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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,