Bong-Kee Lee
School of Mechanical Systems Engineering Chonnam National University
Engineering Mathematics I
2. Second-Order Linear ODEs
2.1 Homogeneous Linear ODEs of 2 nd Order
2계 선형상미분방정식
– 표준형(standard form)
x y q
xy r
x py'' ' nonhomogeneous
'
0 ''px yqx y
y homogeneous
0 ' ''
0 1 ' ''
cos 25
'' example
2
y y y
y xy y
x e y
y x 2nd order nonhomogeneous linear ODE 2nd order homogeneous linear ODE 2nd order homogeneous nonlinear ODE
School of Mechanical Systems Engineering Engineering Mathematics I
2.1 Homogeneous Linear ODEs of 2 nd Order
중첩의 원리(superposition principle) 또는 선형성의 원리(linearity principle)
– 주어진 해에 어떤 상수를 더하거나 곱함으로써 추가적인 해를 얻을 수 있음
– (정리) 제차 선형상미분방정식에 대하여, 어떤 열린 구간 I 에서 두 개 해의 일차결합은 다시 구간 I에서 주어진 미분 방정식의 해가 된다. 특히 그러한 방정식에 대해서는 해들 의 합과 상수곱도 다시 해가 된다.
2 2 1 1
2 1&
0 ' ''
y c y c y
y y y y
qy py y
2.1 Homogeneous Linear ODEs of 2 nd Order
중첩의 원리(superposition principle) 또는 선형성의 원리(linearly principle)
– 비제차 선형상미분방정식 및 비선형 상미분방정식에는 적 용되지 않음
0 sin 2 cos 7 . 4 sin 2 cos 7 . 4
sin 2 cos 7 . 4 '' sin 2 cos 7 . 4
sin 2 cos 7 . 4
0 sin sin sin '' sin 2
0 cos cos cos '' cos 1
sin
&
cos 0
'' example
x x x
x
x x x
x
x x y
x x x x
x x x x
x y x y y
y
1 cos
'' 21 cos
2cos 2 2cos 2 2cos 1 2
1 sin 1 sin sin 1 '' sin 1 2
1 cos 1 cos cos 1 '' cos 1 1
sin 1
&
cos 1 1 '' example
x x
x x
x y
x x
x x
x x
x x
x y
x y
y y
2 2 2 22
2 2 2 2 2
2
4 ' ''
0 ' 1 1 '' 1 2
0 2 2 ' ''
1
1
&
0 ' '' example
x x x x x x y
x
x x x x x x
y x y xy y y
School of Mechanical Systems Engineering Engineering Mathematics I
2.1 Homogeneous Linear ODEs of 2 nd Order
초기값 문제(initial value problem, IVP)
– 2계 제차 상미분방정식의 일반해(general solution)
– 2계 제차 상미분방정식에는 두 개의 초기조건이 요구됨
– 2계 제차 상미분방정식의 초기값 문제
2 2 1
1y C y
C
y
x0 K0 & y'
x0 K1y
0 0& '
0 1with
0 '
' '
K x y K x y
y x q y x p y
2.1 Homogeneous Linear ODEs of 2 nd Order
초기값 문제(initial value problem, IVP)
x x
y
C y x C x C y
C y x C x C y
x C x C y x y x y
y y
y y
sin 5 . 0 cos 0 . 3
5 . 0 0
' cos sin
'
0 . 3 0 sin cos
step 2nd
sin cos
sin
&
cos step 1st
5 . 0 0 ' , 0 . 3 0 , 0 '' example
2 2
1
1 2
1
2 1
general solution
particular solution
School of Mechanical Systems Engineering Engineering Mathematics I
2.1 Homogeneous Linear ODEs of 2 nd Order
기저(basis) 또는 기본시스템(fundamental system)
– 일반해를 구성하는 해(y
1, y
2)는 구간 I에서 1차 독립 (linearly independent)이어야 함
– 비례(proportional)
• 주어진 두 초기조건을 만족시키는 해를 정의할 수 없음
– 1차 독립
– 1차 종속(linearly dependent)
0 const
2
1ky k
y
0
0 1 2
2 2 1
1y k y k k
k
2.1 Homogeneous Linear ODEs of 2 nd Order
기저(basis) 또는 기본시스템(fundamental system)
– 기저: 1차 독립인 해의 쌍(pair)
x x
x x x x x
x x x x
x x x x
x x
e e y
C C y
C C y
e C e C y C y C y
e e e y y
e e e e y y
e e e e y y
e y e y
y y y y
4 2
2 0
' 2
6 0
1
const
0 ''
'' 2
0 ''
'' 1
?
&
2 0 ' , 6 0 , 0 '' example
2 1
2 1
2 1 2 2 1 1
2 2
1 2 2
1 1
2 1
general solution
particular solution
School of Mechanical Systems Engineering Engineering Mathematics I
2.1 Homogeneous Linear ODEs of 2 nd Order
차수 축소(reduction of order)
– 하나의 해를 알고 있을 때, 두 번째 독립해를 1계 상미분 방정식을 이용하여 구할 수 있음
y uy y Udx
y e U
pdx y U
dx y p
y U U dU y p
U y
U u U u py
y u y u
qy py y u py y u y u
uy q uy y u p uy y u y u
y x q y x p y
uy y u y u uy y u y u y u y y
uy y u y y uy y y
y y
x q y x p y
pdx
1 1 2 2
1
1 1
1 1
1
1 1 1
1 1 1 1 1 1
1 1 1 1
1 1
1 1 1 1 1 1 1 2
1 1 2 1
2
1
&
1
ln 2 ' ln
0 2 '
' 2
' '' , ' 0 '
2 ' ''
0 ' '' '
2 ' ''
0 '
' '' ' ' 2 ''
0 '
''
'' ' ' 2 '' '' ' ' ' ' '' '' ''
' ' ' '
solution, known e with th 0 '
''
2.1 Homogeneous Linear ODEs of 2 nd Order
차수 축소(reduction of order)
1 1 ln
ln
1 1 1 1
ln ln 2 1 ln ln
2 1 1 0 2
2 ' : '
0 ' 2 '' 0
' ' 2 ''
' 2 '' ' ' '' '' '
'
? then , intuition, by
0 ' '' example
2
2 2
2
2 2
2 2
1 2
2 1
2
x x ux x y
x u
x x x x dx v du x x x x
v
x dx dx x
x x
x v v dv x v x x u v
u x u x x ux u x u x u x u x x
u x u u u x u y u x u y
ux uy y y
y x y
y xy y x x
separating variables
School of Mechanical Systems Engineering Engineering Mathematics I
2.2 Homogeneous Linear ODEs: Const. Coffs.
상수계수를 갖는 2계 제차 선형상미분방정식
– 특성방정식(characteristic equation), 보조방정식(auxiliary equation)
0 ' ''
' ''
by ay y
x r y x q y x p y
0' '' :
''
&
' assuming
by
2 2
2
x x
x x
x x
x
e b a be
e a e by ay y
e y e y e y
2ab0
1&2y1e1x&y2e2x
2.2 Homogeneous Linear ODEs: Const. Coffs.
특성방정식의 근에 따른 세 경우
– (a)>0: 두 개의 서로 다른 실근(two real roots) – (a)=0: 실 이중근(real double root)
– (a)<0: 공액 복소근(complex conjugate roots)
2 4 0
2 2
, 1 2
b a b b a
(a)
School of Mechanical Systems Engineering Engineering Mathematics I
2.2 Homogeneous Linear ODEs: Const. Coffs.
특성방정식의 근에 따른 세 경우(I)
– 두 개의 서로 다른 실근(two real roots)
x
x Ce
e C
y 1 1 2 2
2
1&
general solution
x x
x x
e e y
C C C
C y
C C y
e C e C y
y y y y y
2
2 1 2
1 2 1
2 2 1
2
3
3 , 5 1
2 0 '
4 0
solution particular
solution general
2 , 1 0 2 equation
stic characteri
5 0 ' , 4 0 , 0 2 ' '' example
2.2 Homogeneous Linear ODEs: Const. Coffs.
특성방정식의 근에 따른 세 경우(II)
– 실 이중근(real double root)
x
x Cxe
e C
a y
1 2 1 2
2
general solution
xx x
x
e x y
C C C
C y
e xC C e
C x y
C y
e xC C y
y y y y y
5 . 0
2 1 1
2
5 . 0 2 1 5 . 0 2 1
5 . 0 2 1
2
2 3
2 , 3 5
. 3 5 . 0 0 '
5 . 0 '
3 0
solution particular
solution general
5 . 0 0
25 . 0 equation
stic characteri
5 . 3 0 ' , 3 0 , 0 25 . 0 ' '' example
School of Mechanical Systems Engineering Engineering Mathematics I
2.2 Homogeneous Linear ODEs: Const. Coffs.
특성방정식의 근에 따른 세 경우(III)
– 공액 복소근(complex conjugate roots)
x e
C x e
C y a i
a i ax ax
cos sin
, 2 2
2 / 2 2
/ 1 2
1
general solution
x e
y
C C C
y
x e
C x e
C x y
C y
x e
C x e
C y i
y y y y y
x
x x
x x
3 sin
1 , 0 3
3 0 '
3 cos 3
3 sin 2
'
0 0
solution particular
3 sin 3
cos solution
general 3 2 . 0
0 04 . 9 4 . 0 equation stic characteri
3 0 ' , 0 0 , 0 04 . 9 ' 4 . 0 '' example
2 . 0
2 1 2
2 . 0 2 2
. 0 2 1
2 . 0 2 2
. 0 1
2
2.2 Homogeneous Linear ODEs: Const. Coffs.
특성방정식의 근에 따른 세 경우(III)
– 공액 복소근(complex conjugate roots)
x e
C x e
C y a i
a i ax ax
cos sin
, 2 2
2 / 2 2
/ 1 2
1
x i x
e e e e
e y
x i x e
e e e
e y
a i i
a a
a b b a b
a
b a b a
a
ax x x i x a a i x
ax x x i x a ai x
sin cos
sin cos
2 2
2 2
4
4 4 4
0 4
2 0 4
2 2 2
2
2 2 2
1
2 2 2
, 1
2 2
2 2
2 2
, 1 2
2 1
1 2
2
1 2
1 2
, 1 2
1 y y
Y i y y
Y
sin
cos i
ei
Euler formula
School of Mechanical Systems Engineering Engineering Mathematics I
2.3 Differential Operators
연산자(operator)
– 함수를 다른 함수로 바꾸는 변환을 의미 – 미분연산자(differential operator)
– 항등연산자(identity operator)
– 선형연산자(linear operator)
' y dxy y d dx
d
D
D
y y I
a b
y y a y by y ay byy
b a
'
2 ''
2 2
I D D I D D
I D D L L
2.4 Modeling: Free Oscillations
자유진동(free oscillation): 질량-용수철 시스템 (mass-spring system)
2 ''
2
dt my y md ma
F
Newton’s second lawky
F Hooke’s law
0 0
0
FW ks mg ks static equilibrium0 ''
''
ky mymy ky F
undamped system
School of Mechanical Systems Engineering Engineering Mathematics I
2.4 Modeling: Free Oscillations
자유진동(free oscillation): 질량-용수철 시스템 (mass-spring system)
– 비감쇠 시스템(undamped system)
• 조화진동(harmonic oscillation)
A B B
A C t C t y
m t k
B t A t y ky my
1 2
2 0
0 0 0
tan ,
, cos
, sin cos
0 ''
amplitude
(진폭) phase angle (위상각)
2
0
natural frequency (고유주파수, Hz) (a)
(b) (c)
1 ) c (
0 ) b (
1 ) a (
1 ,
0 1
B B B
A
2.4 Modeling: Free Oscillations
자유진동(free oscillation): 질량-용수철 시스템 (mass-spring system)
– 감쇠 시스템(damped system)
' cy
F damping force
0 ' ''
'' '
ky cy my
my cy ky F
damped system
homogeneous 2nd order linear ODE
(case I) overdamping (case II) critical damping (case III) underdamping
m mk c k c
c
m 2
0 4
2
2
School of Mechanical Systems Engineering Engineering Mathematics I
2.4 Modeling: Free Oscillations
자유진동(free oscillation): 질량-용수철 시스템 (mass-spring system)
– 감쇠 시스템(damped system): 과감쇠(overdamping)
m mk c m
e c C e
C t y
mk c ky cy my
t t
2
& 4 2
, 0 4 0 ' ''
2 2
1
2
2.4 Modeling: Free Oscillations
자유진동(free oscillation): 질량-용수철 시스템 (mass-spring system)
– 감쇠 시스템(damped system): 임계감쇠(critical damping)
m e c
t C C t y
mk c ky cy my
t
2 ,
0 4 0 ' ''
2 1
2
School of Mechanical Systems Engineering Engineering Mathematics I
2.4 Modeling: Free Oscillations
자유진동(free oscillation): 질량-용수철 시스템 (mass-spring system)
– 감쇠 시스템(damped system): 저감쇠(underdamping)
m c mk m
t c Ce
t B t A e t y
mk c ky cy my
t t
2
* 4
&
2 ,
* cos
* sin
* cos
0 4 0 ' ''
2 2
frequency
2 1 2 4 2
* 2
m c mk
as c→0
2 1 2
2
* 0
m
k
2.4 Modeling: Free Oscillations
자유진동(free oscillation): 질량-용수철 시스템 (mass-spring system)
– 감쇠 시스템(damped system)
c↓
School of Mechanical Systems Engineering Engineering Mathematics I
2.4 Modeling: Free Oscillations
자유진동(free oscillation): 질량-용수철 시스템 (mass-spring system)
– 감쇠 시스템(damped system)
c=10 c=5 c=1 c=0
2.5 Euler-Cauchy Equations
오일러-코시 방정식(Euler-Cauchy equation)
– 2계 제차 상미분방정식의 한 형태
0 ' '
2y'axyby x
1
00 1
0 1
1 ''
'
, assuming
by
2
1 2
2 2 1
b m a m
b am m m
bx mx ax x m m x x m m y
mx y
x x y
m m m
m m
m
characteristic equation
School of Mechanical Systems Engineering Engineering Mathematics I
2.5 Euler-Cauchy Equations
오일러-코시 방정식(Euler-Cauchy equation)
– (case I) 서로 다른 두 실근
– (case II) 이중근
– (case III) 공액복소근
2 1
2 1
2 1
2 1 0 ,
m
m Cx
x C y
m m b m a m
mm
m C x x C C xx
x C y
m b m a m
ln ln
0 1
2 1 2
1 2
A x B x
x y
i m b m a m
ln sin ln
cos 0
2 1
2.5 Euler-Cauchy Equations
오일러-코시 방정식(Euler-Cauchy equation)
x x y x u x u U xU
U u U
xu u xu u x xu m mxu u x
y xu m y
mx u y u x
u y m xy m y
x y xu m y
u y u x
y m uy y u x m uy
y u y u uy x
y u y u y y
uy y u y y
uy y y mx mx y x y
a m b m a a m
b a
b a m a
b m a m x y by axy y x
m m
m
m
ln ln
1 ' 1 0
' : '
0 1 ' '' 0 ' '' 0 ' 2 1 ' 2 ''
0 ' 2 1 '
2 ''
0 '
2 1 '' '
2 1 ' ' 2 ''
0 '
' 2 1 '' ' ' 2 '' ''
' ' 2 '' '' ''
' ' ' '
' ,
4
& 1 2 1 21
then 1 , 0 4 1 if
2 4 1 0 1
1 0
' '' example
2 2
2
1 1
1 1
2
1 2 1 1
2 1 1
1 2
2 2 1 1 1
1 1 2 1 1 1 2
1 1 2
1 2 1 1 1 1 1
2 2 2
2 2
2
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2.5 Euler-Cauchy Equations
오일러-코시 방정식(Euler-Cauchy equation)
A x B x
x y
Y C Y C y x x
y i y Y
x x
y y Y
x i x x
e x e
x x x x y
x i x x
e x e x x x x y
x y x y i m b
a
b a m a
b m a m x y by axy y x
x i i
x i
i
x i i
x i
i
i i
m
ln sin ln
cos
ln 2 sin
1
ln 2 cos
1
ln sin ln cos
ln sin ln cos
&
then , 0 4 1 if
2 4 1 0 1
1 0
' '' example
2 2 1 1 2
1 2
2 1 1
ln ln
2
ln ln
1
2 1
2
2 2
2
2.6 Existence and Uniqueness of Solutions
해의 존재성과 유일성
– 연속적인 가변변수를 가지는 제차 선형미분방정식의 해
0 0, '
0 1with
0 '
' '
K x y K x y
y x q y x p y
2 2 1
1y c y
c y
solution ?
Theorem 1
Theorem 4
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2.6 Existence and Uniqueness of Solutions
해의 존재성과 유일성
. interval on the
solution unique a has problem value initial ing correspond then the
, in is and interval open some on functions continuous
are and If
1.
Theorem
0
I
x y
I x I x
q x p
constants.
suitable are , and on ODE the of solutions of
basis any is , where
form the of is on solution
every then ,
interval open some on and ts coefficien continuous
has ODE given the If
4.
Theorem
2 1 2
1
2 2 1 1
C C I y
y
x y C x y C x Y
I x Y y I
x q x p
2.6 Existence and Uniqueness of Solutions
해의 존재성과 유일성
– Wronski 행렬식(determinant) 또는 Wronskian
' ''
, ' 1 2 2 1
2 1
2 1 2
1 yy y y
y y
y y y
y
W
y1,y2
0 y1,y2:linearly dependentW
x x
y y
y W y
x y
x y
y y
2 2
2 1
2 1
2 1
2
sin cos
' '
sin , cos
0 ''
example
x x xx x
e xe e x
y y
y W y
xe y e y
y y y
2 2 2
2 1
2 1
2 1
1
' ' ,
0 ' 2 '' example
School of Mechanical Systems Engineering Engineering Mathematics I
2.7 Nonhomogeneous ODEs
2계 비제차 상미분방정식
– 일반해
• 제차 상미분방정식의 일반해 + 비제차방정식의 어떤 해
– 제차 방정식의 해와 비제차 방정식의 해 사이의 관계
• 비제차 방정식의 두 해의 차는 제차 방정식의 해
• 비제차 방정식의 해와 제차 방정식의 해의 합은 비제차 방정식의 해
'
0'
'px yqxyrx y
x r y x q y x p y x y
y x q y x p y y C y C x y
x y x y x y
p p p
p
h h h
h
p h
' ''
0 '
1 ''
2 1 1
2.7 Nonhomogeneous ODEs
미정계수법(method of undetermined coefficients)
– 비제차 방정식의 어떤 해를 구하기 위한 방법
– 일반적인 방법(매개변수 변환법, method of variation of parameter; Chap. 2.10)보다 간단하게 적용이 가능
– 상수계수를 가지는 선형상미분방정식에 적합kex Cex
xr yp
x
0,1,
n
kxn KnxnKn1xn1K1xK0 x
k x
kcos or sin KcosxMsinx x
ke x
kexcos or xsin ex
KcosxMsinx
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2.7 Nonhomogeneous ODEs
미정계수법(method of undetermined coefficients)
– 선택 규칙(I): 기본 규칙(basic rule)
• 비제차 상미분방정식의 r(x)가 표의 함수 중 하나라면, 같은 줄의 yp를 선택하고, yp와 그 도함수를 비제차 상미분방정식에 도입하 여 미정계수를 결정함
002 . 0 001 . 0 002
. 0 , 0 , 001 . 0
2 2
001 . 0 '' 2
'' 2 '
~ 001 . 0 solution particular :
step 2nd
sin cos 0
'' solution s homogeneou :
step 1st
5 . 1 0 ' , 0 0 with 001 . 0 '' example
2 0
1 2
0 2 1 2 2 0 1 2 2 2
2
2 1 2
0 1 2 2
2 2
x y
K K K
K K x K x K K x K x K K
x y
y K
y
K x K y
K x K x K y
kx x x
r
x B x A y y y
y y x y y
p p p p
p p
n h
2.7 Nonhomogeneous ODEs
미정계수법(method of undetermined coefficients)
002 . 0 001 . 0 sin 5 . 1 cos 002 . 0
5 . 1 0 '
002 . 0 cos sin '
002 . 0 0 002 . 0 0
conditions initial
: step 3rd
002 . 0 001 . 0 sin cos
002 . 0 001 . 0
&
sin cos
2 2 2
x x x
y B y
x x B x A y
A A
y
x x B x A y y y
x y
x B x A y
p h
p h
x y
x yh
x ypSchool of Mechanical Systems Engineering Engineering Mathematics I
2.7 Nonhomogeneous ODEs
미정계수법(method of undetermined coefficients)
– 선택 규칙(II): 변형 규칙(modification rule)
• 만약 yp로 선택된 항이 비제차 상미분방정식에 대응하는 제차 상 미분방정식의 해가 된다면, 선택된 yp에 x(또는 x2)를 곱함
x p
x p
p p x p
x p
x x p
x x
x h
x
e x y
C Cx
x x C x x C
e y y y e
x x C y
e x x C y
Ce e Cx y
ke e x r
e x C C y y y y
y y e
y y y
5 . 1 2
2 2
2
5 . 1 5
. 1 2 5 . 1 2
5 . 1 5 . 1 2
5 . 1
5 . 1 2 1 5
. 1
5
5 10
25 . 2 5 . 1 2 3 25 . 2 6 2
10 25 . 2 ' 3 '' 25
. 2 6 2 ''
5 . 1 2 '
~ 10 solution
particular :
step 2nd
0 25 . 2 ' 3 '' solution s homogeneou :
step 1st
0 0 ' , 1 0 with 10
25 . 2 ' 3 '' example
2.7 Nonhomogeneous ODEs
미정계수법(method of undetermined coefficients)
xx x
x p
h
x p
x h
e x x y
C C
y
e x x C C e
x C y
C y
e x x C C y y y
e x y e x C C y
5 . 1 2
2 2
5 . 1 2 2 1 5 . 1 2
1
5 . 1 2 2 1
5 . 1 2 5
. 1 2 1
5 5 . 1 1
5 . 1 0
5 . 1 0 '
5 5
. 1 10
'
1 0
conditions initial
: step 3rd
5 5
&
xy
x yh
xyp
School of Mechanical Systems Engineering Engineering Mathematics I
2.7 Nonhomogeneous ODEs
미정계수법(method of undetermined coefficients)
– 선택 규칙(III): 합 규칙(sum rule)
• 만약 r(x)가 표의 함수들의 합이라면, 이에 대응하는 함수들의 합 으로 yp를 선택함
x e
y
K M C
x x
e y y y x M x K Ce y
x M x K y Ce y
x x
y e y y y y
x M x K ke x x
e x r
x B x A e y y y y
y y
x x
e y y y
x p
x p p p x
p
p x p
p x p p p p
x x
x h x
10 sin 2 16 . 0
0 , 2 , 16 . 0
10 sin 190 10 cos 40 5
' 2 '' 10 sin 10 cos
10 sin 10 cos
&
10 sin 190 10 cos 40
~
&
~
sin cos
~ 10 sin 190 10 cos 40 solution
particular :
step 2nd
2 sin 2 cos 0
5 ' 2 '' solution s homogeneou :
step 1st
08 . 40 0 ' , 16 . 0 0 with 10 sin 190 10 cos 40 5
' 2 '' example
5 . 0
5 . 0 5
. 0
2 5 . 0 1
1 5 . 0 1 2 1
5 . 0 5
. 0
2.7 Nonhomogeneous ODEs
미정계수법(method of undetermined coefficients)
x y
x yh
x yp
x e
x e y
B B
y
x e
x B e x B e y
A A
y
x e
x B x A e y y y
x e
y x B x A e y
x x
x x
x
x x
p h
x p
x h
10 sin 2 16 . 0 2 sin 10
10 08
. 40 20 08 . 0 2 0 '
cos 20 08 . 0 2 cos 2 2 sin '
0 16 . 0 16 . 0 0
conditions initial
: step 3rd
10 sin 2 16 . 0 2 sin 2 cos
10 sin 2 16 . 0
&
2 sin 2 cos
5 . 0
5 . 0
5 . 0 5 . 0