ODE - Phil Dybvig
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Transcript ODE - Phil Dybvig
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
Fin500J Topic 6
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1
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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2
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
0
2
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
Exam ples:
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
Exam ples:
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.
Exam ple:
dx(t )
x(t ) 0
dt
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Solution x(t ) e t
P roof:
dx(t )
t
e
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) an1 ( x) y n1 ( x) a1 ( x) y' ( x) a0 ( x) y( x) g ( x)
Exam ples:
dy
y ex
dx
d2y
dy
2
5
2
x
y cos(x)
2
dx
dx
Linear ODE
3
d y dy
2 y 1
dx dx
2
Linear ODE
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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
y' '2 y' y e
The auxiliary conditions are not at
one point of the independent
variable
More difficult to solve than initial
value problem
y' '2 y' y e 2 x
2 x
y(0) 1, y' (0) 2.5
same
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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)
T hesolution ot a linear first order differential equationis 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:
y ( x) e
p ( x ) dx
e
( 1) dx
e p ( x ) dx g ( x)dx c
e ( 1) dx e 2 x dx c
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:
1
xy '2 y x x, y (1)
2
2
y '
2
y x 1
x
y ( x) e
p ( x ) dx
e p ( x ) dx g ( x)dx c
2
x dx
2
2
e
e
(
x
1
)
dx
c
x
x
( x 1)dx c
1
1
1
1
x 2 x 4 x 3 c x 2 x cx 2
3
3
4
4
Apply theinitialconditionto get c,
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2
dx
x
1 1 1
7
cc .
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
erx (ar2 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 c1er1x c2er2 x
• Example
y' '5 y'6 y 0
r1 2
r2 3
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r 2 5r 6 0
y c1e2 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' erxV 'rVe rx
and y' ' erxV ' '2rerxV 'r 2Ve rx
where V is a
function of x
ay' 'by'cy 0
ar2 br c 0
2ar b 0
V ' ' ( x) 0
V cx d
y berx (cx d )erx c1erx 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 ex (cos(x) i sin(x)),
1
y2 e( i ) x ex (cos(x) i sin(x)).
1
1
u ( x) ( y1 y2 ) ex cos(x), v( x) ( y1 y2 ) ex sin(x)
2
2i
It is easy tosee thatu and v are two solutionsto thedifferential
equation.T herefore,thegenealsolution ot thed.e. is :
y(x) c1e λx cos(x) c2e 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
2
n
When g(x) is a polynomial of the form a0 a1x a2 x ... an x where
all the coefficients are constants. The form of a particular solution is
y 0 1 x 2 x2 ... n xn
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Example
Solve
y' '4 y'4 y 4x 8x3
complementary function
y' '4 y'4 y 0
y p qx rx 2 sx 3
y' q 2rx 3sx 2
y' ' 2r 6sx
r 2 4r 4 0
characteristic equation
(2r 6sx) 4(q 2rx 3sx 2 ) 4( p qx rx 2 sx3 ) 4x 8x3
equating coefficients of equal powers of x
r2
2r 4 q 4 p 0
6s 8r 4q 4
4r 12s 0
4s 8
yc (c1 c2 x)e2 x
y yc y p
y p 7 10x 6x2 2x3
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(c1 c2 x)e 2 x 7 10x 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 cosnx
(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 c2e2 x
C 3
y yc y p
y p 3x
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3x c1 c2e 2 x
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Example
Solve
complementary function
3 y' '10y'8 y 7e4 x
3 y' '10y'8 y 0
y Cxe4 x
y' (1 4x)Ce 4 x
y' ' (16x 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 Be4 x
1
2
y yc y p
1 4 x
xe Ae2 x / 3 Be4 x
2
1 4 x
y p xe
2
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Example
Solve
complementary function
y' ' y'6 y 52cos2 x
y' ' y'6 y 0
y C cos2 x D sin 2 x
y' 2(C sin 2 x D cos2 x)
y' ' 4(C cos2 x D sin 2 x)
r 2 r 6 (r 2)(r 3) 0
characteristic equation
10C 2 D 52
2C 10D 0
r1 2, r2 3
yc Ae2 x Be3x
C 5
D 1
y yc y p
y p 5 cos2x sin 2x
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Ae2 x Be3 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 xr1 c xr2
1
2
Example:
2 x 2 y ' '3xy'15y 0, thecharacteristic equationis :
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, thecharacteristic equation is :
r(r-1 ) 7 r 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 thecase of complexroots,thegeneralsolution will be :
y(x) x λ(c1 cos(μ ln x) c2 sin (μ ln x))
Example:
x 2 y' '3xy'4 y 0, thecharacteristic equationis :
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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