1

Chapter 8 Ordinary differential equation

Mathematical methods in the physical sciences 3rd edition Mary L. Boas

Lecture 5 Introduction of ODE

2

1. Introduction (differential equation)

- A great many applied problems involve rates, that is, derivatives. An equation containing derivatives is called a differential equation.

- If it contains partial derivatives, it is called a partial differential equation; otherwise it is called an ordinary differential equation.

ex 1) Newton’s equation

2

2

dt

dm

dt

dmm

rvaF

ex 2) Heat transfer

dx

dTkA

dt

dQ

ex 3) RLC circuit

dt

dV

C

I

dt

dIR

dt

IdLV

C

qRI

dt

dIL

2

2

RIVR

C

qVC

dt

dILVL

3

- order of a differential equation : order of the highest derivative in the equation

order 2nd :

order1st : 1

2

2

2

krdt

rdm

xyy

- (non)Linear differential equation

equationnonlinear ,, 1 ,cot

. offunction aor constant either are and Here,

equationlinear

2

3210

xyyyyyy

xba

byayayaya

n

Note 1 : A solution of a differential equation (in the variable x and y) is a relation between x and y which, if substituted into the differential equation, gives an identity. If you come up with a function to give an identity, that should be a solution of the differential equation.

Example 1) xyCxy cos (DE),equation aldifferenti theofsolution a is sin

Example 2)xxx BeAeyeyyy Solution DE

In order to verify if your solutions are correct, put the solutions into the equations and check the identity.

4

5

Example 3)Find the distance which an object falls under gravity in t seconds if it starts from rest.

solution)r (particula 2

1 ),0(condition initial With the

solution) (general 2

1

200

002

2

2

gtxxv

xtvgtxgdt

xd

Note 2 - First order DE one arbitrary integration constant (IC) - Second order DE two ICs - N-th order DE # of ICs is n

- General solution with arbitrary IC- Particular solution determined by the boundary condition or initial condition

6

Example 4)Find the solution which passes through the origin and (ln2, 3/4)

xx BeAeyyy

yy

xeey

BABABeAe

BA

xx sinh)(

.2

0

condition,given esatisfy th To

21

21

212ln2ln

43

7

2. Separable equations

- Separable equation

ex) dxxfdy )(

y terms in one side and x terms in the other side the equation is separable.

Example 1)Radioactive substance decay rate

0at for

const.ln

00

tNNeNN

tNdtN

dN

Ndt

dN

t

8

Example 2) .1 Solve yyx

axy

axaxconstxy

x

dx

y

dy

xy

y

1

lnlnln.ln1ln

1or ,

1

1

9

- Orthogonal trajectories: ex) lines of force intersect the equipotential curves at right angles.

121

)1(

s)reciprocal (negative 1

1

1

22

2212

21

y trajectororthogonal

Cyx

Cxyy

xdxdyy

y

xy

x

yy

ayaxy

10

Chapter 8 Ordinary differential equation

Mathematical methods in the physical sciences 3rd edition Mary L. Boas

Lecture 6 First order ODE

11

3. Linear first-order equations

- Linear first-order equation . of functions are , where, xQPQPyy

PdxIAeAeey

cPdxyPdxy

dy

Pydx

dyPyy

Q

IPdxcPdx

where,

ln ,

or 0

.0let First,

12

IIIIIIIatingdifferenti

I

QePyyePyeeydx

dIyeeyye

dx

d

Aye

)(

PdxIcedxQeey

cdxQeye

III

II

whereor ,

13

Example 1) ./12 Solve 2 xxyyx

PdxI

cedxQeey

cdxQeyeQPyy

III

II

whereor ,

,For

.4

1

,4

111

,1

,ln22

22

45

322

2ln2

cxx

y

cx

dxxdxxxx

yye

xeexdx

xI

I

xI

.1

,2

Here, .12

/1233

2

xQ

xP

xy

xyxxyyx

14

Example 2)Radium decays to radon which decays to polonium. If at t=0, a sample is pure radium, how much radon does it contain at time t (created and simultaneously decay)?

N_0 = # of radium atoms at t=0N_1= # of radium at time tN_2 = # of radon atoms at time t,lambda_1, lambda_2 = decay constants for Ra and Rn.

tt

t

eNQPeNNNdt

dNNN

dt

dN

eNNNdt

dN

11

1

0120111222

22112

0111

, Here,.or ,

. ,

PdxI

cedxQeey

cdxQeyeQPyy

III

II

whereor ,

,For

15

. if ,

,

2112

0101012

22

1212212

ceN

cdteNcdteeNeN

tdtI

ttttt

.

.or 0 ,0)0( Since

21

12

012

12

01

12

012

tt eeN

N

N cc

NtN

16

4. Other methods for first-order equations1) Bernoulli equation

nQyPyy

It is not linear, but, is easily reduced to a linear equation by making the change of the variable.

equationorder -firstlinear : 11

, Using

111

11

,1by equation original thegMultiplyin

1

1

1

1

QnPznz

yz

QnPynyyn

QyynPyyyn

yn

yynz

yz

n

yn

nnn

n

n

n

17

2) Exact equations; integrating factors

.or 0

).,(),( where,0 cf.

. if al,differentiexact an is ,,

dyAdxAdFF

QPAA

x

Q

y

PdyyxQdxyxP

yx

yx

rAAA

AA

.,0

. if exact, called is or 0

. ,,

),,(For

constyxFdFQdyPdx

x

Q

y

P

Q

PyQdyPdx

dFQdyPdxy

FQ

x

FP

yxF

18

factor gintegratin :1

.exact. is 01

exact.not is 0

2

22

x

constx

y

x

yddx

x

ydy

xx

ydxxdy

ydxxdy

ex. 1)

ex. 2)

factor gintegratin : where, IIII eQePyeyeQPyy

19

3) Homogeneous equations

equation. homogenous a isIt

functions homogenous :, where0,, QPdyyxQdxyxP

- The above equation can be reduced to a separable equation in variable v=y/x and x.

xyfxxfyxzxf

zyxfttztytxf

n

n

/),/ln()( ex)

),,,(),,( :function shomogeneou cf.

2

x

yf

yxQ

yxP

dx

dyy

dyyxQdxyxP

,

,

0,,

20

Prob. 8)

dxx

dvvv

v

dxvvvxdv

dxvdxvvxdv

dxvxdxxvxxvdxxdvvx

vdxdvdyvxy

dxyxxydy

1

11 , variables theSeparating

11

11

11)(

eq., original theinto thisPutting

eq.) us(homogeneo

22

22

22

2222

22

21

4) Change of variables

: If a differential equation contains some combination of the variable x, y, we try replacing this combination by a new variable.

cf. Problem 11. yuyxuyxy 1 :Hint cos

dxu

du

uyu

cos1

cos11

22

Chapter 8 Ordinary differential equation

Mathematical methods in the physical sciences 3rd edition Mary L. Boas

Lecture 7 Second order ODE I

23

5. Second order linear equations with constant coefficients and zero right-hand side

usinhomogeno :

homogenous :0

012

2

2

012

2

2

xfyadx

dya

dx

yda

yadx

dya

dx

yda

Example 1.

04114

.045or 045045 Then,

operator aldifferenti.,Let

045

22

2

22

yDDyDD

yDDyDyyDyyy

ydx

yd

dx

dy

dx

dyDy

dx

dyDy

yyy

1) Auxiliary equation

24

. is DE theofequation general The

,04 and 01

solvefirst will we this,solve To

24

1

24

1DE separable thesesolving

xx

xx

ececy

ecyecyyDyD

. if ,0))(( ofsolution general theis 21 baybDaDececy bxax

Comment. Here, we can see that solving the second-order linear differential equation (y’’+my’+ny=0) is quite similar to solving the second order normal equation (D2+mD+n=0). We know that there are three cases for the solutions of the second order equation, two real numbers, single real number, and two complex numbers. The first case of DE corresponds to the equation with the two real solutions. How about the others cases?

25

2) Equal roots of the auxiliary equation

? isother then the, issolution One

0

1axecy

yaDaD

.0))(( ofsolution general theis yaDaDeBAxy ax

BAxAdxdxAeeye

AeayyAeyaD

AeuuaDyaDaD

yaDu

axaxax

axax

ax

equation)linear order (first or ,

00

26

3) Complex conjugate roots of the auxiliary equations

- The roots of auxiliary equations are complex.

xcexcxce

BeAeeBeAeyxx

xixixxixi

sincossin 21

Example 2) .096 yyy

xeBAxyyDDyDD 32 03396

27

Example 3) motion of a mass oscillating at the end of a spring

motion harmonic simple : sincossin

0

for

21

2222

222

2

2

2

tctctcBeAey

iDyDyyD

m

ky

dy

ydky

dy

ydm

titi

‘We can determine two unknown constants using initial conditions.’

Example 4. Initial condition: at t=0, y=-10, y’=0

ty

cccc

cc

tctcy

cos10

.10,0010

,1010 condition, initial For the

sincos

21

21

21

21

28

Example 5. Considering the friction,

2222

222

2

2

2

02

)0(2 ,for ,02

)0(

bbDbDD

m

lb

m

ky

dt

dyb

dt

yd

ldt

dylky

dt

ydm

290 cf.

2

2

C

I

dt

dIR

dt

IdL

ibtcey

b

eBtAy

b

bb

bbBeAey

b

bt

bt

tt

22

22

22

22

22

22

where,sin

ify oscillatoror dunderdampe-

)(

if damped critically-

where,

if doverdamtpe-

30

Undamped Underdamped Envelope

- Underdamped oscillator

31

- Critically/over-damped

32

REVIEW

homogenous :0012

2

2 yadx

dya

dx

yda

. if ,0))(( ofsolution general theis 21 baybDaDececy bxax

.0))(( ofsolution general theis yaDaDeBAxy ax

., where,0))(( ofsolution theis

sincossin 21

ibiabDaD

xcexcxce

BeAeeBeAeyxx

xixixxixi

0)(00

,

212

212

212

2

2

22

yaDaDyaDyayDyaDyayDa

ydx

yd

dx

dy

dx

dyDy

dx

dyDy

33

Chapter 8 Ordinary differential equation

Mathematical methods in the physical sciences 3rd edition Mary L. Boas

Lecture 8 Second order ODE II

34

6. Second-order linear equations with constant coefficient and right-hand side not zero

equation nsinhomogeou ofsolution particular : 2sin

equation) homogenous ofsolution (generalary complement :

2cos45

101

4

2

xy

BeAey

xyDD

p

xxc

“The general solution of an inhomogeneous DE is the combination of y_c and y_p”

1) solution of an inhomogeneous DE

xBeAeyyy

xyyDDyDD

xyDD

xxcp

cp

c

p

2sin

2cos45045

2cos45

1014

2

2

2

35

2) Inspection of particular solutions

xxp

x

p

eAeyeyyy

yyyy

2896

532 35

: To find a simple particular solution, we may be able to guess and verify it.

simple.not ?? 2 xp

x Aeyeyyy

36

3) Successive integration of two first-order equations

simple.not 2 xeyyy

xx

x

euueuDyDu

eyDD

or 12

21

xxxxx

xxxxxxx

ececxeycecxe

cecexedxecxeeye

xdxI

2213

12

31

331

23

1313

913

31

122

22

DEorder first theofn integratio second

.2or 2

.,

DEorder first theofn integratiofirst

11

11

xxxx

xxxxx

ecxeyyecxeyD

ecxeucxdxeeuexdxI

37

4) Exponential right-hand side

cxcx kexFybDaDkexFyadx

dya

dx

yd 012

2

First, suppose that c is not equal to either a or b. Solving the DE by the successive integration of two first-order equation gives the particular solution, ecx.

ex)

xxx

xxxxppp

xp

x

eBeAey

CeeeeCyyy

Cey

eyDD

25

2222

2

2

1758454

2) (1,-5 751

bac if

b)(a bor a equals c if

bor aeither toequalnot is c if

2 cx

cx

cx

cx

eCx

Cxe

Ce

keybDaD

3/1 bor a c 2 CCxeyeyyy xp

x

Backing to the previous DE,

38

5) Use of complex exponentials

ixexxxF cosor sin

xeYYYxeYYY

eYYYxyyyix

IIIix

RRR

ix

2sin44Im2 ,2cos44Re2

422sin4222

2

part) (imaginary 2sin2cos

want.esolution w theispart imaginary theHere,

)2sin2(cos33

340

624

62

4

4224

53

51

512

51

51

22

2

xxy

xixieiY

ii

iC

eCei

CeY

p

ixp

ixix

ixp

39

solution. theaspart imaginary and real thee then tak

, solvefirst

,cos

sin solve To

xikeybDaD

xk

xkybDaD

40

6) Method of undetermined coefficients

The method of assuming an exponential solution and determining the constant factor C is an example of the method of undetermined coefficients.

t.coefficien edundetermin with as degree same theof lpolynomina theis

bac if

b)(a bor a equals c if

bor aeither toequalnot is c if

2

nn

ncx

ncx

ncx

ncx

PQ

xQeCx

xQCxe

xQCe

xPeybDaD

Example)

1222222

2 Solve

221

22

2

2

xy

xxCBxAxBAxAyyy

CBxAxy

xxyyy

p

ppp

p

41

7) Several terms on the right-hand side; principle of superposition

12sin2cos

121

2sin2cos2sin421

21

2sin4212

221

53

51

31

321

221

32

53

51

2

31

1

2

xxxxeyyyy

xyxxyDD

xxyxyDD

xeyeyDD

xxxeyDDyyy

xpppp

p

p

xp

x

x

- Solve a separate equation and add the solutions. principle of superposition (working only for linear equations)

42

8) Forced vibrations (steady state motion)

titi

tip

ti

FeCebi

CeYFeYdt

dYb

dt

Yd

constFtFydt

dyb

dt

yd

22

22

2

22

2

2

2

.)( sin2

.sin

44

of angle ,4

4

2

2

22222222

2222

2222

22

22

tb

Fye

b

FY

Cb

FC

reb

Fbi

bi

FC

pti

p

i

43

222p

p

2at occurs y of maximum the,Given 2)

at occurs y of maximum the,Given )1

b

.sin

4 2222

t

b

Fyp

9) Resonance

44

10) Use of Fourier series in Finding particular solutions

ion.superposit of principle theuse then and ,012

2

2

012

2

2

inxn

n

inxn

ecyadx

dya

dx

yda

ecxfyadx

dya

dx

yda

Example)

2 0,

0 1, where,102

2

2

t

ttftfy

dt

dy

dt

yd

tBtAey

iDDDt

c 3sin3cos

310102

equation,auxiliary For 2

45

.1

2

1

expansion seriesFourier

3331 itititit eeee

itf

222

2

2

2

2

410

2101

210

11

1102 :First term)1

kk

ikk

ikikkikCCey

eik

ydt

dy

dt

yd

ikt

ikt

.3cos63sin373

2cos2sin9

85

2

20

1

237

12

237

1

3

2

285

4

285

92

20

1

37

61

3

1

37

61

3

1

85

291

85

291

20

1

3333

33

tttt

ee

i

eeee

i

ee

ei

ie

i

ie

i

ie

i

iy

itititititititit

ititititp

46

PROBLEM

5-38. Solve the RLC circuit equation with V=0. Write the conditions and solutions for overdamped, critically damped, and underdamped electrical oscillations.

6-11 & 6-25xexyDD

xyyy

216)1)(3(

4cos100102

6-42

.01 ,0

,10 ,9

x

xxyy

47

7. Other second-order equations

DE)order (first )( ,

)( missing,y variabledependent :(a) Case

12

12

xfpapapypy

xfyaya

y) t variableindependen as p with DEorder (first ,

missing, x t variableindependen :b Case

dy

dpp

dx

dy

dy

dp

dx

dpypy

48

Example 1.

equation) (Bernoulli .4

1

2

1or 024

,

missing.) is(t 024 :case special

0 0

12

2

2

2

2

2

2

2

2

yppdy

dpyp

dy

dpp

dy

dpp

dt

ydp

dt

dy

ydt

dy

dt

yd

lkydt

dyl

dt

ydm

y

yyy

ceyz

cyedyyeze

yzdy

dz

dy

dpp

dy

dzpz

1

1

equation)linear order (first 2,

21

21

21

212

(plus or minus sign much be chosen correctly at each stage of the motion so that the retarding force opposes the motion.)

(describe the motion…)

49

.

0or 0:by multiply this,solve To

0 (c) Case

221 constdyyfy

dyyfydyyyfyyy

yfy

Example)

energy) ofion (conservat .2

1

or

2

2

2

constdxxFmv

dxxFmvdvdt

dxxF

dt

dvmv

xFdt

xdm

50

DE) linear''order (second

and

equation)Cauchy or (Euler )( :d Case

012

2

2

2

2

2

22

012

22

2

z

z

efyadz

dya

dz

dy

dz

yda

dz

dy

dz

yd

dx

ydx

dz

dy

dx

dyx

ex

xfyadx

dyxa

dx

ydxa

51

Chapter 8 Ordinary differential equation

Mathematical methods in the physical sciences 3rd edition Mary L. Boas

Lecture 9 Laplace transform

52

8. The Laplace transform

.0

pFdtetffL pt

Example 1. f(t)=1

.0,11

10

0

p

pe

pdtepF ptpt

Example 2. f(t)=e^(-at)

.0Re,1

00

apap

dtedteepF tpaptat

cf. Laplace transform are useful in solving differential equations.

- Laplace transform

53

.

)1

00

0

gLfLdtetgdtetf

dtetgtftgtfL

ptpt

pt

- Some properties of Laplace transform

.)200

fcLdtetfcdtetcftcfL ptpt

54

Example 3. Let us verify L3. L(cos at)

iaaatatetf iat sincos

.0Re,1

L Start with00

ap

apdtedteee tpaptatat

.0Re,sincos

.sincossincos

.0Re,1

, transformLaplace Taking

2222

2222

22

iapap

ai

ap

patiatL

ap

ai

ap

patiLatLatiatL

iapap

iap

iappFeL iat

4:2sin2sincossincos

3:2cos2sincossincos

results, above theUsing

22

22

Lap

aiatiLatiatLatiatL

Lap

patLatiatLatiatL

55

Example 4. Let us verify L11. L(t sin at)

11:.

2sin

2sin

, respect towith relation above theateDifferenti

.coscos

22202220

220

Lap

padtatteor

ap

apdtatte

a

ap

pdtateatL

ptpt

pt

56

9. Solution of differential equations by Laplace transforms

.Here.0 0

000

YyLypYypLy

dtetyptyedtetyyL ptptpt

- Laplace transforms can reduce an linear DE to an algebraic equation and so simply solving it. Also since Laplace transforms automatically use given values of initial conditions, we find immediately a desired particular solution.

Note) The relations already include the initial conditions.

.000

result, above theUsing

0022 ypyYpypyyLpyypLyL

57

Example 1.

.0,044 0022 yyforetyyy t

).6(

2

244

equation, in the each term of s transformLaplace theTaking 1)

322

0002 L

petLYypYypyYp t

.

2

2

2

244

condition, initial the Using2)

532

pY

pYpp

.12!4

2

).6( m transforinverse the Using3)

2424 tt etety

L

58

.0,102sin4 00 yyfortyyExample 2.

.2cos4

12sin

8

12cos102cos22sin

8

12cos10

),17,4( m transforinverse theUsing

.4

2

4

10

4

2104

condition, initial theUsing

.4

22sin4

equation, in the each term of s transformLaplace theTaking

22222

2002

tttttttty

LL

pp

pY

ppYp

ptLYypyYp

59

Example 4. .0,1,02,02 00 zyforzyzzyy

.02,02

)( equation, in the each term of s transformLaplace theTaking

00

ZYzpZZYypY

ZzL

.02,12

condition, initial theUsing

ZpYZYp

.cos),14( m transforinverse theUsing

12

2212)1

2

2

2

teyL

p

pYorpYp

t

.02 1), ofresult ith theequation w original theusecan ely weAlternativ

.sin12

1,)2 2

2

zyy

tezp

ZSimiarly t

60

10. Convolution

- Method to write a formula for y

Example 1.

8,7 ex. function, some of LT11

.1

.0),(

2

22

00

LLbpapACBpAp

pT

pFCBpAp

YpFfLCYBpYYAp

yytfCyyByA

In this case, y is the inverse Fourier Transform of a product of two functions whose inverse transforms we know.

61

Example 2.

.

23

1

23

.0,23

22

2

00

pHpGeLeeLeLpp

Y

eLYpYYp

yyeyyy

tttt

t

t

.1

1

example, In this

2

00

0

2

0

ttttt

tttt

ttt

eeteete

eedee

deeedthgy

t

t

dhtgy

yLYdhtgLpHpG

0

0

62

- Fourier Transform of a Convolution

xfxfgg TransformFourier2121 ,,

duufuxfff

dxduufuxfe

dvdxuvxdxduufuxfe

dvduufvfe

dueufevfgg

xi

xi

uvi

uivi

2121

0 0 21

2

0 0 21

2

0 0 21

2

2121

* :nConvolutio

.2

1

,2

1

2

1

2

1

2

1

63

ansforms.Fourier tr ofpair a:&*

ransforms.Fourier t ofpair a:*2

1&

.* of ransformFourier t2

1*

2

1

2

1

2121

2121

212121

ffgg

ffgg

ffdxeffgg xi

64

11. The Dirac delta function

- Impulse: impulsive force f(t), t=t_0 to t=t_1

01

1

0

1

0

1

0

)( vvmmdvdtdt

dvmdttf

v

v

t

t

t

t

- We are not interested in the shape of f(t). What we think important is the value of the integration of f(t) during t_1 – t_0 = t.

65

- The above functions have the same integration value, 1.

function' Delta' called is ,0 of caseIn tft

66

- Laplace Transform of a Function

Figure in the :Prove

otherwise.,0

,

0000

0

0

ntdttttdtttt

btatdtttt

b

a

b

a

b

a

67

- Example 2.

.0,0

aedteatatL papt

- Example 3.

).28,3(,sin1

.0,

0022

022

0002

0

0

LLttttyp

eY

ettLYp

yyttyy

pt

pt

68

- Fourier Transform of a Function

mechanics quantumin useful.2

1

.2

1

2

10

deax

edxexg

axi

aixi

69

- Another physical application of functions

axqaxm , charge)(or massPoint

Example 4. 2 at x=3, -5 at x=7, and 3 at x=-4

437532 xxx

70

- Derivative of the function

.1

,

.

adxaxx

Similarly

adxaxxaxxdxaxx

nnn

71

- Some formulas involving functions

..

,0

,1.

axaxub

ax

axaxua

.00

;,1

.

;0,1

.

;.

;.

iii i

i xfandxfifxf

xxxf

babxaxba

bxaxd

axa

axc

xaaxandxxb

xaaxandxxa

72

- functions in 2 or 3 dimensions

00003

0

000

00

,,,,

,,,

zzyyxx

zyxdxdydzzzyyxxzyx

yxdxdyyyxxzyx

ooo

oo

rrrr

Spherical coord.

Cylindrical coord.

000000

0

,,,,

,,

zrfdr

zzrrzrf

dzrf

rr

0002000

0

,,sin

,,

,,

rfdr

rrrf

drf

rr

73

re

re

411

:4

22

2

rrr

r

r

r

rD

eee

qcf

ddrr

dr

dr rsurfaceclosed

r

volume

r

.

.4sin12

0 0

2222

cf. divergence theorem

74

12. A brief introduction to Green Functions

- Example 1

tftdtftttdtfttGdt

d

tdtfttGdt

dy

dt

dyy

tdtfttGty

ttttGttGdt

d

tdtttftf

yytfyy

00

22

2

0

22

22

2

22

0

22

2

0

002

,

,

,

.,,

.

0,