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EEngineering Mathematics Ingineering Mathematics I

Prof. Dr. Yong-Su Na g(32-206, [email protected], Tel. 880-7204)

Text book: Erwin Kreyszig, Advanced Engineering Mathematics,

9th Edition, Wiley (2006)

Ch 6 Laplace Ch 6 Laplace TransfomsTransfomsCh. 6 Laplace Ch. 6 Laplace TransfomsTransfoms6.1 Laplace Transform. Inverse Transform. Linearity.

s-Shifting

6.2 Transforms of Derivatives and Integrals. ODEsg

6.3 Unit Step Function. t-Shifting

6 4 Short Impulses Dirac’s Delta Function Partial Fractions6.4 Short Impulses. Dirac s Delta Function. Partial Fractions

6.5 Convolution. Integral Equations

6.6 Differentiation and Integration of Transforms

6.7 Systems of ODEs

6.8 Laplace Transform: General Formulas

6.9 Table of Laplace Transforms6 ab o ap a a o

2

Ch. 6 Laplace TransformsCh. 6 Laplace TransformsC 6 ap ace a s o sC 6 ap ace a s o s((라플라스라플라스 변환변환))

제어 시스템 해석, 설계 등 미분방정식을 풀 수 있는 도구로 공학전반에 널리 이용

Laplace Transform

상미분방정식을 라플라 변환하여 방정식 변환• 상미분방정식을 라플라스 변환하여 보조방정식으로 변환

• 대수적인 연산을 통하여 보조방정식을 푼다.

• 보조방정식의 해를 역변환하여 상미분방정식의 해를 구한다.

장점

• 비제차 상미분방정식의 해를 구할 때, 제차 상미분방정식의 일반해를따로 구할 필요가 없다.따 구할 필 가 없다

• 초기값은 보조방정식을 만드는 과정 중에서 자동적으로 고려된다.

• 불연속성, 순간적인 충격량, 또는 복잡한 주기함수를 입력으로 갖는 상미분방정식도쉽게 해를 찾을 수 있다쉽게 해를 찾을 수 있다.

Ch. 6 Laplace TransformsCh. 6 Laplace TransformsC 6 ap ace a s o sC 6 ap ace a s o s((라플라스라플라스 변환변환))

< Time domain과 s-Domain과의 관계>

66.1 .1 LapalceLapalce Transform. Inverse Transform.Transform. Inverse Transform.Shifti Shifti ss--Shifting Shifting

((라플라스라플라스 변환변환. . 역변환역변환. . 선형성선형성 그리고그리고 ss--이동이동))

Laplace Transform (라플라스 변환): ( ) ( ) ( )dttfefsF st∫∞

−==0

L

Inverse Transform (역변환):

0

( ) ( )tfF =−1L

Ex. 1 Let when . Find .0≥t( ) 1=tf ( )sF

( ) ( ) ( )0111 >====∞

−∞

−∫ sedtef ststLL( ) ( ) ( )0 100

>=−=== ∫ ss

es

dtef LL

Ex. 2 Let when . Find .0≥t( ) atetf = ( )fL

( ) ( ) 11

00 ase

sadteee tasatstat

−=

−==

∞−−

∞−∫L



Linearity of the Laplace Transform: 라플라스 변환은 선형연산이다.

( ) ( )( ) ( )( ) ( )( )tbtftbtf LLL ( ) ( )( ) ( )( ) ( )( )tgbtfatbgtaf LLL +=+

Ex.3 Find the transforms of andatcosh atsinh

( ) ( )atatatat eeateeat −− −=+=21sinh ,

21cosh

( ) ( ) 11( ) ( )as

e,as

e atat

+=

−= − 1 1

LL



Linearity of the Laplace Transform: 라플라스 변환은 선형연산이다.

( ) ( )( ) ( )( ) ( )( )tbtftbtf LLL ( ) ( )( ) ( )( ) ( )( )tgbtfatbgtaf LLL +=+

Ex.3 Find the transforms of andatcosh atsinh

( ) ( )atatatat eeateeat −− −=+=21sinh ,

21cosh

( ) ( ) 11( ) ( )as

e,as

e atat

+=

−= − 1 1

LL

( ) ( ) ( )[ ] 1111cosh seeat atat =⎟⎞

⎜⎛ +=+=⇒ −LLL( ) ( ) ( )[ ]

( ) ( ) ( )[ ] 22

22

1121

21sinh

22cosh

aeeat

asasaseeat

atat =⎟⎠⎞

⎜⎝⎛ −=−=

−⎟⎠

⎜⎝ +

+−

+⇒

−LLL

LLL

( ) ( ) ( )[ ] 2222 asasas −⎟⎠

⎜⎝ +−



First Shifting Theorem (제 1이동정리), s-Shifting (s-이동)

( )( ) ( ) ( )( ) ( ) ( ) ( ){ }asFtfeasFtfesFtf -atat −=−=⇒= 1 , LLL ( )( ) ( ) ( )( ) ( ) ( ) ( ){ }fff ,

Ex. 5 ( ) ( )( ) 2222 cos cos

ωω

ωω

+−−

=⇒+

=as

astes

st atLL

Use these formulas to find the inverse of the transform

( )

( ) ( )( ) 2222 sin sin

ωωω

ωωω

+−=⇒

+=

aste

st atLL

Use these formulas to find the inverse of the transform

( )4012

13732 ++

−=

sssfL

( )( ) ( ) ( )

( )ttess

ss

sf t 20sin720cos34001

2074001

13400114013

21

21

21 −=⎟⎟

⎠

⎞⎜⎜⎝

⎛

++−⎟⎟

⎠

⎞⎜⎜⎝

⎛

+++

=⎟⎟⎠

⎞⎜⎜⎝

⎛

++−+

= −−−− LLL



Existence Theorem for Laplace Transforms (라플라스 변환의 존재정리)

함수 가 영역 상의 모든 유한구간에서 구분적 연속인 함수.

( )( )tf 0≥t

어떤 상수 와 에 대해 (너무 빠른 속도로 값이 커지지 않음)

모든 에 대해 의 라플라스 변환 가 존재

( ) ktMetf ≤k M⇒ ks > ( )tf ( )fL



PROBLEM SET 6.1

HW 23HW: 23

66.2 Transforms of Derivatives and Integrals..2 Transforms of Derivatives and Integrals.ggODEs ODEs ((도함수와도함수와 적분의적분의 변환변환. . 상미분방정식상미분방정식))

Laplace Transform of Derivatives (도함수의 라플라스 변환):

( ) ( ) ( ) ( ) ( ) ( ) ( )0'0''0' 2 fsffsfffsf −−=−= LLLL( ) ( ) ( ) ( ) ( ) ( ) ( )000 fsffsf, ffsf LLLL

( )( ) ( ) ( ) ( ) ( ) ( )00'0 121 −−− −−−−= nnnnn ffsfsfsf LLProve!

Ex. 1 Let . Find ( ) tttf ωsin= ( )fL


Laplace Transform of Derivatives (도함수의 라플라스 변환):

( ) ( ) ( ) ( ) ( ) ( ) ( )0'0''0' 2 fsffsfffsf −−=−= LLLL( ) ( ) ( ) ( ) ( ) ( ) ( )000 fsffsf, ffsf LLLL

( )( ) ( ) ( ) ( ) ( ) ( )00'0 121 −−− −−−−= nnnnn ffsfsfsf LLProve!

Ex. 1 Let . Find ( ) tttf ωsin= ( )fL

( ) ( ) ( ) ( ) ttttffttttff ωωωωωωω sincos2'' ,00' ,cossin' ,00 2−==+==

( ) ( ) ( ) ( ) ( )22 2ωss( ) ( ) ( ) ( ) ( ) ( )22222

222sin 2'' ω

ωωωω

ω+

==⇒=−+

=⇒s

sttffsfs

sf LLLLL


Laplace Transform of Integral (적분의 라플라스 변환):

( )( ) ( ) ( ) ( ) ( ) ( )⎟⎞⎜⎛==⎟⎟

⎞⎜⎜⎛

⇒= ∫∫ sFdfsFdfsFtf -tt 11 1LLL ττττ( )( ) ( ) ( ) ( ) ( ) ( )⎟

⎠⎜⎝

==⎟⎟⎠

⎜⎜⎝

⇒= ∫∫ sFs

df, sFs

dfsFtf 00

LLL ττττ

1 1Ex. 3 Find the inverse of and( )22

1ω+ss

( )t ωτ 1sin111 11 ⎟

⎞⎜⎛⎞⎛ ∫

( )2221ω+ss

( ) ( )tdss

ts

ωω

τωωτ

ωω

ωωcos11sin1 sin11

20

221

221 −==⎟⎟

⎠

⎞⎜⎜⎝

⎛+

⇒=⎟⎠⎞

⎜⎝⎛

+ ∫−− LL

( )1 sin11 ωttt

⎟⎞

⎜⎛

∫( ) ( ) 320

22221 sincos111

ωω

ωτωτ

ωωttd

ss−=−=⎟⎟

⎠

⎞⎜⎜⎝

⎛+ ∫−L


Differential Equations. Initial Value Problems

( ) ( ) ( ) 10 0' ,0 ,''' KyKytrbyayy ===++

• Step 1. Setting up the subsidiary equation (보조방정식의 도출): ( ) ( )rRyY LL == ,

( ) ( )[ ] ( )[ ] ( )sRbYysYaysyYs =+−+−− 00'02

• Step 2 Solution of the subsidiary equation by algebra:

( ) ( ) ( ) ( ) ( )sRyyasYbass +++=++ 0'02

Step 2. Solution of the subsidiary equation by algebra:

Transfer Function (전달함수): ( )2

22

41

21

11

abasbass

sQ−+⎟

⎠⎞

⎜⎝⎛ +

=++

=

solution of the subsidiary equation:

Step 3 Inversion of Y to obtain

42 ⎠⎝

( ) ( ) ( ) ( )[ ] ( ) ( ) ( )sQsRsQyyassY +++= 0'0

( )Yy -1L• Step 3. Inversion of Y to obtain ( )Yy L=


Ex 4 Solve the initial value problemEx. 4 Solve the initial value problem

( ) ( ) 10' ,10 ,'' ===− yytyy

Step 1 subsidiary equation

( ) ( ) ( ) 22

22 111 10'0 sYsYysyYs ++=−⇒=−−−

Step 2 transfer function

( ) ( ) ( ) 22 ss

11

2 −=

sQ

( ) ( ) ⎟⎠⎞

⎜⎝⎛ −

−+

−=

−+

−+

=++= 2222221

11

11

11

1111

sssssssQ

sQsY

Step 3 inversion

( ) ( ) tteYty t +=⎟⎞

⎜⎛

⎟⎞

⎜⎛+⎟

⎞⎜⎛== −−−− sinh111 1111 LLLL( ) ( ) tte

sssYty −+=⎟

⎠⎜⎝

−⎟⎠

⎜⎝ −

+⎟⎠

⎜⎝ −

== sinh11 22 LLLL


Ex 2 Derive the formulas in three different waysEx. 2 Derive the formulas in three different ways

( ) cos 22 ωω

+=

sstL

( ) 22sinω

ωω+

=s

tL


PROBLEM SET 6.2

HW 24HW: 24

66.3 Unit Step Function. t.3 Unit Step Function. t--Shifting Shifting 66 3 U S ep u c o3 U S ep u c o S gS g((단위계단함수단위계단함수. t. t--이동이동))

Unit Step Function(단위계단함수)의 라플라스 변환

• Unit step function or Heaviside function: ( ) ( )( )⎩

⎨⎧

><

=−atat

atu10

• 단위계단함수의 라플라스 변환 :

( )⎩ > at1

( ){ }s

eatuas−

=−L Prove!

S d Shif i Th (제 2이동정리) Ti Shif i ( 이동)Second Shifting Theorem (제 2이동정리), Time Shifting (t-이동)

( )( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ){ }sFeatuatfsFeatuatfsFtf as-as −− =−−=−−⇒= 1 , LLL


( )( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ){ }sFeatuatfsFeatuatfsFtf as-as −− =−−=−−⇒= 1 , LLL

( ) ( ) ( )dfedfeesFe assasas

0

)(

0== ∫∫

∞ +−∞ −−− ττττ ττ

( ) ( )

( ) ( ) dtatuatfesFe

dtatfesFe

stas

a

stas

)(=

−=

∫∫∞ −−

∞ −−

( ) ( ) dtatuatfesFe )(0

−−= ∫


E 1 W it th f ll i f ti i it t f ti d fi d it t fEx. 1 Write the following function using unit step functions and find its transform.

( )( )( )1

10 2 2⎪⎪

⎨

⎧

<<

<<

= πttt

tf ( ) ( )( )

2 cos21 2

⎪⎪⎩

⎨

>

<<=π

π

tt

tttf

( ) ( )( ) ( ) ( ) ⎟⎞

⎜⎛⎟

⎞⎜⎛

⎟⎞

⎜⎛ 111 2Step 1 단위계단함수의 식:

Step 2 항별 라플라스 변환

( ) ( )( ) ( ) ( ) ⎟⎠⎞

⎜⎝⎛ −+⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ −−−+−−= ππ

21cos

211

21112 2 tuttututtutf

p

( ) ( ) ( ) ( ) 2322

21111

2111

211

21 se

ssstutttut −⎟

⎠⎞

⎜⎝⎛ ++=

⎭⎬⎫

⎩⎨⎧

−⎟⎠⎞

⎜⎝⎛ +−+−=

⎭⎬⎫

⎩⎨⎧ − LL

22

23

222

821

21

821

221

21

21

21 s

esss

tutttutπππππππππ−

⎟⎟⎠

⎞⎜⎜⎝

⎛++=

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎠⎞

⎜⎝⎛ −⎟

⎟⎠

⎞⎜⎜⎝

⎛+⎟⎠⎞

⎜⎝⎛ −+⎟

⎠⎞

⎜⎝⎛ −=

⎭⎬⎫

⎩⎨⎧

⎟⎠⎞

⎜⎝⎛ − LL


( ) 22 11

21

21sin

21cos

se

stuttut

ππππ

−

+−=

⎭⎬⎫

⎩⎨⎧

⎟⎠⎞

⎜⎝⎛ −⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ −−=

⎭⎬⎫

⎩⎨⎧

⎟⎠⎞

⎜⎝⎛ − LL

( ) 22

22

2323 11

821

211122

ssss es

esss

esss

ess

fLππππ −−−−

+−⎟⎟

⎠

⎞⎜⎜⎝

⎛++−⎟

⎠⎞

⎜⎝⎛ +++−=⇒

1822 sssssssss +⎠⎝⎠⎝


Ex 2 Find the inverse transform f (t) ofEx. 2 Find the inverse transform f (t) of

( )( )2

3

22

2

22 2++

++

+=

−−−

se

se

sesF

sss

ππ

ππ

πt

ssin1

221 =⎟

⎠⎞

⎜⎝⎛

+−L

( )tte

st

s2

21

21

21 1 −−− =⎟⎟

⎠

⎞⎜⎜⎝

⎛

+⇒=⎟

⎠⎞

⎜⎝⎛ LL (제 1이동정리)

( ) ( )( ) ( ) ( )( ) ( ) ( ) ( ) ( )−−+−−+−−=⇒ −− 3322sin111sin1 32 t tuettuttuttf ππ

ππ

( )( ) ( )

( )⎪

⎪⎪⎨

⎧

<<<<−

<<

=3t202t1 sin1t0 0

tπ

π

( )( ) ( ) ( )⎪

⎪⎩ >−

<<−− 3t 3

3t2 032 tet


PROBLEM SET 6.3

HW 35HW: 35

66.4 .4 Short Impulses. Dirac’s Delta Function.Short Impulses. Dirac’s Delta Function.ppPartial FractionsPartial Fractions ((짧은짧은 충격충격. . 디랙의디랙의 델타함수델타함수))

Dirac Delta Function의 라플라스 변환Dirac Delta Function의 라플라스 변환

• Dirac delta function or unit impulse function (단위충격함수):

( ) ( )⎧ =∞ at( ) ( )( )⎩

⎨⎧

=−경우 밖의그 0

atatδ

( ) ( ) ( ) ( )kata⎧ +≤≤1( ) ( )

( )( ) ( )atfatkatakatf kkk −=−⇒

⎩⎨⎧ +≤≤=−

→ 0 lim

0

1δ

경우 밖의그

( ) ( ) 111⇒ ∫∫∫

∞+∞

dttdtdttfka

δu(t )와 δ(t ) 관계?

• Dirac delta function의 라플라스 변환:

( ) ( ) 1 100

=−⇒==− ∫∫∫ dtatdtk

dtatfa

k δ

( ) ( ) ( )( )[ ]

ks

k katuatuk

atf

−

+−−−=−

11

1

( )( ) ( )[ ] ( ){ } asks

asskaask eat

kseeee

ksatf −−+−− =−⇒

−=−=−⇒ δLL 11


Ex. 1 Determined the response of the damped mass-spring system under a square wave.

( ) ( ) ( ) 00' ,00 ),2(1)(2'3'' ==−−−==++ yytututryyy ( ) ( ) ( ),),()( yyyyy

( )⎧ ( )

( ) 2121

21

100)1(2)1(

⎪

⎪⎪

⎨

⎧

<<+−

<<

=−−−−

tee

t

ytt

)2(21

21 )2(2)1(2)2()1(⎪

⎪

⎩>

−++− −−−−−−−− teeee tttt


Ex. 2 Find the response of the system with a unit impulse at time t = 1.

( ) ( ) ( ) 00' ,00 ,12'3'' ==−=++ yytyyy δ

상미분방정식: seYsYYs −=++ 232

보조방정식:

( ) ( )( )s

se

ssssesY −−

⎟⎠⎞

⎜⎝⎛

+−

+=

++=

21

11

21

( ) ( ) ( )( ) ( ) ( )⎩

⎨⎧

><<

== −−−−−

1t1t0 0

1211

tt eeYty L

( )( ) ⎠⎝

( ) ( ) ( )⎩ >− 1t ee

( ) ( ) ( ){ }sFeatuatf as- −=−− 1 L( ) ( ) ( ){ }f


PROBLEM SET 6.4

HW 16 (b)HW: 16 (b)

66.5 .5 Convolution. Integral Equations Convolution. Integral Equations 66 55 Co o u o eg a qua o sCo o u o eg a qua o s((합성곱합성곱. . 적분방정식적분방정식))

( ) ( ) ( ) ( ) ( )( )1 , ex)

? ==

=⇒≠

gefgfgffg

t

LLLLLL -1


( ) ( ) ( ) ( ) ( )( )1 , ex)

? ==

=⇒≠

gefgfgffg

t

LLLLLL -1

Convolution (합성곱):

Properties of Convolution

( )( ) ( ) ( ) τττ dtgftgft

−=∗ ∫0

• Commutative law:

• Distributive law:

fggf ∗=∗

( ) 2121 gfgfggf ∗+∗=+∗

• Associative law:

•

( ) ( )vgfvgf ∗∗=∗∗

000 =∗=∗ ff

• 합성곱의 특이성질:

Convolution Theorem: ( ) ( ) ( )gfgf LLL =∗

ff ≠∗1



−=∗ ∫Ex. 1 Let . Find( ) ( )[ ]sassH −= 1 ( )th

⎞⎛⎞⎛

( )( ) ( ) ( )gfgf ∫0

( ) ( ) ( )gfgf LLL =∗

( ) ( )1111

11 ,1 11

==∗=⇒

=⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛

−

∫

−−−

att

aat

at

edeeth

se

as

ττ

LL

( ) ( )111 0

−=⋅=∗=⇒ ∫ ea

deeth τ



−=∗ ∫Ex. 2 Let . Find( ) ( )222

1ws

sH+

= ( )th( )( ) ( ) ( )gfgf ∫

0

( ) ( ) ( )gfgf LLL =∗



−=∗ ∫Ex. 2 Let . Find( ) ( )222

1ws

sH+

= ( )th

⎞⎛

( )( ) ( ) ( )gfgf ∫0

( ) ( ) ( )gfgf LLL =∗

( ) ∗⇒

=⎟⎠⎞

⎜⎝⎛

+−

∫ dtwtwtth

wwtws

t

)(ii1sinsin

/)(sin122

1L

( )

[ ]+−=

−=∗=⇒

∫

∫

dwwtw

dtwwwww

th

t

coscos2

1

)(sinsin

2

02

ττ

τττ

⎥⎦⎤

⎢⎣⎡ +−=

=

∫

wwwt

w

wt

r

sincos2

1

2

02

0

ττ

⎥⎦⎤

⎢⎣⎡ +−=

wwtwtt

wsincos

21 2


Ex. 4 In an undamped mass-spring system, resonance occurs if the frequency of the driving force equals the natural frequency of the system.

( ) ( ) 00'00sin'' 2+ tK ( ) ( ) 00' ,00 ,sin'' 020 ===+ yytKyy ϖϖ

( ) ( ) ( ) ( ) ( ) ( ) ( )0'0'' 0' 2 fsffsf, ffsf −−=−= LLLL


Ex. 4 In an undamped mass-spring system, resonance occurs if the frequency of the driving force equals the natural frequency of the system.

( ) ( ) 00'00sin'' 2+ tK ( ) ( ) 00' ,00 ,sin'' 020 ===+ yytKyy ϖϖ

( ) ( ) ( ) ( ) ( ) ( ) ( )0'0'' 0' 2 fsffsf, ffsf −−=−= LLLL

KY 2220ϖ

=

( ) ( )tttKty

sY

0002

220

2

sincos2

)(

ϖϖϖ

ϖ

+−=⇒

+

( ) ( )y 000202ϖ


Integral Eq ation (적분방정식)Integral Equation (적분방정식):

미지의 함수 y(t )가 적분기호 안 또는 밖에 나타나는 방정식

Ex 6 Solve the Volterra integral equation of the second kindEx. 6 Solve the Volterra integral equation of the second kind.

( ) ( ) ( ) tdtytyt

=−− ∫ τττ sin0

합성곱을 이용

0

ttyy =∗− sin

라플라스 변환 ( ) ( ) 221

11

sssYsY =

+−

32( ) ( )

6 111 3

424

2 tttysss

ssY +=⇒+=+

=


PROBLEM SET 6.5

HW 25HW: 25

66.6 Differentiation and Integration of.6 Differentiation and Integration ofT ansfo ms ODEs ith Va iable CoefficientsT ansfo ms ODEs ith Va iable CoefficientsTransforms. ODEs with Variable CoefficientsTransforms. ODEs with Variable Coefficients((변환의변환의 미분과미분과 적분적분. . 변수계수의변수계수의 상미분방정식상미분방정식))

( ) ( ) ( )∫∞

−==0

dttfefsF stL

( ) ( ) ( )∫∞

− −=−==⇒0

0

' tfdtttfedsdFsF st L

0

( )( ) ( ) ( ){ } ( )ttfsFsFttf −=−=⇒ − ' ,' 1LL


Ex 1 Derive the following three formulasEx. 1 Derive the following three formulas.

( )( ) ( )Fttf 'L ( )( ) ( )sFttf '−=L

( ) 22sinβ

ββ+

=s

tL ( ) ( )222

2sinβ

ββ+

=s

sttL미분에 의하여

β ( )β+s

( )222sin

2 ββ

β=⎟⎟

⎠

⎞⎜⎜⎝

⎛ sttL ( )222 ββ +⎠⎝ s


( ) 22cosβ

β+

=s

stL ( ) ( )( ) ( )222

22

222

222 2cosβ

β

β

ββ+

−=

+

−+−=

s

s

s

ssttL미분에 의하여

β ( ) ( )ββ ++ ss

( )( ) ( )

( )222

2222

22222

22 1sin1cos βββ

βββ

β +±−=±

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛±

ssstttL ( ) ( )22222222 ββββ

ββ

+++⎟⎠

⎜⎝ sss

( ) ( ) ( ) ( )tfsdsFsdsFtf=⎬

⎫⎨⎧

⇒=⎬⎫

⎨⎧

∫∫∞

−∞

~~~~ 1LL

Integration of Transforms (변환의 적분):

( ) ( ) ( ) ( )t

sdsFsdsFt ss

=⎭⎬

⎩⎨⇒=

⎭⎬

⎩⎨ ∫∫ LL


222 ⎞⎛Ex. 2 Find the inverse transform of .2

22

2

2ln1ln

ss

sωω +

=⎟⎟⎠

⎞⎜⎜⎝

⎛+

( )( ) 22d222 +⎞⎛ 미분 ( )( ) 222222 22lnln

ss

ssss

dsd

−+

=−+ω

ω2

22

2

2ln1ln

ss

sωω +

=⎟⎟⎠

⎞⎜⎜⎝

⎛+

미분

Case 1) 변환의 미분이용

( ) ( ) ( )( ) ( )ttftss

sssF

sfsF −=−=⎟

⎠⎞

⎜⎝⎛ −

+=⇒⎟⎟

⎠

⎞⎜⎜⎝

⎛+== −− 2cos222' 1ln 222

112

2ω

ωω

LLLsss ⎠⎝ +⎠⎝ ω

( ) ( )tt

tf ωcos12 −=∴

22Case 2) 적분이용 ( ) ( ) ( ) ( )1cos2 22 1

22 −==⇒−+

= − tGtgss

ssG ωω

L

( ) ( ) ( )tgdGω 12~~1l 12

1 ⎟⎞

⎜⎛

⎟⎞

⎜⎛

⎟⎞

⎜⎛

∫∞

LL ( ) ( ) ( )ttt

tgsdsGs s

ωω cos12~~1ln 12

1 −=−=⎟⎟⎠

⎜⎜⎝

=⎟⎟⎠

⎜⎜⎝

⎟⎟⎠

⎜⎜⎝+ ∫−− LL


Variable Coefficient (변수계수)를 가진 상미분방정식Variable Coefficient (변수계수)를 가진 상미분방정식

( ) ( )[ ]0'dsdYsYysY

dsdty −−=−−=L

( ) ( ) ( )[ ] ( )020'0'' 22 ydsdYssYysyYs

dsdty +−−=−−−=L

Ex 3 Laguerre’s Equation Laguerre PolynomialsEx. 3 Laguerre s Equation. Laguerre Polynomials.

( ) ( ), , , nnyytty 210 0'1'' ==+−+

( ) ( ) ( ) ( )22 010002 ++⇒+⎟⎞

⎜⎛+⎥

⎤⎢⎡ + YsndYs snYdYsYysYydYssY ( ) ( ) ( ) ( )

( )111

01 0002

−=⇒⎟

⎞⎜⎛ +

=−+

=⇒

=−++⇒=+⎟⎠

⎜⎝

−−−−+⎥⎦⎢⎣+−−

nsYdsnndssndY

Ysnds

s-snYds

sYysYyds

ssY

( )12

1 +=⇒⎟

⎠⎜⎝

−−

=−=⇒ nsYds

ssds

s-sY

( ) ( ) ( )⎪⎨⎧ =

== −0 1

1 de n,

Ytl ntL( ) ( ) ( )⎪⎩⎨ =

== − ,2 ,1 ,!

netdtd

neYtl tn

nn L


PROBLEM SET 6.6

HW 15HW: 15

66.7 Systems of ODEs.7 Systems of ODEs ((연립연립상미분방정식상미분방정식))66 Sys e s o O sSys e s o O s ((연립연립상미분방정식상미분방정식))Ex. 3 The mechanical system consists of two bodies of mass 1 of three

springs of the same spring constant k and of negligibly small massesof the springs Also damping is assumed to be practically zeroof the springs. Also damping is assumed to be practically zero.

( ) ( ) ( ) ( ) kyyyy 300 ,100 '2

'121 =−===

66.7 Systems of ODEs.7 Systems of ODEs ((연립연립상미분방정식상미분방정식))66 Sys e s o O sSys e s o O s ((연립연립상미분방정식상미분방정식))Ex. 3 The mechanical system consists of two bodies of mass 1 of three

springs of the same spring constant k and of negligibly small massesof the springs Also damping is assumed to be practically zeroof the springs. Also damping is assumed to be practically zero.

( ) ( ) ( ) ( ) kyyyy 300 ,100 '2

'121 =−===

( )12112 3 YYkkYksYs −+−=−−

( )1211 '' yykkyy −+−= 라플라스 변환

지배방정식:

( ) 21222 3 kYYYkksYs −−−=+−( ) 2122 '' kyyyky −−−=

Cramer의 법칙

( )( ) ( ) kskskksksY 3323 2+=

−+++=

또는 소거법 적용

( )

( )( ) ( ) kskskksksY

kskskksY

3323

32

2

222221

=−++−

=

++

+=

−+=

( ) ( )

( ) ( ) tktkYty

tktkYty

3sincos

3sincos

21

2

11

1

−==

+==

−

−

L

L 역변환

( )( ) ( )( ) kskskks

Y32

222222 +−

+=

−+=( ) ( )22

66.7 Systems of ODEs.7 Systems of ODEs ((연립연립상미분방정식상미분방정식))66 Sys e s o O sSys e s o O s ((연립연립상미분방정식상미분방정식))

PROBLEM SET 6.7

HW 22HW: 22

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