DIFFERENTIAL EQUATIONS (DE)
DIFFERENTIAL EQUATIONS(DE)
DEFINITION: differential equation
An equation containing the derivative of one or more dependent variables, with respect to one or more independent variables is said to be a differential equation (DE).
Definitions and Terminology
Differential Equations
A differential equation is an algebraic equation that contains some derivatives:
037352
2
ydx
dy
dx
ydty
dt
dy
• Recall that a derivative indicates a change in a dependent variable with respect to an independent variable.
• In these two examples, y is the dependent variable and t and x are the independent variables, respectively.
Why study differential equations?
• Many descriptions of natural phenomena are relationships (equations) involving the rates at which things happen (derivatives).
• Equations containing derivatives are called differential equations.
• Ergo, to investigate problems in many fields of science and technology, we need to know something about differential equations.
Why study differential equations?
• Some examples of fields using differentialequations in their analysis include:
— Solid mechanics & motion— heat transfer & energy balances— vibrational dynamics & seismology— aerodynamics & fluid dynamics— electronics & circuit design— population dynamics & biological systems— climatology and environmental analysis— options trading & economics
Examples of Fields Using Differential Equations in Their Analysis
CoolBath
HotBath
kydt
dyckx
dt
dxc
dt
xdm
2
2
02
2
kxdt
dxc
dt
xdm
ikkxdt
dxc
dt
xdm f
2
2
ghACdt
dhA od 2
70 Hkdt
dH
Recall Calculus
Definition of a Derivative
If , the derivative of or
With respect to is defined as
The derivative is also denoted by or
Definitions and Terminology
)(xfy y )(xfx
h
xfhxf
dx
dyh
)()(lim
0
dx
dfy ,' )(' xf
Recall the Exponential function
dependent variable: y
independent variable: x
Definitions and Terminology
xexfy 2)(
yedx
xde
dx
ed
dx
dy xxx
22)2()( 22
2
Differential Equation: Equations that involve dependent variables and their derivatives with respect to the independentvariables.
Differential Equations are classified bytype, order and linearity.
Definitions and Terminology
Differential Equations are classified bytype, order and linearity.
TYPE
There are two main types of differential equation: “ordinary” and “partial”.
Definitions and Terminology
Ordinary differential equation (ODE) Differential equations that involve only ONE independent variable are called ordinary differential equations.
Examples:
, , and
only ordinary (or total ) derivatives
Definitions and Terminology
xeydx
dy5 06
2
2
ydx
dy
dx
yd yxdt
dy
dt
dx 2
Definition:
A differential equation is an equation containing an unknown function and its derivatives.
32 xdx
dy
032
2
aydx
dy
dx
yd
364
3
3
y
dx
dy
dx
yd
Examples:.
y is dependent variable and x is independent variable, and these are ordinary differential equations
1.
2.
3.
ordinary differential equations
Partial differential equation (PDE)Differential equations that involve two or more independent variables are called partial differential equations.Examples:
and
only partial derivatives
Definitions and Terminology
t
u
t
u
x
u
22
2
2
2
x
v
y
u
Partial Differential Equation
Examples: 02
2
2
2
y
u
x
u
04
4
4
4
t
u
x
u
t
u
t
u
x
u
2
2
2
2
u is dependent variable and x and y are independent variables, and is partial differential equation.
u is dependent variable and x and t are independent variables
1.
2.
3.
Derivatives
Derivatives
dx
dy
Partial Derivatives
u is a function of more than one
independent variable
Ordinary Derivatives
y is a function of one independent variable
y
u
Differential Equations
• Heat transfer
• Mass transfer
• Conservation of momentum, thermal energy or mass
dz
TCd
A
q pz)(
dz
dCDJ A
ABAZ
Rzt
2
2
ODE
PDE
ORDER
The order of a differential equation is the order of the highest derivative found in the DE.
second order first order
Definitions and Terminology
xeydx
dy
dx
yd
45
3
2
2
Definitions and Terminology
xeyxy 2'
3'' xy
0),,( ' yyxFfirst order
second order 0),,,( ''' yyyxF
Written in differential form: 0),(),( dyyxNdxyxM
LINEAR or NONLINEAR
An n-th order differential equation is said to be linear if the function
is linear in the variables
there are no multiplications among dependent variables and their derivatives. All coefficients are functions of independent variables.
A nonlinear ODE is one that is not linear, i.e. does not have the above form.
Definitions and Terminology
)1(' ,..., nyyy
)()()(...)()( 011
1
1 xgyxadx
dyxa
dx
ydxa
dx
ydxa
n
n
nn
n
n
0),......,,( )(' nyyyxF
LINEAR or NONLINEAR
or
linear first-order ordinary differential equation
linear second-order ordinary differential equation
linear third-order ordinary differential equation
Definitions and Terminology
0)(4 xydx
dyx
02 ''' yyy
04)( xdydxxy
xeydx
dyx
dx
yd 53
3
3
LINEAR or NONLINEAR
coefficient depends on y
nonlinear first-order ordinary differential equation
nonlinear function of y
nonlinear second-order ordinary differential equation
power not 1
nonlinear fourth-order ordinary differential equation
Definitions and Terminology
0)sin(2
2
ydx
yd
xeyyy 2)1( '
024
4
ydx
yd
xeydx
dy
dx
yd x cos44)( 23
3
ordinary differential equation :only one independent variable involved: x
)(2
2
2
2
2
2
z
T
y
T
x
Tk
TC p
partial differential equation: more than one independent variable involved: x, y, z,
Differential Equation
An equation relating a dependent variable to one or more independent variables by means of its differential coefficients with respect to the independent variables is called a “differential equation”.
xeydx
dy
dx
yd x cos44)( 23
3
3rd order O.D.E.
1st degree O.D.E.
Order and Degree
The order of a differential equation is equal to the order of the highest differential coefficient that it contains. The degree of a differential equation is the highest power of the highest order differential coefficient that the equation contains after it has beenrationalized.
xeydx
dy x cos44)( 2 1st order O.D.E.
2nd degree O.D.E.
xeydx
dy
dx
yd x cos44)( 23
3
product between two derivatives ---- non-linear
xydx
dycos4 2
product between the dependent variable themselves ---- non-linear
Linear or Non-Linear
Differential equations are said to be non-linear if any products exist between the dependent variable and its derivatives, or between thederivatives themselves.
Terminologies:1) Ordinary vs. partial differential equationsAn ordinary differential equation (ODE) involves only one independent variable and contains only total differentials.
A partial differential equation (PDE) involves more than one independent variable and contains partial differentials.
2) Linear vs. nonlinear differential equationsA linear differential equation contains only terms that are linear in the dependent variable or its derivatives.
A nonlinear differential equation contains nonlinear function of the dependent variable.
First order differential equations
Differential equations
xxyxdx
xdy4)(
)( 2
0),(),(
5),(
yxy
yxxy
x
yx
xxyxdx
xdy4)(
)( 2
ydx
xdyx
dx
xydxy
dx
xdyx
dx
xydxyx
dx
xydsin
)()( ,0)(
)()( ,0)(
)( 22
22
2
222
2
2
3) Order of a differential equationThe order of a differential equation is determined by the highest derivative in the equation.
4) Homogeneous vs. inhomogeneous differential equationsA linear differential equation is homogeneous if every term contains the dependent variable or its derivatives.
• A homogeneous differential equation can be written as where L is a linear differential operator.
• A homogeneous differential equation always has a trivial solution y(x) = 0.• Superposition principle: If y1(x) and y2(x) are solutions to a linear homogeneous
differential equation, then ay1(x) +by2(x) is a solution to the same equation.
order second ,0)(4)()(
orderfirst ,4)()(
22
2
2
xxydx
xdyx
dx
xyd
xxyxdx
xdy
0)(4)()( 2
2
2
xxydx
xdyx
dx
xyd
,0)( xyL
xdx
dx
dx
d4 e.g., 2
2
2
L
An inhomogeneous differential equation has at least one term that contains no dependent variable.
• Inhomogeneous equations are often called “driven” equations. They describe the response of the system to an external force.
• The general solution to a linear inhomogeneous differential equation can be written as the
sum of two parts:
Here yh(x) is the general solution of the corresponding homogeneous equation, and yp(x) is any particular solution of the inhomogeneous equation.
xxyxdx
xdy4)(
)( 2
tAxxx sin20
)()()( xyxyxy ph
Solutions of ODEs
DEFINITION: solution of an ODEAny function , defined on an interval I and possessing at
least n derivatives that are continuous
on I, which when substituted into an n-th order ODE reduces the equation to an identity, is said to be a solution of the equation on the interval.
(Zill, Definition 1.1, page 8).
Definitions and Terminology
Namely, a solution of an n-th order ODE is a function which possesses at least n
derivatives and for which
for all x in I
We say that satisfies the differential equation on I.
Definitions and Terminology
0))(),(),(,( )(' xxxxF n
Verification of a solution by substitution
Example:
left hand side:
right-hand side: 0
The DE possesses the constant y=0 trivial solution
Definitions and Terminology
xxxx exeyexey 2, '''
xxeyyyy ;02 '''
0)(2)2(2 ''' xxxxx xeexeexeyyy
ODE
• Definition
• Example
• A 3rd order differential equation for r = r(t)• Solution
independent
dependent
0)(),...,('),(, )( ttttf n (4.4)
4'''''' 2 tet (4.5)
kQdt
dQ (4.6)
tcetQ kt ,)( (4.7)
Linear Equation (1)
1. Rewrite 4.9
2. Determine
where m(t) is called an integrating factor
)()()()(...)()( 0)0(
1'
2)1(
1)( tgtatatatata n
nn
n (4.8)
)()()(')( 012 tgtatata (4.9)
0)(;)(
)()(
)(
)(' 2
2
0
2
1
tata
tatg
ta
ta for all t (4.10)
dt
ta
tat
)(
)(exp)(
2
1 (4.11)
Linear Equation (2)
3. Multiply both sides of equation 4.10 by m(t)
Observe that the left-hand side of eqn 4.12 can be written as
or
)(tdt
d
)()(
)()()(
)(
)('
2
0
2
1 tta
tatgt
ta
ta
(4.12)
)()(
)(exp
)(
)(
)(
)(exp'
2
1
2
1
2
1 tdt
ddt
ta
ta
ta
tadt
ta
ta
(4.13)
Linear Equation (3)
Equation (4.12) can be rephrase as:
4. Integrate both sides of Equation (4.14) with respect to the independent variable:
dt
ta
ta
ta
tatgdt
ta
ta
dt
d
)(
)(exp
)(
)()(
)(
)(exp
2
1
2
0
2
1
cdtdtta
ta
ta
tatgdt
ta
ta
)(
)(exp
)(
)()(
)(
)(exp
2
1
2
0
2
1
(4.14)
dt
ta
tacdtdt
ta
ta
ta
tatgdt
ta
tat
)(
)(exp
)(
)(exp
)(
)()(
)(
)(exp)(
2
1
2
1
2
0
2
1 (4.15)
where c is the constant of integration
Higher ODE Reduces to 1st Order2
2( ) ( )
Define , we have
( ) ( )
d y dyq x r x
dx dxdy
zdx
dyz
dxdz
r x q x zdx
2
1 2 3
12
23
231 3 2 1
'''( ) ( ) ''( ) 2( '( )) ( ) 0
Define , ', '', we have
2( )
y x y x y x y x y x
y y y y y y
dyy
dxdy
ydxdy
y y y ydx
In general, it is sufficient to solve first-order ordinary differential equations of the form
1( , , , ), 1,2, ,ii N
dyf x y y i N
dx
• Nonlinear equations can be reduced to linear ones by a substitution. Example:
y’ + p(x)y = q(x)yn
and if n ¹ 0,1 then
n(x) = y1-n(x)
reduces the above equation to a linear equation.
(4.16)
(4.17)
First Order Differential Equations
No general method of solutions of 1st O.D.E.s because of theirdifferent degrees of complexity.
Possible to classify them as:-exact equations
-equations in which the variables can be separated
-homogenous equations
-equations solvable by an integrating factor
First order linear differential equations occasionally arise in chemical engineering problems in the field of heat transfer,momentum transfer and mass transfer.
First Order Ordinary Differential equation
38
(1) The order of ODE: the order of the highest derivative
e.g.,
First-order ordinary differential equation
order) (second order),(first 2
2
dx
yd
dx
dy
(2) The degree of ODE: After the equation has been rationalized, the power of
the highest-order derivative.
e.g.,
ODE degree second andorder third the
)( and )()(0)( 23
3
3
332/322/3
3
3
dx
yd
dx
yd
dx
dy
dx
dyyx
dx
dyx
dx
yd
(3) The general solution of ODE contains constants of integration, that may
be determined by the boundary condition.
(4) Particular solution: The general solution contains the constants which are
found by the boundary condition.
(5) Singular solution: Solutions contain no arbitrary constants and cannot be
found from the general solution.
with n parameters satisfies an nth-order ODE in general. The boundary conditions on the solutions determine the parameters.
xaxay cossin 21
First-order ordinary differential equation
General form of solution
),.....,,,,( 321 naaaaxfy
Ex: Consider the group of functions
equationorder -second 0
cossin
sincos
2
2
212
2
21
ydx
yd
xaxadx
yd
xaxadx
dy
First-order ordinary differential equation
Exact equations
dy
ydFdxyxA
yyxB
y
yxU
yFyFdxyxAyxU
cyxU
yxdUx
yxB
y
yxA
y
yxU
xx
yxU
y
y
yxUyxB
x
yxUyxA
dyy
yxUdx
x
yxUyxdUdyyxBdxyxA
yxU
)(]),([),(
),(
by determined be can )( and )(),(),(
ODE of solution the is ),( so,
0),(),(),(
)),(
()),(
(
),(),( and
),(),(
),(),(),(),(),(
satisfies which ),( function aFor
First-order ordinary differential equation
solution the is 2
3
2
3
)(0
)(2
3)()3(),(
exact is equation the1 ,1
),( ,3),(0)3(
03 :Ex
2
12
2
2
1
2
1
cxyx
ccyxx
cyFdy
dFx
dy
dFx
cyFyxx
cyFdxyxyxU
x
B
y
A
xyxByxyxAxdydxyx
yxdx
dyx
Exact differential equations
0),(),( dyyxQdxyxP
A first-order ordinary differential equation can be generally written as
The differential is exact if we can find j (x, y) so that
).,( ),,(or ,0),(),( yxQy
yxPx
dyyxQdxyxPdyy
dxx
d
The solution of the differential equation is thus simply
The condition for the differential to be exact is
.)(),()(),(
)(),()(),(),(
yx
yx
CxgdttxQyfdtytP
xgdyyxQyfdxyxPyx
). ng(consideri . 2
yxx
Q
y
P
.2)(4)(23
exact. is aldifferenti the,4 Since
4
23
:1 Example
2322
22
Cxyxxgdyxyyfdxyx
yx
Q
y
P
xy
yx
dx
dy
yx
0),(),( dyyxNdxyxM
dFdyy
Fdx
x
F
General solution: F (x,y) = C
For example
Exact Equations
Exact?
0 ydx
dyx Cxy
x
N
y
M
Exact Equations Example
0)2(cossin3 dx
dyyxxyx
dyy
Fdx
x
FdF
xyxx
Fsin3
yxy
F2cos
)(cos4
1 4 yfxyxF
)('cos yfxy
F
')( 2 Cyyf Cyxyx 24 4cos4
First-order differential equations
A first-order ordinary differential equation can be generally written as
)(
)(
yQ
xP
dx
dy
Separable variables
If the equation has the form , then
.0),(0)()(
0)()(
yxfdyyQdxxP
dyyQdxxPyx
),(
),(
yxQ
yxP
dx
dy
cxyydyxdx
y
x
dx
dy
yx
sin20cos
cos
:1 Example
2
First-order ordinary differential equation
First-degree first-order equation
0),(),(or ),( dyyxBdxyxAyxFdx
dy
Separable-variable equation
dxxfyg
dyygxf
dx
dy)(
)()()(
1)2
exp(
)2
exp()2
exp(1
2)1ln(
1
)1( :Ex
2
22
2
xAy
xAC
xy
Cx
yxdxy
dy
yxxyxdx
dy
For example
xdx
dyy sin
Separable-variables Equations
In the most simple first order differential equations, the independent variable and its differential can be separated from the dependent variable and its differential by the equality sign, using nothing morethan the normal processes of elementary algebra.
xdxydy sin Cxy cos2
1 2
Example
1
93'
yx
yxySolve mYynXx �,Let�
)1(
)93(3
1)()(
9)()(3'
mnYX
mnYX
mYnX
mYnXY
01
093
mn
mn
3
2
m
nXY
XY
YX
YXY
/1
/33'
XYv /Let �
Xv'vY' v
vXv'v
1
3
X
dXdv
vv
v
223
1C'Xvv lnln)32ln(
2
1 2
C)3()3)(2(2)2(3 22 yyxx2
3
x
yv
First-order ordinary differential equation
Homogeneous equations
x
dx
vvF
dv
vvFdx
dvxvF
dx
dvxv
dx
dy
vxy
x,yfyxf
yxf
yxBxyyxA
yxByxA
x
yF
yxB
yxA
dx
dy
n
)(
)()(
onsubstituti the Making
)(),( obeys
it , anyfor If n. degree shomogeneou of function a is ),( (2)
degree. third the with and e.g., degree,
same the of functions shomogeneou ),( and ),( Where(1)
)(),(
),(
3322
x
yf
dx
dy
is termed a homogeneous differential equation of the first order.
Homogeneous Equations
Homogeneous/nearly homogeneous?
A differential equation of the type,
Such an equation can be solved by making the substitution y=vx and thereafter integrating the transformed equation.
)(vfdx
dvxv
dx
dy
Cvvf
dvx
)(ln
First-order ordinary differential equation
)(sin)sin()ln()ln(sin
ln)ln(sinsin
cos
lncot
)tan(set
)tan( :Ex
1
12
1
AxxyAxx
yAx
x
y
cxcvdvv
v
cxx
dxvdv
vvdx
dvxv
dx
dyv
x
y
x
y
x
y
dx
dy
QPydx
dy
where P and Q are functions of x only
Assuming the integrating factor R is a function of x only, then
dx
dRy
dx
dyRRy
dx
dRQyRP
dx
dyR )(
PdxR exp is the integrating factor
Equations Solved by Integrating Factor
There exists a factor by which the equation can be multiplied so that the one side becomes a complete differential equation. Thefactor is called “the integrating factor”.
PRdx
dR
RQdx
Ryd
)( dxxPdxxPdxxP dxxQy )()()( Cee)(e
Linear first order ODEs:
The general form of a linear first order ODE is
Linear first order ODEs
).()( xqyxpdx
dy
First-order differential equations
Cdssqdttpdttpx
Cdttqtxy
dttqtxg
yf
xgyxxgdtx
yfdttqtydttptyfdttqytptC
dttpCxdxxpd
xpxxqyxpxdy
d
dx
d
dyxdxxqyxpxyxdx
dydxxqyxpxqyxpdx
dy
x sx
x
x
y
x xx
x
)()(exp)(exp)(
)()()(
)()()(
0)(
)()()()(
)()()()()()()()()(
)(exp)()()()()()()(
0)()()()(),( then ,factor ngintergrati thebe )(Let
0)()()()(
2
2
1
All linear first order ODEs have a formulated solution.
2
322
2
2
23
5
1)(
.)ln2exp(2
exp)(
2 ,2
:1 Example
x
CxCdsss
xxy
xxdtt
x
xyxdx
dyxy
dx
dyx
x
x
.2
1)(
.exp)(
:2 Example
xxx ssx
xx
x
CeeCdseeexy
edtx
eydx
dy
.)1ln(1
1)(
.11
exp)(
1
:3 Example
CxxCdss
s
sxxy
xdt
tx
x
x
x
y
dx
dy
x
x
Cdssqdttpdttpxyxqyxp
dx
dy x sx)()(exp)(exp)()()(
Solve
2
3exp
24 x
ydx
dyxy
Let z = 1/y3
4
3
ydy
dz
dx
dy
ydx
dz4
3
2
3exp33
2x
dx
dzxz
integral factor
2
3exp3exp
2xxdx
32
3exp
2
3exp3
22
x
dx
dzxxz
32
3exp
2
xz
dx
dCx
xz
3
2
3exp
2
Cxx
y
3
2
3exp
1 2
3
Example
Example
0cossin ydyydxSolve yN
yM
cos
sin
x
N
y
M
integral factor = F(x))cos()sin( yF
xyF
y
dx
dFyyF coscos
dxF
dF
xeF
0cossin ydyydx 0cosesine ydyydx xx
dyy
F'dx
x
F'dF'
ye
x
F' x sin
yey
F' x cos
CyeF x sin'
0),(),( dyyxNdxyxM)(xF
0),()(),()( dyyxNxFdxyxMxF
)()( FNx
FMy
dx
dFN
x
NF
y
MF
dx
dFN
x
N
y
MF
F
dFdx
x
N
y
M
N
1
)(function1
xx
N
y
M
N
Equations Solved by Integrating Factor
0),(),( dyyxNdxyxM)(yG
0),()(),()( dyyxNyGdxyxMyG
)()( GNx
GMy
x
NG
dy
dGM
y
MG
dy
dGM
y
M
x
NG
G
dGdy
y
M
x
N
M
1
)(function1
yy
M
x
N
M
Equations Solved by Integrating Factor
Example
0)63()6( 22 dyxxydxxyySolve
2
2
63
6
xxyN
xyyM
xyx
N
xyy
M
123
62
x
N
y
M
yxyy
xy
y
xyxy
y
M
x
N
M
1
)6(
6
xy6
)62()123(12
G
dGdy
y
1yyG )(
0)63()6( 22 dyxxydxxyy 0)63()6( 22 dyxxyydxxyyy
dyy
F'dx
x
F'dF'
)6( xyyyx
F' 2
)(3)()6(' 223 yfyxxyyfdxxyyyF 2
dy
ydfyxxy
y
F' )(63 22
yxxyy
F' 2 263
C)(0)(
yfdy
ydfCyxxy 223 3
Example
First-order ordinary differential equation
Bernoulli’s equation
ODElinear )()1()()1(
)()1()()1(
)()()1
(
1)1(
linear nonlinear onsubstituti a make
1or 1 where )()(
11
1
xQnvxPndx
dvv
yxQnyxPn
dx
dv
v
yxQyxP
dx
dv
n
y
dx
dv
n
y
dx
dy
dx
dyyn
dx
dv
yv
nnyxQyxPdx
dy
nn
n
nn
n
n
343
33333
13
3
33
33
434
44341
43
6 is solution the
6)6(1
)1
()()()(
1)(
1}
3exp{})(exp{)(..
6)( ,3
)( with ODElinear 63
23
12
3
33let
2 :Ex
cxxy
ycxxdxxxx
dxxQxx
xv
xdx
xdxxPxFI
xxQx
xPxvxdx
dv
xx
y
dx
dvyx
x
y
dx
dvy
dx
dvy
dx
dy
dx
dyy
dx
dvyyv
yxx
y
dx
dy
First-order ordinary differential equation
Bernoulli Equation yxRyxQyxP )()(')( 0)())()(( dyxPdxyxRyxQ
byxfyx,F )()( 0)()())()(()( dyxPyxfdxyxRyxQyxf bb
])()([])()()()([ b1 bb yxPxfx
yxRxfyxQxfy
bbbb yxP'xfyxPxf'yxRxfbyxQxfb )()()()()()()()()()1( 1
)()()()()()()1( xP'xfxPxf'xQxf b
dyy
F'dx
x
F'dF'
Bernoulli Equation
)(
)()1(
)(
)(
)(
)(
xP
xQ
xP
xP'
xf
xf'
dxxP
xQxPxf
)(
)()1()(ln)(ln
dxxP
xQ
exP
xf )(
)()1(
)(
1)(
ye
xPyx,F
dxxP
xQ
)(
)()1(
)(
1)(
Example
232' yxyxy Solve 2�,3)(�,2)(�,)( xRxxQxxP
xdxx
xdxxP
xQ
ex
ex
exP
xf 22
)21()(
)()1( 11
)(
1)(
221)( ye
xyx,F x
0])32[(1 222 xdydxyxyyex
x dyy
F'dx
x
F'dF'
)3
2( 12
xye
x
F' x
22
yey
F' x )()(dy 1222 xfyexfyeF' xx
dx
xdfye
x
F' x )(2 12
Example
x
e
dx
xdf x23)(
dxx
exf
x
23
)(
dxx
eyeF'
xx
212 3
Cdxx
eye
xx
212 3
Riccati Equation
)()()(' 2 xRyxQyxPy
)(1
)()(1
)()(1
)('2
2xR
zxSxQ
zxSxPz'
zxS
z
xSy1
)(
zxQ
zxSxP
zxP
xRxSxQxSxPz'z
xS
1)(
1)()(2
1)(�����������
)]()()()()([1
)('
2
22
)()()()()(')( 2 xRxSxQxSxPxS )(xSy
zxQ
zxSxP
zxPz'
z
1)(
1)()(2
1)(
122
)())()()(2( xPzxQxSxPz'
dxxQxSxP
exf)]()()(2[
)(
actoregrating f�int
)(dx)()(
)(
1)(
xf
CxfxP
xfxz
Riccati Equation
Example
xx eyyey 2'SolvexexS )(
ze
zxSy x 11)(
xx exRxQexP )(�,1)(�,)(
xdxdxeeeeexf
xx
33]1)(2[)(actoregrating f�int
xx Ceexf
CdxxfxP
xfxz 3
2
1
)()()(
)(
1)(
13
2
1)(
xxx Ceeexy
First-order ordinary differential equation
Linear equations
solution the is )()()(
1)()()(
)()(])([)(
)()()()((1) Eq.
})(exp{)()()()(
ODEexact an is 0))()()(()(
))()()(()(
)1()()()()()(
)(factor gintegratin multiply )()(
dxxQxx
ydxxQxyx
xQxyxdx
d
dx
xdy
dx
dyxyxPx
dx
dyx
dxxPxxPxdx
xd
dxxQyxPxdyx
xPxQxdx
dyx
xQxyxPxdx
dyx
xxQyxPdx
dy
First-order ordinary differential equation
)exp(2
)exp())(exp(2)2ln(
22
22
)2(2
method separated-Variable (2)
)exp(2
)exp(2)exp(4)exp(
)exp(}2exp{)( (1)
42 :Ex
2
222
2
222
2
xky
xkcxycxy
xdxy
dyxdx
y
dyyx
dx
dy
xcy
cxdxxxxy
xxdxx
xxydx
dy
First-order ordinary differential equation
Isobaric equations
mvxydxxm
dyy
yxB
yxA
dx
dy
onsubstituti a make then , and to relative weight
a given each are and if consistent llydimensiona is equation The
),(
),(
cxxycxyxcxv
x
dxvdv
vxdx
dv
x
v
dx
dvvx
xx
v
vxdx
dvxvx
dx
dy
vdxxdvxdyvxym
yxdydxx
y
xy
yxdx
dy
ln2
1ln)(
2
1ln
2
1
122
)2
(2
1RHS
2
1LHS
2/2/1
lyrespective 1,2m 0, 1,2m is litydimensiona the 02)2
(
)2
(2
1 :Ex
222/12
21
2
2/12/12/3
2/32/12/1
2
2
Miscellaneous equations
First-order ordinary differential equation
)(
onsubstituti a make
)( (1)
xbFadx
dyba
dx
dv
cbyaxv
cbyaxFdx
dy
11
11
2
2
2
)1(tan
tan1
111
)1( :Ex
cxyx
cxvdxv
dv
vdx
dy
dx
dvyxv
yxdx
dy
ODE shomogeneou a
0)((
0)()(
shomogeneou is RHS and let (2)
fYeX
bYaX
dX
dY
gfefYeXgYfX e
cbabYaXcYbXa
YyXxgfyex
cbyax
dx
dy
First-order ordinary differential equation
2223
123
1
2
2
)32)(34()21
1)(1
1
14()1(
)3exp()2)(14()2ln(3
2)14ln(
3
1ln
23
2
143
4
472
42
42
472
42
52
42
52let
42
52
10642 and 0352
,let 642
352 :Ex
cxyxycx
y
x
yx
cvvXcvvX
dX
dX
v
dv
v
dvdv
vv
v
v
vv
dX
dvX
v
v
vXX
vXX
dX
dvXv
dx
dvXv
dX
dYvXY
YX
YX
dX
dY
YyXxyx
yx
dx
dy
Second Order Differential Equations
Purpose: reduce to 1st O.D.E.
Likely to be reduced equations:
-Non-linear Equations where the dependent variable does not occur explicitly Equations where the independent variable does not occur explicitly Homogeneous equations
-Linear The coefficients in the equation are constant The coefficients are functions of the independent variable
)(2
2
xfdx
yd
Second order Ordinary Differential Equation
77
Higher-degree first-order equation
First-order ordinary differential equation
0),().....,(),(),( is solution general The
,...2,1for ),( equation of solution the is 0),(
),( and ),(
0))........()((
for 0),(),(...........),(
321
21
011
1
yxGyxGyxGyxG
niyxFdx
dypyxG
yxFpyxFF
FpFpFpdx
dypyxapyxapyxap
n
ii
iii
n
nn
n
0)]1()][1([ is solution general The
0)1()1ln(ln1
202)1( (2)
0)1(1lnln1
0)1( (1)
0]2)1][()1[(
for 02)123()1( :Ex
221
222
22
2
11
2
22223
xkyxky
xkycxyx
xdx
y
dyxy
dx
dyx
xkyc)(xyx
dx
y
dyy
dx
dyx
xypxypx
dx
dypxyypxxpxxx
dy
dp
p
F
y
F
pdy
dx
dx
dyppyFx
x
1
for ),(
for soluable Equation
First-order ordinary differential equation
xyyxy
pxypyxpypy
ypyp
kkxyykxkyy
kx
y
ky
y
kpkpyc
ypcyp
dyyp
dpp
dy
dpy
dy
dpyp
dy
dpypypyp
dy
dpy
dy
dp
p
y
ppdy
dxpy
p
yx
dxdypyxppy
p
solutionsingular 038)6/1(94
9)16()3()6( equation. origional the Change
6/1061(2)
solution general 6303603)(6
1lnlnln2ln
2202 (1)
0)2)(61(12613
363
/for 036 :Ex
2322
22222222
22
23322
22
2
22
2
222
2
22
First-order ordinary differential equation
dx
dp
p
F
x
Fp
dx
dypxFy
y
),(
for soluable Equation
solutionsingular 0021 (2)
solution general 4)(
4)2()(eq. origion the intoput
1lnln
202 (1)
0)2)(1(
0)1(2)1(0)1(2
2222
02 :Ex
2
222
2
2
22
2
yxyxxp
kxky
kxxpykkxp
cx
px
dx
p
dp
dx
dpxp
dx
dpxpp
pdx
dpxpppp
dx
dpxp
dx
dpxp
dx
dpxppp
dx
dyxpxpy
yxpxp
First-order ordinary differential equation
Clairaut’s equation
ODE origional the in eliminate 0),(0 (2)
solution general )()(
)(eq. origional the intoput
0 (1)
0)(
1112
12211
212
2
ppxGxdp
dF
cFxcycFc
cFxccxccdx
dyp
cxcydx
yd
dx
dp
xdp
dF
dx
dp
dx
dp
dp
dF
dx
dpxpp
dx
dy
)( pFpxy
solutionsingular 04442
2
02 (2)
solution general ))(()( (1)
0)2(2
:Ex
2222
22
2
yxx
yxx
yx
ppx
ccxcFxyppF
pxdx
dpp
dx
dpp
dx
dpxp
dx
dy
ppxy
Non-linear 2nd O.D.E.
They are solved by differentiation followed by the p substitution.
When the p substitution is made in this case, the second derivative of y is replaced by the first derivative of p thus eliminating y completely and producing a first O.D.E. in p and x.
dx
dyp
2
2
dx
yd
dx
dp
- Equations where the dependent variables does not occur explicitly
Solve axdx
dyx
dx
yd
2
2
Let dx
dyp
2
2
dx
yd
dx
dpand therefore
axxpdx
dp
2
2
1exp xintegral factor
222
2
1
2
1
2
1xxx
axexpeedx
dp
22
2
1
2
1
)(xx
axepedx
d
Bx
Aerfaxy
dxxCaxy
2
2
1exp 2
error function
Example
dx
dyxCap )
2
1exp( 2
Let dx
dyp
dy
dpp
dx
dy
dy
dp
dx
dp
dx
yd
2
2
Non-linear 2nd O.D.E.- Equations where the independent variables does not occur explicitly
They are solved by differentiation followed by the p substitution.
When the p substitution is made in this case, the second derivativeof y is replaced as
Solve 22
2
)(1dx
dy
dx
ydy
Letdx
dyp and therefore
21 pdy
dpyp
Separating the variables
dy
dpp
dx
yd
2
2
dyy
dpp
p 1
12
)1ln(2
1lnln 2 pay
)1( 22 yadx
dyp
)sinh(
1
)(sinh1
)1(1
22
caxa
y
bayax
ya
dyx
Example
x
yf
dx
dy
2
2
dx
ydx
n
nn
dx
ydx 1
dx
dy
x
yf
dx
ydx ,
2
2
Non-linear 2nd O.D.E.- Homogeneous equations
The homogeneous 1st O.D.E. was in the form:
The corresponding dimensionless group containing the 2nd differential coefficient is
In general, the dimensionless group containing the nth coefficient is
The second order homogenous differential equation can be expressed in a form analogous to
If in this form, called homogeneous 2nd ODE
y=vxdx
dvxv
dx
dy
2
2
2
2
2dx
vdx
dx
dv
dx
yd
dx
dvxvvf
dx
vdx
dx
dvx ,2 12
22
dx
dvxvf
dx
vdx ,22
22
x=et or t=lnxdt
dv
xdx
dt
dt
dv
dx
dv 1
dt
dv
dx
dvx
2
2
22
2
22
2
11
11
11
dt
vd
xdt
dv
x����
dt
dv
dt
d
dx
dt
xdt
dv
x����
dt
dv
dx
d
xdt
dv
xdx
vd
dt
dv
dt
vd
dx
vdx
2
2
2
22
dt
dvvf
dt
dv
dt
vd,22
2
Non-linear 2nd O.D.E.
Solve2
222
222
dx
dyxy
dx
ydyx
Dividing by 2xy
2
2
2
2
1
2
1
dx
dy
y
x
x
y
dx
ydx
homogeneous
2
2
2
2
dt
dv
dt
vdv
dx
dy
x
yf
dx
ydx ,
2
2
Let vxy 2
22
22 22
dx
dvx
dx
dvvx
dx
vdvx
Let tex
dt
dvp
22 pdv
dpvp
2)ln( CxBxy
Axy
Singular solution
General solution
Example
02 porp�� ��dv
dpv
)(... 11
1
10 xyPdx
dyP
dx
ydP
dx
ydP nnn
n
n
n
where (x) is any function of x.
2nd Order Linear Differential Equations
They are frequently encountered in most chemical engineering fields of study, ranging from heat, mass, and momentum transfer toapplied chemical reaction kinetics.
The general linear differential equation of the nth order having constant coefficients may be written:
)(2
2
xRydx
dyQ
dx
ydP
where P, Q and R are constant coefficientsLet the dependent variable y be replaced by the sum of the two new variables: y = u + vTherefore
)(2
2
2
2
xRvdx
dvQ
dx
vdPRu
dx
duQ
dx
udP
If v is a particular solution of the original differential equation
The general solution of the linear differential equation will be the sum of a “complementary function” and a “particular solution”.
purpose
2nd Order Linear Differential EquationsThe general equation can be expressed in the form
02
2
Ru
dx
duQ
dx
udP
02
2
Rydx
dyQ
dx
ydP
Let the solution assumed to be:mx
meAy
mxmmeA
dx
dy mx
m emAdx
yd 22
2
0)( 2 RQmPmeA mxm
auxiliary equation (characteristic equation)
Unequal rootsEqual rootsReal rootsComplex roots
The Complementary Function
xmeAy 11 xmeAy 2
2
xmxm eAeAy 2121
Unequal Roots to Auxiliary EquationLet the roots of the auxiliary equation be distinct and of values m1 and m2. Therefore, the solutions of the auxiliary equation are:
The most general solution will be
If m1 and m2 are complex, it is customary to replace the complex exponential functions with their equivalent trigonometric forms.
0652
2
ydx
dy
dx
yd0652 mm
3
2
2
1
m
mxx BeAey 32
Solve
mxAey mxmx mVe
dx
dVe
dx
dymxVey Let mxmxmx Vem
dx
dVme
dx
Vde
dx
yd 22
2
2
2
2
where V is a function of x
02
2
Rydx
dyQ
dx
ydP
02
2
dx
VdDCxV
mxeDCxy )(
Equal Roots to Auxiliary EquationLet the roots of the auxiliary equation equal and of value m1 = m2 = m. Therefore, the solution of the auxiliary equation is:
02 RQmPm 02 QPm
Solve096
2
2
ydx
dy
dx
yd
auxiliary equation
0962 mm
321 mm
xeBxAy 3)(
Example
Solve 0542
2
ydx
dy
dx
yd
auxiliary equation
0542 mm
im 2
xixi BeAey )2()2(
)sincos(2 xFxEey x
)(2
2
xRydx
dyQ
dx
ydP
Particular IntegralsTwo methods will be introduced to obtain the particular solution of asecond order linear O.D.E.
The method of undetermined coefficients ~confined to linear equations with constant coefficients and particular form of (x)
The method of inverse operators ~general applicability
When (x) is a polynomial of the form where all the coefficients are constants. The form of a particular integral is
)(2
2
xRydx
dyQ
dx
ydP
RCy /
nn xaxaxaa ...2
210
nn xxxy ...2
210
Method of Undetermined Coefficients
When (x) is constant, say C, a particular integral of equation is
Example
Solve 32
2
8444 xxydx
dy
dx
yd
32 sxrxqxpy 232 sxrxq
dx
dy
sxrdx
yd62
2
2
3322 84)(4)32(4)62( xxsxrxqxpsxrxqsxr
equating coefficients of equal powers of x
84
0124
4486
0442
s
sr
qrs
pqr
32 26107 xxxy p
0442 mm
auxiliary equation
xc eBxAy 2)(
32 26107)( xxxeBxA��
yyy
x
pc
0442
2
ydx
dy
dx
yd
complementary function
2m
Method of Undetermined Coefficients
rxey
When (x) is of the form Terx, where T and r are constants. Theform of a particular integral is
RQrPr
T
2
When (x) is of the form Gsinnx + Hcosnx, where G and H are constants, the form of a particular solution is
nxMnxLy cossin
2222
2
)(
)(
QnPnR
nQHGPnRL
2222
2
)(
)(
QnPnR
nQGHPnRM
Example
Solve1863
2
2
dx
dy
dx
yd
Cxy C
dx
dy
02
2
dx
yd
18)C(6)0(3
3C
xy p 3
063 2 mmauxiliary equation
xc BeAy 2
x
pc
BeAx��
yyy
23
0632
2
dx
dy
dx
yd
complementary function
20 andm� ��m
Example
Solve xeydx
dy
dx
yd 42
2
78103
xCxey 4xCex
dx
dy 4)41(
xCexdx
yd 42
2
)816(
71024 CC
2
1C
xp xey 4
2
1
0)4)(23(8103 2 mmmmauxiliary equation
xxc BeAey 43/2
xxx
pc
BeAexe��
yyy
43/24
2
1
081032
2
ydx
dy
dx
yd
complementary function
43/2 andm� ��m
Example
Solvexy
dx
dy
dx
yd2cos526
2
2
xDxCy 2sin2cos
)2cos2sin(2 xDxCdx
dy
)2sin2cos(42
2
xDxCdx
yd
0102
52210
DC
DC
1
5
D
C
xxy p 2sin2cos5
0)3)(2(62 mmmmauxiliary equation
xxc BeAey 32
xxBeAe��
yyyxx
pc
2sin2cos532
062
2
ydx
dy
dx
yd
complementary function
32 andm� ��m
Example
xydx
dy
dx
yd2cos526
2
2
ixeydx
dy
dx
yd 22
2
526
ixp key 2
ix
ix
kedx
yd
ikedx
dy
22
2
2
4
2
ixixixix ekeikeke 2222 52624
ixix ekei 22 52)210( 52)210( ki
ii
ii
i
ik
5
4100
)210(52
)210)(210(
)210(52
)210(
52
)2sin2)(cos5()5( 2 xixieiy ixp xxy p 2sin2cos5
取實數部分
)(2
2
xRydx
dyQ
dx
ydP
Particular IntegralsTwo methods will be introduced to obtain the particular solution of asecond order linear O.D.E.
The method of undetermined coefficients ~confined to linear equations with constant coefficients and particular form of (x)
The method of inverse operators ~general applicability
n
nn
dx
ydyD
dx
ydyDDyD
dx
dyDy
...
)(2
22
But, 22)(
dx
dyDy
ydx
dy
dx
yd23
2
2
yDDyDDyDyyD )2)(1()23(23 22
Method of Inverse OperatorsSometimes, it is convenient to refer to the symbol “D” as the differential operator:
The differential operator D can be treated as an ordinary algebraicquantity with certain limitations.
(1) The distribution law:A(B+C) = AB + ACwhich applies to the differential operator DD(u+v+w)=Du+Dv+Dw
(2) The commutative law:AB = BAwhich does not in general apply to the differential operator DDxy xDy (D+1)(D+2)y = (D+2)(D+1)y
(3) The associative law:(AB)C = A(BC)which does not in general apply to the differential operator DD(Dy) = (DD)y D(xy) = (Dx)y + x(Dy)
Method of Inverse Operators
pxpx
pxnpxn
pxpx
epfeDf
epeD
peDe
)()(
...
pxpx eppeDD )23()23( 22 ypDfeyeDf
ypDeyeD
ypDeyeD
ypDeyDeDyeyeD
pxpx
npxpxn
pxpx
pxpxpxpx
)())((
)()(
...
)()(
)()(22
More convenient!
Differential Operator Properties
pxpppxD
pxppxD
pxpppxD
pxppxD
pxipxe
nn
nn
nn
nn
ipx
sin)()(cos
cos)()(cos
cos)()(sin
sin)()(sin
sincos
212
22
212
22
ipxnipxnipxnn eipeDeDpxD )Im(ImIm)(sin
where “Im” represents the imaginary part of the function which follows it.
The operator D signifies differentiation, i.e.
)()( xfdxxfD )()( 1 xfDdxxf
D-1 is the “inverse operator” and is an “integrating” operator.It can be treated as an algebraic quantity in exactly the same manner as D
Inverse Operator
Solvexey
dx
dy 24 differential operator
xeyD 2)4(
xeD
y 2
)4(
1
xeDDDy 232 ...])4
1()
4
1()
4
1(1[
4
1
binomial expansion
xey 2
2
1
pxpx epfeDf )()(
x2e)D
41
1(4
1y
2pxey 2
)42(
1
xx eey 2322
2
1...])
2
1()
2
1()
2
1(1[
4
1
Example
Solve xxeydx
dy
dx
yd 42
2
6168
differential operator
xxeyDyDD 422 6)4()168( 01682 mm
xc eBxAy 4)(
xp xe
Dy 4
2)4(
6
xDey xp
246
f(p) = 0
)()( pDfeeDf pxpx
integration
xxp exxDey 432143
Example
01682
2
ydx
dy
dx
yd
4m
x
pc
exBxA��
yyy
43)(
xxeydx
dy
dx
yd 42
2
6168
xeCxy 42xx eCxCxe
dx
dy 424 42
xxx eCxCxeCedx
yd 42442
2
16162
xx xeCe 44 62
xeCxy 43 xx eCxeCxdx
dy 4342 43
xxx eCxeCxCxedx
yd 434242
2
16246
xx xeCxe 44 66
1C
xp exy 43
Example
Solve 232
2
346 xxydx
dy
dx
yd
differential operator232 34)2)(3()6( xxyDDyDD 062 mm
xxc BeAey 23
)34()2)(3(
1 23 xxDD
y p
expanding each term by binomial theorem
)34()2(
1
)3(
1
5
1 23 xxDD
y p
)34(...16842
1...
812793
1
5
1 233232
xxDDDDDD
y p
...01296
2413
216
)624(7
36
612
6
34 223
xxxxx
y p
Example
062
2
ydx
dy
dx
yd
23 and� ��m
108/)5661872( 23
23
xxx����
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Higher-order ordinary differential equation
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Higher-order ordinary differential equation
General ordinary differential equations
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1
21
122
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2
2
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x
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xx
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Higher-order ordinary differential equation
(II) Independent variable absent: An ODE does not contain independent
variable x, except in d/dx, d2/dx2, making a substitution dy/dx=p
22
22
3
3
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Higher-order ordinary differential equation
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2
3
3
62 :Ex
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42
21
3
1
2
1
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2
2
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2
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3
3
2
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224
62
2)2(
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4])(2[ (2)
22)2( (1)
cxcxcx
ycxcx
dx
dyy
cx
dx
dyy
dx
d
cx
dx
dy
dx
ydy
xdx
dy
dx
d
dx
ydy
dx
d
dx
yd
dx
dy
dx
dy
dx
d
dx
yd
dx
dy
dx
ydy
dx
ydy
dx
d
Higher-order ordinary differential equation
Isobaric or homogeneous equations
)(),(:/(c) )1((b) )((a) :Weight
(2) (1)
nWxmW ydxydWxmWy
exvxynn
tm
0 and Let
equation shomogeneou 0)1(
2 set 1m weihght equalfor
1m 2m, 2m, 1,m 1,m
1m 2m, 2m, m),1-(2 m),2-(3 :Weight
0
0)()( :Ex
2
22
2
22
2
2
2
2
2
2
2
2
222
23
222
23
dt
dvv
dt
vd
dt
dve
dt
vde
dx
vd
dt
dve
dx
dvex
dx
dvv
dx
vdx
dx
dv
dx
vdx
dx
yd
dx
dvxv
dx
dyvxy
xyydx
dyxy
dx
dyx
dx
ydx
xyydx
dyxyx
dx
ydx
tttt
Higher-order ordinary differential equation
)ln2
1tan(
)(tanln2
lnfor
)(tan1
2
2
1
)(2
1
2
1
2)
2(
2111
1
121
1
1
1
122
12
21
21
2
1
22
2
2
ddxdxdy
xd
yddx
dxt
d
v
dd
t
dv
dvdt
dvcvdt
dv
cv
dtv
dt
d
dt
dvdt
dt
dvvdt
dt
vd
Higher-order ordinary differential equation
Equation homogeneous in x or y alone
221
21
221
2
21
21122
333
2
2
32
2
2
22
2
2
32
22
ln2
1
)1(2
)1
(2
)1
(2 1
2
1
2
2 0
2
Set 02
)( and Let
in shomogeneou 02
:Ex
cxctc
ycct
yc
ydyct
yc
dy
yc
dt
dypc
yp
dyy
pdpydy
dpp
ydy
dpp
dy
dpp
dy
dp
dt
dy
dt
dp
dt
yd
dt
dyp
ydt
yd
dt
dy
dt
yde
dx
yd
dt
dye
dt
dy
dx
dt
dx
dydtedxex
xydx
dyx
dx
ydx
tttt
Higher-order ordinary differential equation
Equations having as a solution:
If any general nth-order ODE is satisfied identically by assuming
xAey
solution a is )exp( then /......// 22 xAydxyddxyddxdyy nn
solution a is y.identicall satisfied is Which
0)( Set
0)()( :Ex
222222
2
222
22
xAey
xyyxyxxdx
yd
dx
dyy
dx
dyx
dx
dyyx
dx
yd
dx
dyxx
Simultaneous Diffusion and Reaction
A tubular reactor of length L and 1 m2 in cross section is employed to carry out a first order chemical reaction in which a material A isconverted to a product B,
A B
The specific reaction rate constant is k s-1. If the feed rate is u m3/s, the feed concentration of A is Co, and the diffusivity of A is assumed to be constant at D m2/s. Determine the concentration of A as a function of length along the reactor. It is assumed that there is no volume change during the reaction, and that steady state conditions are established.
A material balance can be taken over the element of length x at a distance x from the inlet
The concentration will vary in the entry sectiondue to diffusion, but will not vary in the sectionfollowing the reactor. (Wehner and Wilhelm, 1956)
x x+x
Bulk flow of A
Diffusion of A
uC
xdx
dCD
dx
d
dx
dCD
dx
dCD
xdx
dCuuC
Input - Output -Reaction = Accumulation
0
xkCx
dx
dCD
dx
d
dx
dCDx
dx
dCuuC
dx
dCDuC
分開兩個 section
uC0
x
L
x
uC
C
Simultaneous Diffusion and Reaction
0
xkCx
dx
dCD
dx
d
dx
dCDx
dx
dCuuC
dx
dCDuC
dividing by x
0
kC
dx
dCD
dx
d
dx
dCu
rearranging
02
2
kCdx
dCu
dx
CdD
auxiliary equation
02 kumDm
a
D
uxBa
D
uxAC 1
2exp1
2exp
In the entry section0
''2
2
dx
dCu
dx
CdD
02 umDm
D
uxC exp'
2/41 ukDa
Simultaneous Diffusion and Reaction
auxiliary equation
a
D
uxBa
D
uxAC 1
2exp1
2exp
D
uxC exp'
B. C.
0
'0
dx
dCLx
dx
dC
dx
dCx
B. C.
CCx
CCx
'0
' 0
Simultaneous Diffusion and Reaction
012
exp)1(12
exp)1(
)1()1(2
0
aD
uxaBa
D
uxaA
aBaA
BA
C
)1()1(2 0 aBaAC
xL
D
uaaxL
D
uaa
D
ux
KC
C
2exp1
2exp1
2exp
2
0
if diffusion is neglected (D0)
u
kx
C
CCexp1
0
0
DuLaaDuLaaK 2/exp12/exp1 22
D
uLa
K
aCB
D
uLa
K
aCA
2exp
)1(2
2exp
)1(2
0
0
Simultaneous Diffusion and Reaction
1.017 kg/s of a tallow fat mixed with 0.286 kg/s of high pressure hot water is fed into the base of a spray column operated at a temperature 232 C and a pressure of 4.14 MN/m2. 0.519kg/s of water at the same temperature and pressure is sprayed into the top of the column and descends in the form of droplets through the rising fat phase. Glycerine is generated in the fat phase by the hydrolysis reaction and is extracted by the descending water so that 0.701 kg/s of final extract containing 12.16% glycerine is withdrawn continuously from the column base. Simultaneously 1.121 kg/s of fatty acid raffinate containing 0.24% glycerine leaves the top of the column. If the effective height of the column is 22 m and the diameter 0.66 m, the glycerine equivalent in the entering tallow 8.53% and the distribution ratio of glycerine between the water and the fat phase at the column temperature and pressure is 10.32, estimate the concentration of glycerine in each phase as a function of column height. Also find out what fraction of the tower height is required principally for the chemical reaction. The hydrolysis reaction is pseudo first order and the specificreaction rate constant is 0.0028 s-1.
Continuous Hydrolysis of Tallow
tallow fat hot water G’’ kg/s
glycerine (extract)
fatty acid (raffinate)L kg/s
L kg/sxH
zH
x0
z0
y0
G kg/syH
h
xz
x+xz+z y+y
y
h
Continuous Hydrolysis of Tallow
hot water G’ kg/s
Consider the changes occurring in the element of column of height h:
glycerine transferred from fat to water phase, hyyKaS )*(
S: sectional area of towera: interfacial area per volume of towerK: overall mass transfer coefficient
rate of destruction of fat by hydrolysis, hSzk rate of production of glycerine by hydrolysis, whSzk /
k: specific reaction rate constant: mass of fat per unit volume of column (730 kg/m3)w: kg fat per kg glycerine
x = weight fraction of glycerine in raffinatey = weight fraction of glycerine in extracty*= weight fraction of glycerine in extract in equilibrium with xz = weight fraction of hydrolysable fat in raffinate
Continuous Hydrolysis of Tallow
A glycerine balance over the element h is:
hyyKaSw
hSzkh
dh
dxxLLx
*
A glycerine balance between the element and the base of the tower is:
00 Gy
w
LzLxGy
w
Lz
L kg/sxH
zH
x0
z0
y0
G kg/syH
h
xz
x+xz+z y+y
y
h
The glycerine equilibrium between the phases is:
mxy *
hyyKaSGyhdh
dyyG *
in the fat phase
in the extract phase
Continuous Hydrolysis of Tallow
hyyKaSGyhdh
dyyG *
dh
dyGKaSyKaSmx
00 Gy
w
LzLxGy
w
Lz
hyyKaSw
hSzkh
dh
dxxLLx
*
hyyKaSGyhdh
dyyG *
dh
dyG
dh
dxL
w
Szk
dh
dyG
dh
dxLxyy
L
G
w
zSk
0
0multiply by KaSm/LG
xyyL
G
w
z
w
z )( 0
0
02
2
00
2
dh
dy
L
KaSm
dh
yd
dh
dy
G
KaS
dh
dyy
G
KaS
L
Skyy
L
mG
w
mz
LG
KaSk
Continuous Hydrolysis of Tallow
dh
dy
KaSm
Gy
mx
1
02
2
00
2
dh
dy
L
KaSm
dh
yd
dh
dy
G
KaS
dh
dyy
G
KaS
L
Skyy
L
mG
w
mz
LG
KaSk
L
mGr
L
Skp
)1( rG
KaSq
w
mzry
r
pqpqy
dh
dyqp
dh
yd 002
2
1)( 2nd O.D.E. with constant coefficients
Complementary function Particular solution
0)(2 pqmqpm Constant at the right hand side
)exp()exp( qhBphAyc Cpqw
mzry
r
pqy p
/
10
0
Continuous Hydrolysis of Tallow
w
mzry
rqhBphAy 0
01
1)exp()exp(
B.C.
0,
0,0
yHh
xh
dh
dy
q
rymx
1 )exp()exp(
1qhqBphpA
q
rymx
Continuous Hydrolysis of Tallow
0,0 xh 0)(1
CqBpA
q
rBA
0, yHh 0 CBeAe qHpH
KaL
Gk
q
prpqv
1
)()(
)()(qHpHpH
pHpHpH
erCreveA
veCreveB
=C
dh
dyGKaSyKaSmx
qH
pHqHqh
qH
pHph
er
revee
er
vee
vrw
mzy
)(0
2115.0403.0
1216.0701.0~
403.02
519.0286.0~
069.12
121.1017.1~
0
y
G
L
0,0 yyh
pHqHqhpHphqHpHqH reveeveeerw
mzry
rrevey
)()(
1
1)( 0
0
Continuous Hydrolysis of Tallow
Ka=0.0632 kg glycerine per sec per m3
H=22 my=0
730
66.04
0028.0
32.10
2
S
k
m
qHpH eνr
ve
νr
r
erw
mzy
111
)( pH0
0
L
Skp
L
mGr rortryander� �~~~)1( r
G
KaSq
Continuous Hydrolysis of Tallow
It shows that the chemical reaction is virtually completed in the bottom 9 m of the column, or 40% of the column.
qH
pHqHqh
qH
pHph
er
revee
er
vee
vrw
mzy
)(0
)exp()exp(1
* qhqBphpAq
rymxy
xyyL
G
w
z
w
z )( 0
0
)(2
2
xRydx
dyQ
dx
ydP )()( 2211 xycxycyc
Variation of Parameters
)()()()( 2211 xyxuxyxuy p 22221111 ''''' yuyuyuyuy p
0'' 2211 yuyu''' 2211 yuyuy p
'''''''''' 22112211 yuyuyuyuy p
)()()''()''''''''( 2211221122112211 xyuyuRyuyuQyuyuyuyuP
)()'''()'''()''''( 222211112211 xuRyQyPyuRyQyPyyuyuP
0'''
0'''
222
111
RyQyPy
RyQyPy
Pxyuyu /)('''' 2211
0'' 2211 yuyu
),(
)(]/)([
)'()'(
)()(
)'(/)(
)(0
)('21
2
21
21
2
2
1 yyW
xyPx
xyxy
xyxy
xyPx
xy
xu
),(
)(]/)([
)'()'(
)()(
/)()'(
0)(
)('21
1
21
21
1
1
2 yyW
xyPx
xyxy
xyxy
Pxxy
xy
xu
Variation of Parameters
dxyyW
xyPxxu
),(
)(]/)([)(
21
21
dx
yyW
xyPxxu
),(
)(]/)([)(
21
12
ExampleSolve )3sec(39
2
2
xydx
yd
092 mm
xBxAyc 3sin3cos
33cos33sin3
3sin3cos),( 21
xx
xxyyW
xxxxy p 3sin3cos3cosln3
1
092
2
ydx
yd
im 3
xxxx����
xBxA��
yyy pc
3sin3cos3cosln3
1
3sin3cos
xxy
xxy
3sin)(
3cos)(
2
1
xdxxx
xu 3cosln3
1
3
3sin3sec3)(1
xdxxx
xu 3
3cos3sec3)(2
Euler Equation
02
22 By
dx
dyAx
dx
ydx mxxy )( 2
1
)1(')'(
)'(
m
m
xmmxy
mxxy
0)1( mmm BxAmxxmm
0)1(2 BmAm
21, m�mm
21 mm
21
21)( mm xcxcxy
21 mm
1]ln[)( 21mxxccxy
Our purpose: Use algebraic elimination of the variables until only one differential equation relating two of the variables remains.
Simultaneous Differential EquationsThese are groups of differential equations containing more than one dependent variable but only one independent variable. In these equations, all the derivatives of the different dependent variables are with respect to the one independent variable.
Independent variable or dependent variables?
),(
),(
2
1
yxfdt
dy
yxfdt
dx
elimination of independent variable
),(
),(
2
1
yxf
yxf
dy
dx
Elimination of one or more dependent variablesIt involves with equations of high order and it would be better to make use of matrices.
Solve
0)23()103(
0)96()6(
22
22
zDDyDD
and
zDDyDD
0)1)(2()5)(2(
0)3()2)(3( 2
zDDyDD
and
zDyDD)5( D
)3( D0)1)(2)(3()5)(2)(3(
0)3)(5()5)(2)(3( 2
zDDDyDDD
and
zDDyDDD
0)23()158()3(
0)1)(2)(3()3)(5(
22
2
zDDDDD
zDDDzDD
0)1311)(3( zDD
Elimination of Dependent Variable
0)1311)(3( zDD
xxBeAez 311
13
0)96()6( 22 zDDyDD
xAeyDD 11
1322 9
11
136
11
13)6(
= E
x
p EeDD
y 11
13
2 )6(
1
xxc JeHey 32
pxpx epfeDf )()( 11
13p
x
p Eey 11
13
2 )6)1113
()1113
((
1
xx
p AeEey 11
13
11
13
7
4
700
121
Elimination of Dependent Variable
J=2B
xxxHeJeAey 2311
13
7
4
0)23()103( 22 zDDyDDxx
BeAez 311
13
xxxHeJeAey 2311
13
7
4
xxxHeBeAey 2311
13
27
4
1.25 kg/s of sulphuric acid (heat capacity 1500 J/kg C) is to be cooled in a two-stage counter-current cooler of the following type. Hot acid at 174 C is fed to a tank where it is well stirred in contact with cooling coils. The continuous discharge from this tank at 88 C flows to a second stirred tank and leaves at 45 C. Cooling water at 20 C flows into the coil of the second tank and thence to the coil of the first tank. The water is at 80 C as it leaves the coil of the hot acid tank. To what temperatures would the contents of each tank rise if due to trouble in the supply, the cooling water suddenly stopped for 1h? On restoration of the water supply, water is put on the system at the rate of 1.25 kg/s. Calculate the acid discharge temperature after 1 h. The capacity of each tank is 4500 kg of acid and the overall coefficient of heat transfer in the hot tank is 1150 W/m2 C and inthe colder tank 750 W/m2 C. These constants may be assumed constant.
Example of Simultaneous O.D.E.s
1.25 kg/s
0.96 kg/s
88 C
45 C174 C
20 C40 C80 CTank 1 Tank 2
When water fails for 1 hour, heat balance for tank 1 and tank 2:
dt
dTVCMCTMCT 1
10
dt
dTVCMCTMCT 2
21
Tank 1
Tank 2
M: mass flow rate of acidC: heat capacity of acidV: mass capacity of tankTi: temperature of tank i
dt
dTTT
dt
dTTT
221
110
B.C.t = 0, T1 = 88 teT 861741
t = 1, T1=142.4 C
dt
dTTe t 2
286174 integral factor, et tetT )12986(1742
t = 1, T2 = 94.9 C
B.C.t = 0, T2 = 45
Example of Simultaneous O.D.E.s
When water supply restores after 1 hour, heat balance for tank 1 and tank 2:
dt
dTVCMCTtWCMCTtWC ww
22213
Tank 1
Tank 2
W: mass flow rate of waterCw: heat capacity of watert1: temperature of water leaving tank 1t2: temperature of water leaving tank 2t3: temperature of water entering tank 2
dt
dTVCMCTtWCMCTtWC ww
11102
1 2T0 T1 T2
t3t2t1
Heat transfer rate equations for the two tanks:
)ln()ln(
)()()(
2111
21111121 tTtT
tTtTAUttWCw
)ln()ln(
)()()(
3222
32222232 tTtT
tTtTAUttWCw
4 equations have to besolved simultaneously
Example of Simultaneous O.D.E.s
dt
dTVCMCTtWCMCTtWC ww
22213
dt
dTVCMCTtWCMCTtWC ww
11102
)ln()ln(
)()()(
2111
21111121 tTtT
tTtTAUttWCw
)ln()ln(
)()()(
3222
32222232 tTtT
tTtTAUttWCw
211 )1( ttT
322 )1( ttT
wWC
AU
e11
wWC
AU
e22
Example of Simultaneous O.D.E.s
)1(121 Ttt
)1(232 Ttt
30975.708.6 22
22
2
Tdt
dT
dt
Td
B.C. t = 0, T1=142.4 C, T2 = 94.9 C
dt
dTCtCTCCTCTtC www
121102 )1(
dt
dTCTCCTCTtCTC www
111032 )1()1()1)(1(
dt
dTCtCTCCTCTtC www
232213 )1(
322
1 )1()( tCTCCCdt
dTCCT www
dt
dTCCC
dt
TdC
dt
dTC ww
222
21 )(
0
23
222
222
22
)1)(1()1(
)1)(1()1(
)22(
TCtCCC
TCCCC
dt
dTCCCCC
dt
TdC
ww
ww
www
t = 1, T2 = 48.8 C
9.3982.126.42 tt BeAeT
Example of Simultaneous O.D.E.s
Eliminating T1
A=0.6; B=54.4