} =


0
e  st f(t) dt ≦

≦ 0 Me ct e  st dt =


0
M
sc
|f(t)| e  st dt
if (s - c) > 0
Question: Does LT exists for the following functions? (exponential
order?)
2
t
e 5t , cosh
t , e ?
et  e  t
2
124
Answer:
Linearity of L.T.:
L{a f(t) + bg(t)} = aL{ f(t)} + bL{ g(t)}

L{a f + b g} = 0 e  st {a f + b g}dt


=a 0 e  st f(t)dt + b 0 e  st g(t)dt
=aL{f} + bL{g}
Inverse Transform
L{f(t)} = F(s)
L 1 {F(t)} = f(t)
L{a f(t) + bg(t)} = aF(s) + bG(s)
L 1 { aF(s) + bG(s)} = a f(t) + bg(t)
Ex:
L 1 {
 2s  6
}
s2  4
= L 1 {
 2s
6
}+ L 1 { 2
} = -2 cos 2t + 3 sin 2t
2
s 4
s 4
1 st shifting Theorem : (s-shift)

L{e at f(t)} = 0 e  st e at f(t)dt

= 0 e ( sa)t f(t)dt
L{f(t)} = F(s)
=F(s - a) , s –a > β
,
s>β
, s > a+β
L 1 {F(s - a)} = e t f(t)
125
F(s)
F(s-a)
a
s
Ex:
L{te t }=?
Ex:
1
( s  1) 2
L{sin t} =
1
,
s 1
1
=
( s  1) 2  1
L{e t sin t}
L{e 5t t 3 } , L{ t 3 } =
, s>1
L 1 {
s>0
2
3!
s4
, s>1
∴L{e 5t t 3 } =
L{e 2 t cos4t} , L{cos4t} =
∴L{e 2 t cos4t} =
1
s2
L{t} =
L{ te t } =
3!
( s  5) 4
s
s  16
2
( s  2)
( s  2) 2  16
2s  5
A
B

} = L{
}
2
s  3 ( s  3) 2
( s  3)
=2 L 1 {
s
2
11
1
1
}+11L{
}
s3
( s  3) 2
=2e 3t +11e 3t t
Partial Fractions:
L 1 {
s 2  6s  9
}
( s  1)( s  2)( s  4)
= L 1 {
A
B
C


}
s 1 s  2 s  4
16
25
1
16
25
1
= L 1 { 5  6  30 }= - e t + e 2t + e 4 t
s 1 s  2 s  4
5
6
30

126
Ex: F(s) =
L 1 {
3
s  3s  10
, f(t) = ?
2
3
A
B
} = L 1 {
}

s5 s2
s  3s  10
2
3
3
= L 1 { 7  7 }
s5 s2

=
3
1
3
1
L 1 {
}+
L 1 {
}
7
7
s5
s2
=
3 5 t
3
e + e 2t
7
7
OR
L 1 {
b
} = e at sin bt
(s  a) 2  b 2
(s - a) 2 + b 2 = s 2 – 2as + a 2 + b 2
a=-
｝
=s 2 + 3s-10
b=±
( take b =
L 1 {
3
2
7i
2
7i
7i
b= -  same)
2
2
7i
3
3
1
2
}
=
L
{
}
2
2
3
7i
s  3s  10
( s  )  (7i ) 2
2
2
2
=
6 
e
7i
=
3
t
2
sin(
7i
e
6
3
t
2
e
7it
)
2
i(
7 it
)
2
e
2i
i (
7 it
)
2
3
7
= (-e 5t + e 2t )
Ex:
L 1 {
4 s  10
A
B

} = L 1 {
}
s  3 s 1
s  2s  3
2
partial fraction expansion
4s – 10 = A (s +1) +B(s - 3)
127
s=3
2 = 4A A =
1
2
, B=
7
2
1 3t 7  t
e + e
2
2
∴y =
Ex:
L 1 {
OR:
1
-1
1
} = L 1 {
} = - e 2t + e 3t

s-2 s 3
s  5s  6
2
1
=
s  5s  6
2
t
L{sinh } =
2
1
Ex: L {
1
5
1
s 2  5s  ( ) 2 
2
4
1
2
=
1
5
1
(s  ) 2  ( ) 2
2
2
5
t
2
1
s  ( )2
2
t
2
5
t
2
t
2

t
2
L{2e sinh } = L{2e (e -e )/2}
2
s
}?
( s  1)( s 2  4)
s
A
Bs  C

 2
2
( s  1)( s  4) s  1 s  4
Question:
Why not
(Why not
B
?)
s 4
2
B
?
s2  4
Answer:
s  A( s 2  4)  ( Bs  C )( s  1)
s  1,  1  A(1  4),  A 
1
5
128
s  A( s 2  4)  ( Bs 2  ( B  C ) s  C )
s 2  4, s  4 B  ( B  C ) s  C
 B  C  1,  4 B  C  0
1
4
B , C
5
5
1 1
4
s
s
L1{
}  L1{ 5  5 2 5 }
2
s 1 s  4
( s  1)( s  4)
1
1
2
  e t  cos 2t  sin 2t
5
5
5

Partial fractions :
1). Unrepeated factor (s-a)
2). Repeated factor (s-a) m
Y ( s) 
F ( s)
A
B



G ( s ) ( s  a ) ( s  b)
A= s lim a
Ex: y=
( s  a) F ( s)
G( s)
or
A=
F ( s)
G' ( s)
A
A
A
s 1
= 1+ 2 + 3
2
s  s  6s s
s2 s3
3
A 1 = s lim 0
s ( s  1)
1
=2
6
s  s  6s
A 2 = s lim 2
( s  2)( s  1)
s 3  s 2  6s
3
or
a 1
3

3a  2a  6 10
2
(a = 2)

Y ( s) 
F ( s)
Am
Am 1
A


 1
m
m 1
G ( s) ( s  a)
( s  b)
sa
 y(t) =e at ( Am
A m = s lim a
t m1
t m2
t
 Am1
   A2  A1 )
(m  1)!
(m  2)!
1!
( s  a) m F ( s)
G( s)
Others:
Ak =
d ( m  k ) ( s  a) m F ( s)
1
[
]
lim
a
(m  k )! s
ds ( m  k )
G( s)
, k = 1 , 2 , … , m-1
129
s 3  4s 2  4
F ( s)
= 2
=
G( s)
s ( s  2)( s  1)
Ex: Y(s) =
A2 A1
B
C
+ +
+
2
s s  2 s 1
s
s 2 (s 2
s 2 ( s 3  4s 2  4)
4
A 2 = s lim 0 2 2
=
=2
2
s ( s  3s  2)
-3s+2)
(m = 2 , k = 1)
d s 3  4s 2  4
(
)
ds s 2  3s  2
( s 2  3s  2)(3s 2  8s)  ( s 3  4s 2  4)( 2s  3)
12
=
=
=3
2
2
4
( s  3s  2)
A 1 = s lim 0
B=
F (s)
| s 2  1
s ( s  1)
2
∴y(t) = L 1 {
, C=
F (s)
| s 1  1
s ( s  1)
2
2 3
1
2
 

}
2
s s  2 s 1
s
=2t+3-e 2t -e t
Transforming a derivative (why ?)
d2y
dy
L{ } , L{
}…
dt
dt 2
If f’(x) is continuous ( no discontinuity) for all t ≧ 0


L{f’(t)} = 0 e  st f’(t) dt = e-stf(t)
|  dt + s 0 e  st f(t) dt
 0
0 a s t 
= - f(0) + sL{f(t)}
∴L{f’(t)} = sF(s) –f(0)
Let g = f’(t)
L{f”(t)} = L(g’(t)) = sL{g(t)} – g(0)
130
=sL{f’(t)} – f’(0)
=s[sL{f(t)}-f(0)] – f’(0)
=s 2 F(s) – sf(0) – f’(0)
L{f (n) (t)} = s n F(s) - s n 1 f(0) - s n2 f’(0) - … - f n 1 (0).
 hence , the LT replaces operations of Calculus by operations of algebra
 Laplace’s basic idea.
Theorem: What if f’(x) has discontinuers?
If f is continues and dominated by e at , and f’ is piecewise
continues on every interval and of exponential order .
Then for s > a,
L{f’(t)} = sL{f(t)} – f(0  ),
Pf:
L{f’(t)} = Blim



f(0+) = lim f(t)
t 0 
f’(t)e  st dt
0
If f’(t) has discontinuities at tj on [ 0 , B ]
t1
L{f’(t)}= 0 f’(t)e  st dt+ t
t2
B
f’(t)e  st dt+…+ B lim  t f’(t)e  st dt
1
n
t1
=f(t)e  st | t0 +s 0 f(t)e  st dt + f(t)e  st | tt +s t
1
2
1
B
lim  f(t)e  st | tBn + B lim  s 
B
t2
f’(t)e  st dt+…+
1
f(t)e  st dt
tn
t1
= f(t 1 ) e  st - f(0  ) + s 0 f(t)e  st dt
t2
+ f(t 2 ) e  st - f(t 1 ) e  st + s t
2
f(t)e  st dt + …
1
B
+ B lim  f ( B)e  st - f(t n )e  st +s t f(t)e  st dt
n

n
0

=s 0 f(t)e  st dt - f(0  )
2 nd derivative:
131
L{f”} = L{(f’)’} = sL{(f’)} - f(0  )
=s(sL{f}) - f(0  )- f’(0  )
=s 2 L{f}- sf(0  ) – f’(0  )

L{f (n) } = s n {f(t)} - s n 1 f(0  ) - s n2 f’(0  ) - … - f ( n 1) (0  )
Ex : f(t)=t 2
L{t 2 } = ?
f(0) = 0 , f’(0)=0 , f”(0) = 2
L{f”}=s 2 L{f} - s 
f ( 0) - f ' ( 0 )

0
L{2} = 2L{1} =
0
2
2
= s 2 L{t 2 } ∴ L{t 2 } = 3
s
s
Ex: L{sin 3t} f(t) = sin 3t
f’(t) = 3cos 3t
f”(t) = 9sin 3t
L{-9sin 3t} = s 2 L{sin 3t} – 3
(s 2 + 9)L{sin 3t} = 3
∴L{sin 3t} =
3
s 9
2
Transforms of derivatives and integrals:
Differentiation of functions corresponds to the multiplication of
transforms by s, and integration of functions corresponds to the
division of transforms by s.
L{ f ' (t )}  sF ( s)  f (0)
t
1
L{ f (u )du}  F ( s)
s
0
132
LT of Integral of a function:
If f is piecewise continuous and of exponential order, dominated
by e at , Then for s > a
L{ 0 f(t’)dt’} = 1 L{f(t)}
t
s
Pf: Let g(t) = 0 f(t’)dt’
t
∴g’(t) = f(t) and
g(0) = 0
∴L{g’(t)} = L{f(t)} = sL{ 0 f(t’)dt’ }
t
∴L{ 0 f(t’)dt’} =
t
1
L{f(t)}
s
F(s) = L{f(t)}
1
F(s)
s
= L{ 0 f(t’)dt’}
t
∴L 1 { 1 F(s)} = L 1 L{ 0 f(t’)dt’} = 0 f(t’)dt’
s
t
Ex:
t
t
L{ 0 u 2 e 3u du} = ?
L{ t 2 e 3t } =
2
( s  3) 3
t
∴L{ 0 u 2 e 3u du} =
Ex:
L{f} =
L 1 {
1
s(s   2 )
1
}
(s   2 )
L 1 { 1 (
f(t) = ? f = L 1 {F(s)}
2
2
= 1 sin ωt
1
)}
s s 2
2
2
s( s  3)3

= 1 0 sin ωt’ dt’

t
133
Or : f = L 1 {
1
}
s(s   2 )
2
= L 1 { A 
s
B
}
s 2
2
Differentiation of Laplace Transform

L{f(t)} = F(s) = 0 e  st f(t)dt
dF ( s )
=
ds

- 0 te  st f(t)dt = -L{t f(t)}
d 2 F (s)
=
ds 2

+ 0 t 2 e  st f(t)dt = L{t 2 f(t)}

L{t n f(t)} = (-1) n F (n) (s)
Ex:
L{t 2 e 3t } = ?
L{e 3t } =
1
s3
L{t 2 e 3t } = L{t 2 e 3t } =
Ex:
d
ds
d2
ds 2
1
)
s3
=+
1
)
s3
=
(
(
1
( s  3) 2
2
( s  3) 3
L{f(t)} = ln[ s  2 ] = F(s)
s3
-
d
ds
F(s) =
1
- 1 =
s3 s2
L{tf(t)}
L 1 {
1
- 1 }
s3 s2
= L 1 L{tf(t)}
e 3t - e 2t = tf(t)
∴f(t)
e 3t
=
t
e 2 t
t
134
Linear DE of initial value problems
dny
d n 1 y
+
a
+ … + a 0 y = g(t)
n 1
dt n
dt n 1
an
IC’s : y(0)=y 0 , y’(0) = y 1 , … , y n 1 (0) = y n 1
〝
L{
} = L{g(t)}
a n [s n Y (s) - s n 1 y(0) - … - y n 1 (0)] +
a n 1 [s n 1 Y (s) - s n2 y(0) - … - y n2 (0)] + … + a 0 Y (s) = G(s)
Ex:
dy
+3y = 13 sin 2t,
dt
y(0)=6
sY (s) – 6 + 3Y (s) =
26
s 4
2
Therefore, the L.T. of a linear DE with constant coefficients becomes an
algebraic equation in y (s).
Y(s) =
6
26
8
 2s  6


 2
2
s  3 ( s  3)( s  4) s  3 s  4
∴y(t) = 8e 3t -2cos 2t + 3sin 2t
Find unknown
y(t) with IC’s
be difficult)
Find
original y(t)
L.
DE becomes an
algebraic eq for Y (s)
T.
L .T.
1
(may be
Solve transformed eq
forY(s)
difficult)
Differential eqs of initial value problem
135
y” + ay’+ by =
r(t)

, y(0) = k 0
, y’(0) = k 1
input (driving force)
y(t) : output (response of the system)
L{y”+ay”+by} = L{r(t)}
Y= L(y) R = L{r}
[s 2 Y- sy(0) – y’(0)] + a[sY- y(0)] + bY = R(s)
( s2

as
b )Y = (s + a) y(0) + y’(0) + R(s)
yQ ( s )
Y (s) = [(s + a) y(0) + y’(0)]Q(s) +R(s)Q(s)
If y(0) = y’(0) = 0
Q(s) =
∴ y =RQ
L(output )
Y ( s)
=
R( s)
L(input )
transfer functions
last step: y(t) = L 1 {Y (s)}
Ex: y”- y = t
, y(0) = 1 , y’(0) = 1
1
s2
s 2 Y - sy(0) – y’(0) - y =
1
s2
s 1
1
∴Y= 2
+ 2 2
s 1
s ( s  1)
(s 2 -1)Y = s + 1 +
how ?
1
1
1
+( 2  2 )
s 1
s 1 s
=
y(t) = L 1 {Y (s)} = L 1 {
=e t + sinh t – t
1
1
1
} + L 1 { 2 } - L 1 { 2 }
s 1
s 1
s
#
What if y(0) and y’(0) are unknown ?

4
y( ) =

2
s 2 Y- sy(0) –y’(0) +Y=
2
s2
Ex: y” + y = 2t ,
∴Y=

4
,y’( ) =2- 2
5
1
2
+ y(0) 2 + y’(0) 2
2
s 1
s 1
( s  1) s
2
y = L 1 {Y} = 2t – 2sin t + y(0)cos t + y’(0)sin t
136
=2t + y(0)cos t +[y’(o)-2]sin t
=2t + Acos t + bsin t
A
B 

+


2
2
2 2
A
B

y’( ) = 2- 
=24
2
2

4
y( ) =
∴A=1 , B=1
2
y(t) = cos t – sin t +2t
A A
B
C
s 3  4s 2  4
F ( s)
= 2
= 1  22 

G( s)
s ( s  3s  2)
s s
s  2 s 1
Ex: Y(s) =
A 2 = s lim 0
s 2 ( s 3  4s 2  4)
=2
s 2 ( s 2  3s  2)
A 1 = s lim 0
d s 2 ( s 3  4s 2  4)
=3
ds s 2 ( s 2  3s  2)
(m =2 , k=1)
B = -1 , C = -1
∴y(t) = 3 + 2t - e 2t - e t
#
Ex: y” – 6y’ +9y = t 2 e 3t ,
y(0) = 2
,
y’(0) = 6
s 2 Y(s) – sy(0) – y’(0) – 6[sY (s) – y(0)] +9Y(s) =
Y(s) =
2
( s  3) 2
2s  5
2

2
( s  3)
( s  3) 5
︷
︸
||
2
11

s  3 ( s  3) 2
∴y(t) = 2e 3t +11t e 3t +
1 4 3t
t e .
12
137
How to transform the following functions?
1
0
a
Unit step function 單位階梯函數 (Heaviside step function)
u(t-a) = { 0 , t
a
<
= H(t-a)
Pulse function
1
1 , t ≧ f(t) = u(t-a)- u(t-b)
a b
a
f(t) = [u(t)- u(t-1)]- [u(t-1)- u(t-2)]
1
1
2
+[u(t-2)- u(t-3)]- [u(t-3)- u(t-4)]
-1

= u(t) - 2 u(t-1) + 2u(t-2) -2 u(t-3) +…

= u(t) + 2  (1)n u (t  n)
n 1
138

L{ u(t-a)} = 0 u(t-a)e  st dt
 st

= a e  st dt = B lim  e | aB =
s
e  as
s
, s>0
L{unit square wave} = L{ u(t)}-2L{ u(t-1)}
+2L{ u(t-2)} – 2L{u(t-3)} + …
=1 2e
s
s
=1 2e
s
s
e 2 s
e 3s
2
s
s
+…
s
1
(
)
s 1  e s
 2e  s
1
s
) = tanh
s
s
2
1 e
= ( 1 )(1  e
s
2
s
2 nd shifting Theorem (t – shift)
F(s) = L{f(t)}
L{ u(t-a)f(t-a)} = e  as F(s)
, a ≧ 0
L 1 { e  as F(s)} = u(t-a)F(t-a) ,
a ≧ 0
Pf:

e  as F(s) = 0 e  s (t ' a ) f(t ' )dt '
t = t ' +a
139

= a e  st f(t-a)dt
=L{u(t-a)F(t-a)}
cos t
cos(t-2)u(t-2
)
π
2
Question: What does f(t-a)u(t-a) mean?
Answer:
Ex:
L 1 { e
L 1 {
3 s
1
s3
s3
}
} = 1 t2
2
∴L 1 { e
3 s
s
3
} = 1 (t  3) 2 u (t  3) ={
0
,
t<3
2
1
(t  3) 2
2
,
t ≧ 3
Alternative form of Second Shifting Theorem:
L{ u(t-a)f(t)} = e  as L{f(t+a)}
Short impulses,
1
k
0 ,
Dirac’s Delta function :
, a ≦ t ≦ a+k
otherwise
140
f k (t) = {
1
Area =
k
 fdt =  (fore)(time) = impulse
1
 f dt = 1
h
t
a a+h
f h (t) = 1 [u(t-a)-u(t-(a+h))]
k
L{ f h (t)} = 1 [ e  as - e ( ah) s ]
ks
= e  as 1  e
 ks
ks
 ks
1 e
k 0
ks
lim
=  se
 ks
=1
s
∴L{ lim
f h (t)} = L{δ(t-a)} = e  as
k 0
δ(t-a) = { ∞
, t=a
0 ,
Properties of
delta function:
otherwise
a
 δ(t-a)dt = 1
a
a
 δ(t-a)f(t)dt = f(a)
a
Ex:
c=?
k=?
141
y” + 3y’ + 2y = δ(t-1) ,
δ
1
Q(s) = 2
s  3s  2
=
,y’(0)=0
s 2 Y + 3sY + 2Y= e  s
Y (s) = e  s (
1
)
( s  1)( s  2)
1
1

)
s

1
s

2


= e s (
F (s)
s
e
Output = ( Input )Q
∴ Input
y(0) = 0
e s
s 2  3s  2
f = L 1 {F(s)} = e  t -e 2t
y(t) = L 1 { e  s F(s)} = f(t-1)u(t-1)
={ 0 , 0 ≦ t < 1
e
-e
,
t
y
≧ 1
0
1
2
t
Transforming of periodic Functions
f
t
F is piecewise continuous with positive period p.
142
f(t) = f(t + p) = f(t + 2p) = … = (t + np)
L{f(t)} =
1
1  e  ps
p
 e f(t)dt , s > 0
 st
0

Pf: L{f(t)} = 0 e  st f(t)dt
p
= 0 e  st f(t)dt + p
2p
e  st f (t )dt



np
+ … + ( n 1) p e  st f(t)dt + …
u t  p
u = t - 2p
p
e  sp 0 e  su f(u+p)du
p
e 2 sp 0 e  su f(u+np)du
p
=( 0 e  st f(t)dt)(1+ e  sp + e 2 sp +…)
=
1
1  e  sp
Ex:

p
e  st f(t)dt ， s > 0
0
f(t) = t
1
1
L{f(t)} =
2
1
1  e s
3
1
 e tdt
 st
0
143
=
1
1  e s
( - 1 e  st | t0 +
1
s
s
e s 1
1
(1  e s
s s2
s
= 12 - e  s
s
s (1  e )
=
t
 e dt)
 st
0
(e  s -1)
Laplace Convolution Theorem 卷積定理:
L{f(t)} = F(s)
Then
,
L{g(t)} = G(s)
L 1 {F(s)G(s)} ≠ f(t)g(t)
=f

g
= 0 f(τ)g(t-τ)dτ
t
or
L{ 0 f(τ)g(t-τ)dτ} = F(s)G(s)
t
Pf: L{ f

g}

= L{ 0 f(τ)g(t-τ)dτ} = 0 { 0 f(τ)g(t-τ)dτ} e  st dt
t
t
144


= 0 [  f(τ)g(t-τ) e  st dt]dτ




= 0 f(τ) dτ  g(t-τ) e  st dt
t-τ= μ
= 0 f(τ) dτ 0 g(μ) e s (   ) dμ


= 0 f(τ) e  st dτ 0 g(μ) e  s dμ= F(s)G(s)
Convolution , Integral Equations
Convolution of f(t) and g(t)≡f
f


g
t
g = 0 f(t-x)g(x) dx
(sin t)
 (cos
t
t) = 0 sin(t-x)cos x dx
= 1 0 [sin t+sin (t-2x)dx
2
t
= 1 xsin t| t0 + 1 cos(t-2x) | t0
2
sinAcosB =
1
2
[sin(A+B)+sin(A-B)]
4
= 1 t sin t
2
f

g=g

t
f = 0 g(t-x)f(x)dx
t
= 0 g(y)f(t-y)dy
t-x = y x =
t-y
145
f
(f
f
{
 (g 1 +

g)

g2) = f
v

=0
f

(sin ft)?
=f
g1 + f


(g

v)
g2
f=0

1=1
 (sin

t) = 0 sin(t-x)sin x dx = - 1 tcos t + 1 sin t > 0 ?
2
2
t
t2
2
t
(t) 
f (t)f= ≧0 0xdx
? =
≠ t2
Convolution Theorem:
L{ f

t
g } = 0 e  st [ 0 f(t-x)g(x)dx]dt

= F(s)G(s) = L{f}L{g}
f1= 1
,
f 2 = f(t)
L{f 1  f 2 } = L{ 0 f(x)dx} = 1 L{f(t)}
t
s
= L{ f 1 }L{ f 2 } = 1 L{f(t)}
s
L{f} = F(s)
L{g} = G(s)
Ex: L 1 {
Method 1:
3
}
s  3s  10
2
=?
3
3
3
3
7
=
=
 7
s 2  3s  10 ( s  5)( s  2) s  5 s  2
146
} = - 3 e 5 t + 3 e 2 t
∴L-1{
7
7
Method 2:
L 1 {
3
}
s  3s  10
2
= L 1 {
b
(s  a) 2  b 2
L 1 {
}
b
(s  a) 2  b 2
} = e at sin bt
s 2 +3s-10 = s 2 -2as+a 2 +b 2
a = -3
, b = ± 7i 2
2
(pick b = 7i 2 )
7i
3
1
2
= L {
2
3
7i
( s  )  (7i ) 2
2
2
2
3t
= 6 e 2 sin( 7i t)
7i
2
= 3 (-e 5t +e 2t )
7
=
}
6 3t 2 e
e
7i
i ( 7 it )
2
e
2i
i ( 7 it )
2
Method 3:
L 1 {
3
}
s  3s  10
2
= 3 [L 1 { (
= 3L 1 { (
1
) }] 
s2
1
1
)(
)}
s2 s5
[L 1 { (
1
) }]
s5
= 3e 2t  e 5t = 3 0 e 2 e 5(t  ) dτ = 3 0 e 5t 7 dτ
t
t
= 3e 5t 0 e 7 dτ = 3e 5t 1 e 7t | t0 =
t
7
3
7
e 5t ( e 7t -1)
= 3 ( e 2t -e 5t )
7
Ex: L 1 {
1
s ( s  1)
2
} = L 1 { 1
1
}
s s 1
f : L 1 {F} = 1
2
= L 1 {F(s)G(s)}
, t ≧ 0
=u (t)
147
g : L 1 {G} = sin t
∴L 1 { F G } = f

g
t
= 0
u (t-x) sin x dx = 1- cos t
Integral Eq :

y(t) = t + 0 sin(t-x)y(x)dx
y (s)
=
1
s2
∴ y (s) = s
+
y (s)
2
1
s
4
∴y(t) = t + t
=
1
s 1
1
s2
2
+
1
s4
3
6
Differential Eqs:
y” + ay’ + by = r(t)
y (s)
= {(s+a) y(0) + y’(0)}Q(s)+R(s)Q(s)
Q(s) = 1 s 2  as  b
R(s) = L{r(t)}
If y(0) = y’(0) =0
y(t) = L 1 {R(s)Q(s)}
t
= 0 q(t-x)r(x)dx
,
q(t) = L 1 {Q(s)}
148
Ex: y” +2y = r(t)
, r(t) = { 1 , 0 < t < 1
y(0) = y’(0) =0
0 ,
otherwise
r(x)
Q(s) = 1 s 2  2
s 2 Y+ 2Y = R(s)
∴q(t) = sin
2t
t
1
2
x
y(t) =
1
t
2
 sin
0
= 1 (1-cos
y(t) =
1
2
 sin
0
= 1 [1-cos
2
for t < 1
2 t)
2
1
2 (t-x)(1)dx
r(x)
2 (t-x)(1)dx
2 (t-1)
-cos
2 t]
for t > 1
1
t
x
f(t)
k
x(t
m )
149
mx” + kx = f(t) let f(t) = F 0 = const.
x(0) = x 0
IC’s{
x’(0) = x 1
L{mx” + kx} = L{ F 0 }
mL{x”} + kL{x} = F 0 L{1}
m[s 2 X(s) - sx 0 - x 1 ] + kX(s) = F 0 1
s
X(s) = sx2 0  x21 +
s 
F0
ms( s 2   2 )
∴x(t) = L 1 { sx2 0  x21 +
ω=
s
s 2
2
} + x 1 L 1 {
s

=1
sin t

1
s 2
2
}+

} = L 1 { 1 } * L 1 {
1
s s 2
2
m
F0
m
L 1 { 1
1
s s 2
2
}
sin t
cos ωt
L 1 { 1
k
F0
}
ms( s 2   2 )
s 
= x 0 L 1 {
,
t
= 0 (1)
1
s 2
2
sin  (t   )

L 1 {F(s)G(s)}=f(t)
}
g(t)
d
= 1  cos2 t

or 1

sin t

t
= 0
sin 
∴x(t) = x 0 cos ωt +

x1

(1)d =
1  cos t
sin ωt +
2
F0
k
(1-cos ωt)
If x(0) = x’(0) = 0
k
150
(mg = F 0 )
If ( F 0 = F(t))
L{f(t)} = F(s)
X(s) =
F ( s)
m( s 2   2 )
x(t) =
1
m
=
L 1 (
F (s)
s  w2
f(t)

1
m
2
)
sin t

 sin ωτ f(t-τ) dτ
t
0
(what if f(t) = sin t?)
Solutions to ODE
R
E
1
e t
C
t
L{Ri + 1 0 i(t’)dt’} = L{ e  t }
c
t
RI(s) + 1 I(s) =
cs
∴I(s) =
1
s 1
sC
=1
( R( s  1)( s  1)) R
s
=1[
A
1
1
R
)( s  1)
(s 
)
RC
RC
RC
1
)
1  ( RC  1) (
RC  1 ]

= [
1
R
( s  1)
(s 
)
RC
(s 

B
]
( s  1)
151
∴i(t) =
t
1
e  RC
RC ( RC  1)
c
e t
RC  1
+
Ex:
#
dq
dt
I=
Lq” + 1 q= E 0 (u(t-2)-u(t-5))
c
E
q(0) = q 0
I(t)
q’(0) = 0
E
2
L(s 2 Q(s)-sq 0 ) + 1 Q(s) = E 0 ( e
c
∴Q(s) =
q0 s
s 2
2
L 1 {
s
s 2
L 1 {
1
}
s(s   2 )
2
+
= L 1 { 1 }
s
sin 
t
1

2
s
E0
1
(e 2 s 2
2
L s(s   )
= 0
=
-e
5 s
s
)
e 5 s ) , ω =
1
LC
} = cos ωt
2
∴ L 1 {
2 s
5
1
(
s(s   2 )
2


L 1 {
1
s 2
2
}=1

sin t

dτ = 1  cos2 t

e 2 s - e 5 s )}
u(t-2)[1-cos ω(t-2)]-
∴q(t) = q 0 cos ωt +
E0
C
1
2
u(t-5)[1-cos ω(t-5)]
{ u(t-2)[1-cos ω(t-2)]-u(t-5)[1-cos ω(t-5)]
If we don’t use LT, how to solve the DE’s?
Lq” + 1 q = 0 , q(0) = q 0 , q’(0) = 0
c
Lq” + 1 E =E 0 , for 2≦t≦5,
c
,
for 0≦t≦2
(What are the IC’s?)
152
Lq” + 1 q = 0 , for 5≦t <∞, (What are the IC’s?)
c
Why does a discontinuous input produce a continuous output?
q(t)
t
E(t) discontinuous
E(t)
 continuous
O/p?
discontinuous I/p → continuous O/p
discontinuous
If q”(t) = u(t-a)
a
integrate: q’(t) = u(t-a)(t-a) + C 1
continuous but not smooth
integrate: q (t) =
Ex:
(t  a ) 2
2
u(t-a) + C 1 t + C 2
C 1 = C 2 =0
continuous
RC circuit
Rq’ +
E(t)
1
q
c
and smooth
= E(t)
50t
q(0) = q 0
I(t)
IC’s:
40
153
q’(0) = 0
E(t) = 50t[1- u(t-2)] + 40 u(t-2)
S2Q + Q = 502 +L{(40-50t) u(t-2)}
s
L{(40-50t)H(t-2)} = L{[-60 - 50(t-2) H(t-2)]}
= -60L{ u(t-2)} - 50L{(t-2) u(t-2)}
= -60 e
∴Q(s) =
2 s
2 s
-50 e 2 s L{t} = -60  2 s -50 e 2
s
s
s
50
- 60 e 2 s - 2 50 e 5 s
s ( s  1) s( s  1)
s ( s  1)
2
L 1 {
1
}
s( s  1)
L 1 {
1
}
s ( s  1)
2
= L 1 { 1 }
s
1
s2
= L 1 {

}
L 1 {

1
}
s 1
L 1 {
= 1  e  t = 0 e  dτ =1- e  t
1
}
s 1
t
=t

e t
= 0 (t-τ)e  dτ = t-1+ e  t
t
L 1 {
1
e 2 s }
s( s  1)
L 1 {
1
e 2 s }
s ( s  1)
2
= (1-e) (t 2) u(t-2)
= [(t-2)-1+e (t 2) ]u(t-2)
∴q(t)= 50(t-1+ e  t ) –60(1-e (t 2) )u(t-2) –50[(t-2)-1+ e (t 2) ]u(t-2)
=50(t-1+ e  t ) + (90-50t+10e 2t )u(t-2)
Convolution and Integral Eqs
t
y(t) = t + 0 sin(t - x) y(x)dx , find y(t) = ?
Y(s) =
1
S2
+ L{ sin t

y(t)}
154
1
S2
Y (s) =
1
s 1
+ Y(s)
so Y(s) =
so y(t) = t
x   px  q(t )
s2 1
s4
t3
+
6
2
1
s2
=
x(0) =Xo
+
1
s4
P = a cos t
sX + Xo + pX =Q(s)
so x(t) = Xo e
 pt
+ e  p (t  ) Q(  ) dt
t
= e  pt [ Xo + 0
Q(  )e pt dt]
other way of looking the DE
t
t
0
0
 (x’ + px ) dt =  q(  ) dt
x(t)
Since
t
0
t
+ p 0

x(t) d =
t
 q(  )dt
0
x(0) = xo
So x(t) – xo + p 0
t
t
x(t) + xo + p 0
t
x(  )dt = 0 q(  )dt
t
x(  )dt = xo + 0 q(  )dt
Integral eq.
x’ +px = q(t)
, x(0) = xo
D.E
155
Solved by L.T.
X(s) + P 1 X(s) = xo
s
x0
Sp
X(s) =
x0
s
+
+
1
Q(s)
s
Q(s)
Sp
same
Ex; x’’-4x = 6δ(t-1) , x(0) = 0 , x’(0) = -3 .
s 2 X+3-4X = F(s)
X=-
3
1
 2
F (S )
s 4 s 4
2
1/2 sinh 2t

6δ(t-1)
x(t) = -3/2 sinh 2t + 1/2 sinh 2t

6δ(t-1)
t
3 0  (  1) sinh 2(t   )d
If t < 0
0
t>1
3sinh 2(t-1)
u(t-1)[3sinh 2(t-1)]
3.u'(t)= δ(t) , u’(t-a)= δ(t-a)
f(t)δ(t-a) =
(a)
Verify
 f (a) (t - a), if f(a)  0

, if f(a)  0
 0
x(t) = u(t-2)sinh (t-2) is the solution.
Of x’’- x =δ(t-2), x(0) = x’(0)=0
x(t) = u(t-2)sinh (t-2)
 x(0)
= u(-2)sinh (-2)= 0
156
x’(t) = u’(t-2)sinh (t-2) + u(t-2)cosh (t-2)
=δ(t-2)sinh (t-2) + u(t-2)cosh (t-2)
= 0δ(t-2) + u(t-2)cosh (t-2)
= u(t-2)cosh (t-2)
 x’(o)
= u(-2)cosh (-2)=0
x”(t) = u’(t-2)cosh (t-2) + u(t-2)sinh (t-2)
=δ(t-2)cosh (t-2) + u(t-2) sinh (t-2)
1
=δ(t-2) + u(t-2)sinh (t-2)
x” – x =δ(t-2)
System of Differential Equations
Ex:
y 1 ’+ y 1 +3y 2 = 1
y 1 (0) = 0 , y 2 (0) = 0
3 y 1 + y 2 ’+2y 2 = t
157
sY 1 +Y 1 +3Y 2 = 1
(s + 1)Y 1 +3Y= 1
s
s
3Y 1 +sY 2 +2Y 2 =
1
s2
3Y+(s + 2)Y 2 =
1
s2
y 1 ” = -k y 1 + k(y 2 - y 1 )
k
1
0
y 2 ” = -k(y 2 - y 1 ) - k y 2
m 1 y 1 (0)=1 ,y 2 (0)=1 ,y 1 ’(0) =
y1
s 2 Y1 - s -
k
0
2
m2 s Y2 - s +
y2
3k
3k
3k
= -k(Y 2 -Y 1 ) - kY 2
-kY 1 + (s 2 + 2k)Y 2 = s Y 1 = (s 
, y 2 ’(0)= -
= -kY 1 +k(Y 2 -Y 1 )
(s 2 + 2k)Y 1 -kY 2 = s +
k2
3k
3k
3k
3k )( s 2  2k )  k ( s  3 )
( s 2  2k ) 2  k 2
( s  3k )( s 2  2k )  k ( s  3 )
Y2 =
( s 2  2k ) 2  k 2
Y1 =
s
s k
+
3k
s  3k
Y2 =
s
s k
-
3k
s  3k
2
2
2
2
y 1 (t) = cos
k
t +sin
3k
t
y 2 (t) = cos
k
t -sin
3k
t
System of eqs
---electrical circuit
E(t)
158
0
0.5
t
 i (o)  o

i ' (o)  o
L di'  Ri  E (t )
dt
1

0.8i1 '1(i1  i2 )  1.4i1  100[u (t )  u (t  )]

2

1‧ i2 '1(i2  i1 )  0

1

i1 '3i1  1.25i2  125[u (t )  u (t  )]

2

i2 'i1  i2  0

S


( s  3) I  1.25I  125[ 1  e 2 ]
1
2

s
s

 I1  ( s  1) I 2  0

125( s  1)
I1 
1
7
s ( s  )( s  )
2
2
(1  e

S
2
)
159
125
I2 
1
7
s ( s  )( s  )
2
2
500

7s
(1  e
S
2
)
125
625

1
7
3( s  ) 21( s  )
2
2
t


125 2 625
e 
e
3
21
7 t
2

500

7s
500
7
e

i1 (t )  
125
e
3
125
(e
3
t
2
250
i2 (t )  
(e
3
t
2
1
( t  )
2
2
e

625
e
21
1
( t  )
2
2
e
)
1
( t  )
2
2
250
250

1
7
3( s  ) 21( s  )
2
2
1
7 ( t  )
2
2
625
(e
21

7 t
2
250
)
(e
21
7t
2
S
2
500
7
7
e 2
e
1
(t  )
2
)
7
1
(t  )
2
2
)
i
160
i1
i2
0.5
why
 i1 (t )


i 2 (t )
why
i1
i2
0
1
as
is not smooth at t =
t
t
1
2

?
while
is continuous and smooth at t =
1
2
?
161