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
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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
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Derivatives
Derivatives
Ordinary Derivatives
u
y
dy
dx
y is a function of one
independent variable
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Partial Derivatives
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u is a function of
more than one
independent variable
3
Differential Equations
Differential
Equations
Ordinary Differential Equations
Partial Differential Equations
2
d y
 6 xy  1
2
dx
involve one or more
Ordinary derivatives of
unknown functions
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u u
 2 0
2
y
x
2
2
involve one or more
partial derivatives of
unknown functions
4
Ordinary Differential Equations
Ordinary Differential Equations (ODE) involve one or
more ordinary derivatives of unknown functions with
respect to one independent variable
Examples :
dy
 y  ex
dx
d2y
dy
 5  2 y  cos( x)
2
dx
dx
y(x): unknown function
x: independent variable
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Order of a differential equation
The order of an ordinary differential equations is the order
of the highest order derivative
Examples :
dy
 y  ex
dx
d2y
dy
 5  2 y  cos( x)
2
dx
dx
 d y

2
dx

2
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First order ODE
Second order ODE
3
 dy
   2 y 4  1
 dx
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Second order ODE
6
Solution of a differential equation
A solution to a differential equation is a function that
satisfies the equation.
Example :
dx(t )
 x(t )  0
dt
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Solution
x(t )  e t
Proof :
dx(t )
 e t
dt
dx(t )
 x(t )  e t  e t  0
dt
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Linear ODE
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
an ( x) y n ( x)  an 1 ( x) y n 1 ( x)    a1 ( x) y ' ( x)  a0 ( x) y ( x)  g ( x)
Examples :
dy
 y  ex
dx
d2y
dy
2

5

2
x
y  cos( x)
2
dx
dx
Linear ODE
Linear ODE
3
 d 2 y  dy
 2    y  1
 dx  dx
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Non-linear ODE
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Boundary-Value and Initial value Problems
Boundary-Value Problems
Initial-Value Problems

The auxiliary conditions are
at one point of the
independent variable
 The auxiliary conditions are not at
one point of the independent
variable
 More difficult to solve than initial
value problem
2 x
y ' '2 y ' y  e
y (0)  1, y ' (0)  2.5
same
Fin500J Topic 6
y ' '2 y ' y  e 2 x
y (0)  1, y (2)  1.5
different
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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
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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
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First Order Linear Differential Equations
 Def: A first order differential equation is
said to be linear if it can be written
y  p ( x) y  g ( x)
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First Order Linear Differential Equations
 How to solve first-order linear ODE ?
y  p( x) y  g ( x) (1)
Sol:
Multiplying both sides by  (x ) , called an integrating factor,
gives
dy
 ( x)   ( x) p( x) y   ( x) g ( x) (2)
dx
assuming
 ( x) p( x)   ' ( x), (3)
we get
dy
 ( x)   ' ( x) y   ( x) g ( x) (4)
dx
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First Order Linear Differential Equations
By product rule, (4) becomes
(  ( x) y ( x))'   ( x) g ( x) (5)
  ( x) y ( x)    ( x) g ( x)dx  c1
 ( x) g ( x)dx  c

 y ( x) 
1
 ( x)
(6)
 (x ) from (3)
 ' ( x)
 ( x) p ( x)   ' ( x) 
 p( x)
 ( x)
Now, we need to solve
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First Order Linear Differential Equations
 ' ( x)
 p( x)  (ln  ( x))'  p( x)
 ( x)
 ln  ( x)   p( x)dx  c2
  ( x)  e
 p ( x ) dx  c2
 c3e
 p ( x ) dx
(7 )
to get rid of one constant, we can use
 ( x)  e  p ( x ) dx (8)
The solution t o a linear first order differenti al equation is then
 ( x) g ( x)dx  c

y ( x) 
e
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p ( x ) dx
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(9)
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Summary of the Solution Process
 Put the differential equation in the form (1)
 Find the integrating factor,  (x ) using (8)
 Multiply both sides of (1) by  (x ) and write the left
side of (1) as (  ( x) y ( x))'
 Integrate both sides
 Solve for the solution y (x)
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Example 1
y  y  e
2x
Sol:
 e  p ( x ) dx g ( x)dx  c 
 

  ( 1) dx 
( 1) dx 2 x


e
e
e
dx

c
 

y ( x)  e 
 p ( x ) dx

 e x  e  x e 2 x dx  c

 ex ex  c
 ce  e
x
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

2x
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Example 2
Sol:
 y '
1
xy'2 y  x  x, y (1) 
2
2
2
y  x 1
x
 y ( x)  e 
 p ( x ) dx
 e  p ( x ) dx g ( x)dx  c 
 


2
2

 x dx   x dx
2
2
e
e
(
x

1
)
dx

c

x
x
( x  1)dx  c





1 3  1 2 1
2  1 4
 x  x  x  c   x  x  cx  2
3
3
4
 4
Apply the initial condition to get c,
1 1 1
7
   cc  .
2 4 3
12

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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
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Second Order Linear Differential Equations
The general equation can be expressed in the form
ay ' 'by 'cy  g ( x)
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
au' 'bu'cu  av' 'bv'cv  g ( x)
If v is a particular solution of the original differential equation
purpose
au' 'bu'cu  0
The general solution of the linear differential equation will be the
sum of a “complementary function” and a “particular solution”.
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The Complementary Function (solution of the
homogeneous equation)
ay ' 'by ' cy  0
Let the solution assumed to be:
dy
 re rx
dx
e rx (ar 2  br  c)  0
y  e rx
d2y
2 rx

r
e
2
dx
characteristic equation
Real, distinct roots
Double roots
Complex roots
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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:
y  e r1x
y  e r2 x
• The general solution will be
y  c1e r1x  c2e r2 x
• Example
y ' '5 y '6 y  0
r1  2
r2  3
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r 2  5r  6  0
y  c1e 2 x  c2e3 x
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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:
y  e rx
Let y  Ve
rx
 y'  e rxV 'rVerx
and y' '  e rxV ' '2re rxV 'r 2Verx
where V is a
function of x
ay ' 'by ' cy  0
ar 2  br  c  0
2ar  b  0
V ' ' ( x)  0
V  cx  d
y  be rx  (cx  d )e rx  c1e rx  c2 xerx
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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: y  e(    i ) x  ex (cos( x)  i sin( x)),
1
y2  e(    i ) x  ex (cos( x)  i sin( x)).
1
1
( y1  y2 )  e x cos( x), v( x)  ( y1  y2 )  e x sin( x)
2
2i
It is easy to see that u and v are two solutions to the differenti al
equation. Therefore, the geneal solution t o the d.e. is :
u ( x) 
y(x)  c1e λx cos (x)  c2 e λx sin( x).
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Examples
(I) Solve y ' '6 y '9 y  0
(II) Solve y ' '4 y '5 y  0
characteristic equation
characteristic equation
r 2  4r  5  0
r 2  6r  9  0
r1  r2  3
y  (c1  c2 x)e
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r1, 2  2  i
3 x
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y  e2 x (c1 cos x  c2 sin x)
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Non-homogeneous Differential Equations (Method of
Undetermined Coefficients)
ay ' 'by 'cy  g ( x)
When g(x) is constant, say k, a particular solution of equation is
y  k /c
When g(x) is a polynomial of the form a0  a1 x  a2 x  ...  an x where
all the coefficients are constants. The form of a particular solution is
2
n
y   0  1 x   2 x 2  ...   n x n
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Example
Solve
y' '4 y'4 y  4 x  8x 3
complementary function
y ' '4 y '4 y  0
y'  q  2rx  3sx 2
y  p  qx  rx 2  sx 3
y ' '  2r  6 sx
r 2  4r  4  0
characteristic equation
(2r  6sx)  4(q  2rx  3sx 2 )  4( p  qx  rx 2  sx 3 )  4 x  8x3
equating coefficients of equal powers of x
r2
2r  4q  4 p  0
yc  (c1  c2 x)e 2 x
6s  8r  4q  4
4r  12s  0
4s  8
y  yc  y p
y p  7  10 x  6 x 2  2 x3
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 (c1  c2 x)e 2 x  7  10 x  6 x 2  2 x 3
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Non-homogeneous Differential Equations
(Method of Undetermined Coefficients)
• When g(x) is of the form Te r x , where T and r are constants. The
form of a particular solution is
rx
y  Ae
A
T
ar 2  br  c
• Whe n g ( x ) i s o f th e fo r m C s i n n x + D cos n x , wh ere C a nd D a re
constants, the form of a particular solution is
y  E sin nx  F cos nx
(c  n 2 a)C  nbD
E
(c  n 2 a) 2  n 2b 2
(c  n 2 a)C  nbD
F
(c  n 2 a ) 2  n 2 b 2
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Example
Solve
complementary function
3 y ' '6 y '  18
3 y ' '6 y '  0
y  Cx
y'  C
3r 2  6r  0
y' '  0
characteristic equation
3(0)  6(C)  18
r1  0, r2  2
yc  c1  c2 e 2 x
C  3
y  yc  y p
y p  3 x
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 3x  c1  c2e 2 x
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Example
Solve
complementary function
3 y' '10 y'8 y  7e 4 x
3 y ' '10 y '8 y  0
y  Cxe4 x
y'  (1  4 x)Ce4 x
y' '  (16 x  8)Ce
 24C  10C  7
C
4 x
3r 2  10r  8  (3r  2)( r  4)  0
characteristic equation
r1  2 / 3, r2  4
yc  Ae2 x / 3  Be 4 x
1
2
y  yc  y p
1 4 x
  xe  Ae2 x / 3  Be 4 x
2
1 4 x
y p   xe
2
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Example
Solve
complementary function
y ' ' y '6 y  52 cos 2 x
y ' ' y '6 y  0
y  C cos 2 x  D sin 2 x
y '  2(C sin 2 x  D cos 2 x)
y ' '  4(C cos 2 x  D sin 2 x)
r 2  r  6  (r  2)( r  3)  0
characteristic equation
 10C  2D  52
 2C  10D  0
r1  2, r2  3
yc  Ae 2 x  Be 3 x
C  5
D 1
y  yc  y p
y p  5 cos 2 x  sin 2 x
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 Ae2 x  Be 3 x  5 cos 2 x  sin 2 x
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Euler Equations
 Def: Euler equations
ax y' 'bxy'cy  0
2
 Assuming x>0 and all solutions are of the
form y(x) = xr
 Plug into the differential equation to get the
characteristic equation
ar (r  1)  b(r )  c  0.
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Solving Euler Equations: (Case I)
• 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:
y( x)  c x r1  c x r2
1
2
Example:
2 x 2 y ' '3xy'15 y  0, the characteri stic equation is :
5
2r(r-1 )  3r-15  0  r1  , r2  3.
2
5
2
 y(x)  c1 x  c2 x 3.
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Solving Euler Equations: (Case II)
• 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:
y( x)  x r (c  c ln x)
1
2
Example:
x 2 y ' '7 xy'16 y  0, the characteri stic equation is :
r(r-1 )  7r  16  0  r  4.
 y(x)  c1 x 4  c2 x 4 ln x.
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Solving Euler Equations: (Case III)
• The characteristic equation has two complex roots r1,2 =
λ±µi.
x    i  e(    i ) ln x  x  cos(  ln x)  ix  sin(  ln x)
So, in the case of complex roots, the general solution w ill be :
y(x)  x λ(c1 cos (μ ln x)  c2 sin (μ ln x))
Example:
x 2 y ' '3xy'4 y  0, the characteri stic equation is :
r(r-1 )  3r  4  0  r1, 2  1  3i.
 y(x)  c1 x 1 cos( 3 ln x)  c2 x 1 sin( 3 ln x).
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