Introduction to ordinary differential equation

PPT on Differential Equation

1. Fin500J Topic 6 Fall 2010 Olin Business School 1
Fin500J: Mathematical Foundations in Finance
Topic 6: Ordinary Differential Equations
Philip H. Dybvig
Reference: Lecture Notes by Paul Dawkins, 2007, page 20-33, page 102-121,
page 137-155 and page 340-344
http://tutorial.math.lamar.edu/terms.aspx
Slides designed by Yajun Wang

2. Introduction to Ordinary
Differential Equations (ODE)
Recall basic definitions of ODE,
order
linearity
initial conditions
solution
Classify ODE based on( order, linearity, conditions)
Classify the solution methods
Fin500J Topic 6 Fall 2010 Olin Business School 2

3. Derivatives
Derivativesdx
dy
Fin500J Topic 6 Fall 2010 Olin Business School
Partial Derivatives
uis a function of
more than one
independent variable
Ordinary Derivatives
yis a function of one
independent variabley
u


3

4. Differential Equations
Differential
Equations16
2
2
xy
dx
yd
Fin500J Topic 6 Fall 2010 Olin Business School
involve one or more
partial derivativesof
unknown functions
Ordinary Differential Equations
involve one or more
Ordinary derivativesof
unknown functions0
2
2
2
2






x
u
y
u
4
Partial Differential Equations

5. Ordinary Differential Equations
Fin500J Topic 6 Fall 2010 Olin Business School )cos(25
:
2
2
xy
dx
dy
dx
yd
ey
dx
dy
Examples
x


Ordinary Differential Equations (ODE)involve one or
more ordinary derivatives of unknown functions with
respect to one independent variable
y(x): unknown function
x: independent variable
5

6. Order of a differential equation
Fin500J Topic 6 Fall 2010 Olin Business School 12
)cos(25
:
4
3
2
2
2
2











y
dx
dy
dx
yd
xy
dx
dy
dx
yd
ey
dx
dy
Examples
x
The orderof an ordinary differential equations is the order
of the highest order derivative
Second order ODE
First order ODE
Second order ODE
6

7. Solution of a differential equation0)(
)(
)(
:Proof
)(






tt
t
t
eetx
dt
tdx
e
dt
tdx
etxSolution
Fin500J Topic 6 Fall 2010 Olin Business School
A solutionto a differential equation is a function that
satisfies the equation.0)(
)(
:
tx
dt
tdx
Example
7

8. Linear ODE
Fin500J Topic 6 Fall 2010 Olin Business School 1
)cos(25
:
3
2
2
2
2
2











y
dx
dy
dx
yd
xyx
dx
dy
dx
yd
ey
dx
dy
Examples
x
An ODE is linear if the unknown function and its derivatives
appear to power one. No product of the unknown function
and/or its derivatives
Linear ODE
Linear ODE
Non-linear ODE
8)()()()(')()()()()(
01
1
1 xgxyxaxyxaxyxaxyxa
n
n
n
n 

 

9. Boundary-Value and Initial value Problems
Boundary-Value Problems
The auxiliary conditions are not at
one point of the independent
variable
More difficult to solve than initial
value problem5.1)2(,1)0(
'2''
2



yy
eyyy
x
Fin500J Topic 6 Fall 2010 Olin Business School
Initial-Value Problems
The auxiliary conditions are
at one point of the
independent variable5.2)0(',1)0(
'2''
2



yy
eyyy
x
same
different
9

10. Classification of ODE
ODE can be classified in different ways
Order
First order ODE
Second order ODE
N
th
order ODE
Linearity
Linear ODE
Nonlinear ODE
Auxiliary conditions
Initial value problems
Boundary value problems
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11. Solutions
Analytical Solutions to ODE are available for linear
ODE and special classes of nonlinear differential
equations.
Numerical method are used to obtain a graph or a
table of the unknown function
We focus on solving first order linear ODE and second
order linear ODE and Euler equation
Fin500J Topic 6 Fall 2010 Olin Business School 11

12. First Order Linear Differential Equations
Def: A first order differential equation is
said to be linearif it can be written)()( xgyxpy 
Fin500J Topic 6 Fall 2010 Olin Business School 12

13. First Order Linear Differential Equations
How to solve first-order linear ODE ?
Sol:)1()()( xgyxpy  )2()()()()()( xgxyxpx
dx
dy
x  
Fin500J Topic 6 Fall 2010 Olin Business School 13
Multiplying both sides by , called an integrating factor,
gives )(x
assuming we get )3(),(')()( xxpx   )4()()()(')( xgxyx
dx
dy
x  

14. First Order Linear Differential Equations
Fin500J Topic 6 Fall 2010 Olin Business School 14
By product rule, (4) becomes
Now, we need to solve from (3) )(
)(
)('
)(')()( xp
x
x
xxpx 


 )6(
)(
)()(
)(
)()()()(
)5()()())'()((
1
1
x
cdxxgx
xy
cdxxgxxyx
xgxxyx









 )(x

15. First Order Linear Differential Equations
Fin500J Topic 6 Fall 2010 Olin Business School 15)7()(
)()(ln
)())'((ln)(
)(
)('
)(
3
)(
2
2 




dxxpcdxxp
ecex
cdxxpx
xpxxp
x
x





to get rid of one constant, we can use )9(
)()(
)(
thenisequation aldifferentiorder first linear a osolution t The
)8()(
)(
)(






dxxp
dxxp
e
cdxxgx
xy
ex



16. Summary of the Solution Process
Fin500J Topic 6 Fall 2010 Olin Business School 16
Put the differential equation in the form (1)
Find the integrating factor, using (8)
Multiply both sides of (1) by and write the left
side of (1) as
Integrate both sides
Solve for the solution )(x )(x ))'()(( xyx )(xy

17. Example 1
Sol:x
eyy
2
  
 
xx
xx
xxx
x
dxdx
dxxpdxxp
ece
cee
cdxeee
cdxeee
cdxxgeexy
2
2
2
)1()1(
)()(
)()(



























Fin500J Topic 6 Fall 2010 Olin Business School 17

18. Example 2
Sol:2
1
)1(,2'
2
 yxxyxy  
.
12
7
3
1
4
1
2
1
c,get tocondition initial Apply the
3
1
4
1
3
1
4
1
)1()1(
)()(
1
2
'
22342
22
22
)()(


































cc
cxxxcxxx
cdxxxxcdxxee
cdxxgeexy
xy
x
y
dx
x
dx
x
dxxpdxxp
Fin500J Topic 6 Fall 2010 Olin Business School 18

19. Second Order Linear Differential Equations
Homogeneous Second Order Linear Differential
Equations
oreal roots, complex roots and repeated roots
Non-homogeneous Second Order Linear Differential
Equations
oUndetermined Coefficients Method
Euler Equations
Fin500J Topic 6 Fall 2010 Olin Business School 19

20. )(''' xgcybyay  wherea,bandcareconstantcoefficients
Let the dependent variable y be replaced by the sum of the
two new variables: y = u + v
Therefore   )('''''' xgcvbvavcubuau 
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
Second Order Linear Differential Equations
The general equation can be expressed in the form 0''' cubuau
Fin500J Topic 6 Fall 2010 Olin Business School 20

21. 0''' cybyay Let the solution assumed to be:rx
ey rx
re
dx
dy
 rx
er
dx
yd
2
2
2
 0)(
2
cbrare
rx
characteristic equation
Real, distinct roots
Double roots
Complex roots
The Complementary Function (solution of the
homogeneous equation)
Fin500J Topic 6 Fall 2010 Olin Business School 21

22. xr
ey
1
 xr
ey
2
 xrxr
ececy
21
21 Real, Distinct Roots to Characteristic Equation
•Let the roots of the characteristic equation be real, distinct
and of values r
1and r
2. Therefore, the solutions of the
characteristic equation are:
•The general solution will be06'5''  yyy 065
2
rr 3
2
2
1


r
r xx
ececy
3
2
2
1
•Example
Fin500J Topic 6 Fall 2010 Olin Business School 22

23. rx
ey rxrx
rVeVey  '' rx
Vey Letrxrxrx
VerVreVey
2
'2'''' and 
where V is a
function of x0''' cybyay 0)(''xV dcxV  rxrxrxrx
xececedcxbey
21)( 
Equal Roots to Characteristic Equation
•Let the roots of the characteristic equation equal and of value r
1
= r
2= r. Therefore, the solution of the characteristic equation is:0cbrar
2
 0bar2 
Fin500J Topic 6 Fall 2010 Olin Business School 23

24. )).sin()(cos(
,))sin()(cos(
)(
2
)(
1
xixeey
xixeey
xxi
xxi







 ).sin(cos
:is d.e. theosolution t geneal theTherefore, equation.
aldifferenti the tosolutions twoare and that see easy to isIt
)sin()(
2
1
)(),cos()(
2
1
)(
21
2121
xecx)(ecy(x)
vu
xeyy
i
xvxeyyxu
λxλx
xx




 Complex Roots to Characteristic Equation
Let the roots of the characteristic equation be complex in the
form r
1,2=λ±µi. Therefore, the solution of the characteristic
equation is:
Fin500J Topic 6 Fall 2010 Olin Business School 24

25. (I) Solve09'6''  yyy
characteristic equation096
2
rr 3
21 rr x
exccy
3
21 )(


Examples
(II) Solve05'4''  yyy
characteristic equation054
2
rr ir 2
2,1 )sincos(
21
2
xcxcey
x

Fin500J Topic 6 Fall 2010 Olin Business School 25

26. When g(x) is a polynomial of the form where
all the coefficients are constants. The form of a particular solution is)(''' xgcybyay  cky/ n
nxaxaxaa  ...
2
210 n
nxxxy   ...
2
210
Non-homogeneous Differential Equations (Method of
Undetermined Coefficients)
Wheng(x)isconstant,sayk,aparticularsolutionofequationis
Fin500J Topic 6 Fall 2010 Olin Business School 26

27. Example
Solve3
844'4'' xxyyy  32
sxrxqxpy  2
32' sxrxqy  sxry 62'' 3322
84)(4)32(4)62( xxsxrxqxpsxrxqsxr 
equating coefficients of equal powers of x84
0124
4486
0442




s
sr
qrs
pqr 32
26107 xxxy
p
 044
2
rr
characteristicequationx
c exccy
2
21 )( 322
21 26107)( xxxexcc
yyy
x
pc

 04'4''  yyy
complementary function2r
Fin500J Topic 6 Fall 2010 Olin Business School 27

28. Non-homogeneous Differential Equations
(Method of Undetermined Coefficients)rx
Aey
•Wheng(x)isoftheformTe
rx
,whereTandrareconstants.The
form of a particular solution iscbrar
T
A


2
•Wheng(x)isoftheformCsinnx+Dcosnx,whereCandDare
constants, the form of a particular solution isnxFnxEy cossin 2222
2
)(
)(
bnanc
nbDCanc
E


 2222
2
)(
)(
bnanc
nbDCanc
F



Fin500J Topic 6 Fall 2010 Olin Business School 28

29. Example
Solve18'6''3 yy Cxy Cy' 0''y 18)C(6)0(3  3C xy
p
3 063
2
rr
characteristic equationx
c eccy
2
21 x
pc
eccx
yyy
2
213 
 0'6''3 yy
complementary function2,0
21 rr
Fin500J Topic 6 Fall 2010 Olin Business School 29

30. Example
Solvex
eyyy
4
78'10''3

 x
Cxey
4
 x
Cexy
4
)41('

 x
Cexy
4
)816(''

 71024  CC 2
1
C x
p xey
4
2
1

 0)4)(23(8103
2
 rrrr
characteristic equationxx
c
BeAey
43/2 
 xxx
pc
BeAexe
yyy
43/24
2
1


 08'10''3  yyy
complementary function4,3/2
21  rr
Fin500J Topic 6 Fall 2010 Olin Business School 30

31. Example
Solvexyyy 2cos526'''  xDxCy 2sin2cos )2cos2sin(2' xDxCy  )2sin2cos(4'' xDxCy  0102
52210


DC
DC 1
5


D
C xxy
p
2sin2cos5  0)3)(2(6
2
 rrrr
characteristic equationxx
c BeAey
32 
 xxBeAe
yyy
xx
pc
2sin2cos5
32


 06''' yyy
complementary function3,2
21 rr
Fin500J Topic 6 Fall 2010 Olin Business School 31

32. Euler Equations
Def: Eulerequations
Assuming x>0 and all solutions are of the
form y(x) = x
r
Plug into the differential equation to get the
characteristic equation0'''
2
 cybxyyax .0)()1(  crbrar
Fin500J Topic 6 Fall 2010 Olin Business School 32

33. Solving Euler Equations: (Case I)
Fin500J Topic 6 Fall 2010 Olin Business School 33
•The characteristic equation has two different real
solutions r
1and r
2.
•In this case the functions y = x
r1and y = x
r2are both
solutions to the original equation. The general solution
is:21
21)(
rr
xcxcxy  .xcxcy(x)
.r,rr-)r(r-
yxyyx
3
2
2
5
1
21
2
3
2
5
015312
:isequation sticcharacteri the,015'3''2




Example:

34. Solving Euler Equations: (Case II)
Fin500J Topic 6 Fall 2010 Olin Business School 34
•The characteristic equation has two equal roots r
1=
r
2=r.
•In this case the functions y = x
r
and y = x
r
lnx are both
solutions to the original equation. The general solution
is:)ln()(
21 xccxxy
r
 x.xcxcy(x)
.rr)r(r-
yxyyx
ln
401671
:isequation sticcharacteri the,016'7''
4
2
4
1
2



Example:

35. Solving Euler Equations: (Case III)
Fin500J Topic 6 Fall 2010 Olin Business School 35
•The characteristic equation has two complex roots r
1,2=
λ±µi. x))(μcx)(μ(cxy(x)
xixxxex
λ
xii
lnsinlncos
:be illsolution w general theroots,complex of case in the So,
)lnsin()lncos(
21
ln)(




 .xxcxxcy(x)
i.rr)r(r-
yxyyx
)ln3sin()ln3cos(
310431
:isequation sticcharacteri the,04'3''
1
2
1
1
2,1
2




Example: