The document covers ordinary differential equations (ODE), detailing their definitions, classifications, and solution methods. It elaborates on concepts such as order, linearity, initial and boundary value problems, and different types of ODEs, including first-order and second-order equations. Additionally, it discusses methods for solving linear ODEs and provides examples to illustrate the processes involved.

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
Derivatives
dx
dy
Fin500J Topic 6 Fall 2010 Olin Business School
Partial Derivatives
u is a function of
more than one
independent variable
Ordinary Derivatives
y is a function of one
independent variable
y
u


3

4. Differential Equations
Differential
Equations
1
6
2
2

 xy
dx
y
d
Fin500J Topic 6 Fall 2010 Olin Business School
involve one or more
partial derivatives of
unknown functions
Ordinary Differential Equations
involve one or more
Ordinary derivatives of
unknown functions
0
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(
2
5
:
2
2
x
y
dx
dy
dx
y
d
e
y
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
1
2
)
cos(
2
5
:
4
3
2
2
2
2
















y
dx
dy
dx
y
d
x
y
dx
dy
dx
y
d
e
y
dx
dy
Examples
x
The order of 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 equation
0
)
(
)
(
)
(
:
Proof
)
(












t
t
t
t
e
e
t
x
dt
t
dx
e
dt
t
dx
e
t
x
Solution
Fin500J Topic 6 Fall 2010 Olin Business School
A solution to a differential equation is a function that
satisfies the equation.
0
)
(
)
(
:

 t
x
dt
t
dx
Example
7

8. Linear ODE
Fin500J Topic 6 Fall 2010 Olin Business School
1
)
cos(
2
5
:
3
2
2
2
2
2
















y
dx
dy
dx
y
d
x
y
x
dx
dy
dx
y
d
e
y
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
)
(
)
(
)
(
)
(
'
)
(
)
(
)
(
)
(
)
( 0
1
1
1 x
g
x
y
x
a
x
y
x
a
x
y
x
a
x
y
x
a 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 problem
5
.
1
)
2
(
,
1
)
0
(
'
2
'
' 2




 
y
y
e
y
y
y x
Fin500J Topic 6 Fall 2010 Olin Business School
Initial-Value Problems
 The auxiliary conditions are
at one point of the
independent variable
5
.
2
)
0
(
'
,
1
)
0
(
'
2
'
' 2




 
y
y
e
y
y
y x
same different
9

10. Classification of ODE
ODE can be classified in different ways
 Order
 First order ODE
 Second order ODE
 Nth order ODE
 Linearity
 Linear ODE
 Nonlinear ODE
 Auxiliary conditions
 Initial value problems
 Boundary value problems
Fin500J Topic 6 Fall 2010 Olin Business School 10

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 linear if it can be written
)
(
)
( x
g
y
x
p
y 


Fin500J Topic 6 Fall 2010 Olin Business School 12

13. First Order Linear Differential Equations
 How to solve first-order linear ODE ?
Sol:
)
1
(
)
(
)
( x
g
y
x
p
y 


)
2
(
)
(
)
(
)
(
)
(
)
( x
g
x
y
x
p
x
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
(
),
(
'
)
(
)
( x
x
p
x 
 
)
4
(
)
(
)
(
)
(
'
)
( x
g
x
y
x
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)
)
(
)
(
)
(
'
)
(
'
)
(
)
( x
p
x
x
x
x
p
x 






)
6
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
5
(
)
(
)
(
))'
(
)
(
(
1
1
x
c
dx
x
g
x
x
y
c
dx
x
g
x
x
y
x
x
g
x
x
y
x















)
(x


15. First Order Linear Differential Equations
Fin500J Topic 6 Fall 2010 Olin Business School 15
)
7
(
)
(
)
(
)
(
ln
)
(
))'
(
(ln
)
(
)
(
)
(
'
)
(
3
)
(
2
2 












dx
x
p
c
dx
x
p
e
c
e
x
c
dx
x
p
x
x
p
x
x
p
x
x





to get rid of one constant, we can use
)
9
(
)
(
)
(
)
(
then
is
equation
al
differenti
order
first
linear
a
o
solution t
The
)
8
(
)
(
)
(
)
(






dx
x
p
dx
x
p
e
c
dx
x
g
x
x
y
e
x



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

))'
(
)
(
( x
y
x

)
(x
y

17. Example 1
Sol:
x
e
y
y 2



 
 
x
x
x
x
x
x
x
x
dx
dx
dx
x
p
dx
x
p
e
ce
c
e
e
c
dx
e
e
e
c
dx
e
e
e
c
dx
x
g
e
e
x
y
2
2
2
)
1
(
)
1
(
)
(
)
(
)
(
)
(











 








 











Fin500J Topic 6 Fall 2010 Olin Business School 17

18. Example 2
Sol:
2
1
)
1
(
,
2
' 2



 y
x
x
y
xy
 
.
12
7
3
1
4
1
2
1
c,
get
to
condition
initial
Apply the
3
1
4
1
3
1
4
1
)
1
(
)
1
(
)
(
)
(
1
2
'
2
2
3
4
2
2
2
2
2
)
(
)
(





































 
















c
c
cx
x
x
c
x
x
x
c
dx
x
x
x
c
dx
x
e
e
c
dx
x
g
e
e
x
y
x
y
x
y
dx
x
dx
x
dx
x
p
dx
x
p
Fin500J Topic 6 Fall 2010 Olin Business School 18

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

20. )
(
'
'
' x
g
cy
by
ay 


where a, b and c are constant coefficients
Let the dependent variable y be replaced by the sum of the
two new variables: y = u + v
Therefore
    )
(
'
'
'
'
'
' x
g
cv
bv
av
cu
bu
au 





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
'
'
' 

 cu
bu
au
Fin500J Topic 6 Fall 2010 Olin Business School 20

21. 0
'
'
' 

 cy
by
ay
Let the solution assumed to be: rx
e
y 
rx
re
dx
dy
 rx
e
r
dx
y
d 2
2
2

0
)
( 2


 c
br
ar
erx
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. x
r
e
y 1
 x
r
e
y 2

x
r
x
r
e
c
e
c
y 2
1
2
1 

Real, Distinct Roots to Characteristic Equation
• Let the roots of the characteristic equation be real, distinct
and of values r1 and r2. Therefore, the solutions of the
characteristic equation are:
• The general solution will be
0
6
'
5
'
' 

 y
y
y 0
6
5
2


 r
r
3
2
2
1


r
r x
x
e
c
e
c
y 3
2
2
1 

• Example
Fin500J Topic 6 Fall 2010 Olin Business School 22

23. rx
e
y 
rx
rx
rVe
V
e
y 

 '
'
rx
Ve
y 
Let rx
rx
rx
Ve
r
V
re
V
e
y 2
'
2
'
'
'
'
and 


where V is a
function of x
0
'
'
' 

 cy
by
ay
0
)
(
'
' 
x
V d
cx
V 

rx
rx
rx
rx
xe
c
e
c
e
d
cx
be
y 2
1
)
( 




Equal Roots to Characteristic Equation
• Let the roots of the characteristic equation equal and of value r1
= r2 = r. Therefore, the solution of the characteristic equation is:
0
c
br
ar2


 0
b
ar
2 

Fin500J Topic 6 Fall 2010 Olin Business School 23

24. )).
sin(
)
(cos(
,
))
sin(
)
(cos(
)
(
2
)
(
1
x
i
x
e
e
y
x
i
x
e
e
y
x
x
i
x
x
i


















).
sin(
cos
:
is
d.e.
the
o
solution t
geneal
the
Therefore,
equation.
al
differenti
the
to
solutions
two
are
and
that
see
easy to
is
It
)
sin(
)
(
2
1
)
(
),
cos(
)
(
2
1
)
(
2
1
2
1
2
1
x
e
c
x)
(
e
c
y(x)
v
u
x
e
y
y
i
x
v
x
e
y
y
x
u
λx
λx
x
x



 









Complex Roots to Characteristic Equation
Let the roots of the characteristic equation be complex in the
form r1,2 =λ±µi. Therefore, the solution of the characteristic
equation is:
Fin500J Topic 6 Fall 2010 Olin Business School 24

25. (I) Solve 0
9
'
6
'
' 

 y
y
y
characteristic equation
0
9
6
2


 r
r
3
2
1 

 r
r
x
e
x
c
c
y 3
2
1 )
( 


Examples
(II) Solve 0
5
'
4
'
' 

 y
y
y
characteristic equation
0
5
4
2


 r
r
i
r 
 2
2
,
1
)
sin
cos
( 2
1
2
x
c
x
c
e
y 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
)
(
'
'
' x
g
cy
by
ay 


c
k
y /

n
n x
a
x
a
x
a
a 


 ...
2
2
1
0
n
n x
x
x
y 


 



 ...
2
2
1
0
Non-homogeneous Differential Equations (Method of
Undetermined Coefficients)
When g(x) is constant, say k, a particular solution of equation is
Fin500J Topic 6 Fall 2010 Olin Business School 26

27. Example
Solve 3
8
4
4
'
4
'
' x
x
y
y
y 



3
2
sx
rx
qx
p
y 



2
3
2
' sx
rx
q
y 


sx
r
y 6
2
'
' 

3
3
2
2
8
4
)
(
4
)
3
2
(
4
)
6
2
( x
x
sx
rx
qx
p
sx
rx
q
sx
r 









equating coefficients of equal powers of x
8
4
0
12
4
4
4
8
6
0
4
4
2









s
s
r
q
r
s
p
q
r
3
2
2
6
10
7 x
x
x
yp 



0
4
4
2


 r
r
characteristic equation
x
c e
x
c
c
y 2
2
1 )
( 

3
2
2
2
1 2
6
10
7
)
( x
x
x
e
x
c
c
y
y
y
x
p
c








0
4
'
4
'
' 

 y
y
y
complementary function
2

r
Fin500J Topic 6 Fall 2010 Olin Business School 27

28. Non-homogeneous Differential Equations
(Method of Undetermined Coefficients)
rx
Ae
y 
• When g(x) is of the form Terx, where T and r are constants. The
form of a particular solution is
c
br
ar
T
A


 2
• When g(x) is of the form Csinnx + Dcosnx, where C and D are
constants, the form of a particular solution is
nx
F
nx
E
y cos
sin 

2
2
2
2
2
)
(
)
(
b
n
a
n
c
nbD
C
a
n
c
E





2
2
2
2
2
)
(
)
(
b
n
a
n
c
nbD
C
a
n
c
F





Fin500J Topic 6 Fall 2010 Olin Business School 28

29. Example
Solve
18
'
6
'
'
3 
 y
y
Cx
y 
C
y 
'
0
'
' 
y
18
)
C
(
6
)
0
(
3 

3
C 

x
yp 3


0
6
3 2

 r
r
characteristic equation
x
c e
c
c
y 2
2
1 

x
p
c
e
c
c
x
y
y
y
2
2
1
3 





0
'
6
'
'
3 
 y
y
complementary function
2
,
0 2
1 
 r
r
Fin500J Topic 6 Fall 2010 Olin Business School 29

30. Example
Solve x
e
y
y
y 4
7
8
'
10
'
'
3 



x
Cxe
y 4


x
Ce
x
y 4
)
4
1
(
' 


x
Ce
x
y 4
)
8
16
(
'
' 


7
10
24 

 C
C
2
1


C
x
p xe
y 4
2
1 


0
)
4
)(
2
3
(
8
10
3 2





 r
r
r
r
characteristic equation
x
x
c Be
Ae
y 4
3
/
2 


x
x
x
p
c
Be
Ae
xe
y
y
y
4
3
/
2
4
2
1 







0
8
'
10
'
'
3 

 y
y
y
complementary function
4
,
3
/
2 2
1 

 r
r
Fin500J Topic 6 Fall 2010 Olin Business School 30

31. Example
Solve
x
y
y
y 2
cos
52
6
'
'
' 


x
D
x
C
y 2
sin
2
cos 

)
2
cos
2
sin
(
2
' x
D
x
C
y 


)
2
sin
2
cos
(
4
'
' x
D
x
C
y 


0
10
2
52
2
10






D
C
D
C
1
5



D
C
x
x
yp 2
sin
2
cos
5 


0
)
3
)(
2
(
6
2





 r
r
r
r
characteristic equation
x
x
c Be
Ae
y 3
2 


x
x
Be
Ae
y
y
y
x
x
p
c
2
sin
2
cos
5
3
2







0
6
'
'
' 

 y
y
y
complementary function
3
,
2 2
1 

 r
r
Fin500J Topic 6 Fall 2010 Olin Business School 31

32. Euler Equations
 Def: Euler equations
 Assuming x>0 and all solutions are of the
form y(x) = xr
 Plug into the differential equation to get the
characteristic equation
0
'
'
'
2


 cy
bxy
y
ax
.
0
)
(
)
1
( 


 c
r
b
r
ar
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 r1 and r2.
• In this case the functions y = xr1 and y = xr2 are both
solutions to the original equation. The general solution
is: 2
1
2
1
)
( r
r
x
c
x
c
x
y 

.
x
c
x
c
y(x)
.
r
,
r
r-
)
r(r-
y
xy
y
x
3
2
2
5
1
2
1
2
3
2
5
0
15
3
1
2
:
is
equation
stic
characteri
the
,
0
15
'
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 r1 =
r2=r.
• In this case the functions y = xr and y = xr lnx are both
solutions to the original equation. The general solution
is: )
ln
(
)
( 2
1 x
c
c
x
x
y r


x.
x
c
x
c
y(x)
.
r
r
)
r(r-
y
xy
y
x
ln
4
0
16
7
1
:
is
equation
stic
characteri
the
,
0
16
'
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 r1,2 =
λ±µi.
x))
(μ
c
x)
(μ
(c
x
y(x)
x
ix
x
x
e
x
λ
x
i
i
ln
sin
ln
cos
:
be
ill
solution w
general
the
roots,
complex
of
case
in the
So,
)
ln
sin(
)
ln
cos(
2
1
ln
)
(




 


 





.
x
x
c
x
x
c
y(x)
i.
r
r
)
r(r-
y
xy
y
x
)
ln
3
sin(
)
ln
3
cos(
3
1
0
4
3
1
:
is
equation
stic
characteri
the
,
0
4
'
3
'
'
1
2
1
1
2
,
1
2















Example: