from Dosen

1. Ordinary Differential
Equations

2. CHAPTER 1
Introduction to Differential Equations
1.1 Definitions and Terminology
1.2 Initial-Value Problems
1.3 Differential Equation as
Mathematical Models

3. Where do ODE’s arise
All branches of Engineering-
Economics-
Biology and Medicine-
Chemistry, Physics etc-
Anytime you wish to find out how
something changes with time (and
(sometimes space

4. Definitions and 1.1
Terminology
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

5. Definitions and 1.1
Terminology
Recall Calculus
Definition of a Derivative
If , the derivative of or
With respect to is defined as
y = f (x) y f (x)
x
f x h f x
= + -
lim ( ) ( )
h
dy
dx
h
0
®
y' , df f ' (x)
The derivative is also denoted by or
dx

6. Definitions and 1.1
Terminology
Recall the Exponential function
y = f (x) = e2x
dependent variable: y
independent variable: x
dy x x
e y
e d x
dx
d e
= = é
dx
dx
( 2
x
) 2 (2 ) ù
2 2 2
= = úû
êë

7. Definitions and 1.1
Terminology
Differential Equation:
Equations that involve dependent variables and
their derivatives with respect to the independent
variables.
Differential Equations are classified
by
type, order and linearity .

8. Definitions and 1.1
Terminology
Differential Equations are classified by
type, order and linearity .
TYPE
There are two main types of differential
equation: . ” “ordinary” and “partial

9. Definitions and 1.1
Terminology
Ordinary differential equation (ODE(
Differential equations that involve only
ONE independent variable are called
ordinary differential equations.
Examples:
dy 2
+ 5 = - + 6 y =
0 2
dx + dy
= 2 +
d y dy
x y
, , and
y ex
dx
dx
dx
dt
dt
only ordinary (or total ( derivatives

10. Definitions and 1.1
Terminology
Partial differential equation (PDE(
Differential equations that involve
two or more independent variables are called
partial differential equations.
Examples:
u
¶
and
u
t
2
2
¶ 2 2
t
u
x
u
- ¶
¶
= ¶
¶
¶
2
= - ¶
only partial derivatives
v
x
y
¶
¶

11. Definitions and 1.1
Terminology
ORDER
The order of a differential equation is the
order of the highest derivative found in
.the DE
3
2
d y - = ÷ø
y ex
dy
+ 5æ 4
ö dx
çè
dx
2
second order first order

12. Definitions and 1.1
Terminology
xy' - y2 = ex
:Written in differential form M(x, y)dx + N(x, y)dy = 0
y'' = x3
first order F(x, y, y' ) = 0
second order F(x, y, y' , y'' ) = 0

13. Definitions and 1.1
Terminology
LINEAR or NONLINEAR
An n-th order differential equation is said to be
linear if the function
is linear in the variables

F(x, y, y' ,......y(n) ) = 0
y, y' ,... y(n-1)
-
1
a x dy
n
n
a x d y n
a x d y
1 a x y g x
n + + + + = -
( ) ( ) 1 ... 1( ) 0 ( ) ( )
dx
dx
dx
n n
-
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

14. Definitions and 1.1
Terminology
LINEAR or NONLINEAR
x dy
4 + ( y - x) = 0
dx
( y - x)dx + 4xdy = 0
or
linear first-order ordinary differential equation
y'' - 2y' + y = 0
linear second-order ordinary differential
equation
3
d y + 3 - 5 = 3
y ex
x dy
dx
dx
 linear third-order ordinary differential equation

15. Definitions and 1.1
Terminology
LINEAR or NONLINEAR
(1- y) y' + 2y = ex
coefficient depends on y
nonlinear first-order ordinary differential equation
2
+ y =
dx
d y
sin( ) 0 2
nonlinear function of y
nonlinear second-order ordinary differential equation
4
+ y =
dx
2 0
d y
power not 1
4
 nonlinear fourth-order ordinary differential equation

16. Definitions and 1.1
Terminology
LINEAR or NONLINEAR
:NOTE
...
3! 5! 7!
sin( )
3 5 7
y = y - y + y - y + -¥ < x < ¥
...
2! 4! 6!
cos( ) 1
2 4 6
y = - y + y - y + -¥ < x < ¥

17. Definitions and 1.1
Terminology
Solutions of ODEs
DEFINITION: solution of an ODE
Any function f
, 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

18. Definitions and 1.1
Terminology
Namely, a solution of an n-th order ODE is a
function which possesses at least n
derivatives and for which
for F(all x, x f in (x), I
f ' (x), f (n) (x)) = 0
We say that satisfies the differential equation
.on I

19. Definitions and 1.1
Terminology
Verification of a solution by substitution
:Example
y'' - 2y' + y = 0 ; y = xex

y' = xex + ex , y'' = xex + 2ex
:left hand side
y'' - 2y' + y = (xex + 2ex ) - 2(xex + ex ) + xex = 0
right-hand side: 0
The DE possesses the constant y=0 trivial solution

20. Definitions and 1.1
Terminology
DEFINITION: solution curve
A graph of the solution of an ODE is called a
solution curve, or an integral curve of
.the equation

21. Definitions and Terminology 1.1
DEFINITION: families of solutions
A solution containing an arbitrary constant
(parameter) represents a set G(x, y,c) = 0
of
solutions to an ODE called a one-parameter
. family of solutions
A solution to an n−th order ODE is a n-parameter
family of
F(x, y, y' ,......y(n) ) = 0
. solutions
Since the parameter can be assigned an infinite
number of values, an ODE can have an infinite
.number of solutions

22. Definitions and 1.1
Terminology
Verification of a solution by substitution
:Example
y' + y = 2
φ(x) = 2 + ke- x
y' + y = 2
φ(x) = 2 + ke-x
φ' (x) = -ke- x
φ' (x) + φ(x) = -ke- x + 2 + ke- x = 2


23. ©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.
y = 2 + ke-x
Figure 1.1 Integral curves of y’ + y = 2 for k = 0, 3, –3, 6, and –6.

24. Definitions and 1.1
Terminology
Verification of a solution by substitution
:Example

for all , 
y y
' = +1
x
φ(x) = x ln(x) + Cx x > 0
φ' (x) = ln(x) +1+C
φ' ( ) = ln( ) + +1 = φ( x
) +1
x
x x x Cx
x

25. ©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.
1 (xe e c)
x
y = x - x +
Figure 1.2 Integral curves of y’ + ¹ y =
ex
for c =0,5,20, -6, and –10. x

26. Second-Order Differential Equation
: Example
is a solution of
:By substitution
φ(x) = 6cos(4x) -17sin(4x)
y'' +16x = 0
x x
= - -
φ '
24sin(4 ) 68cos(4 )
= - +
φ ''
96cos(4 ) 272sin(4 )
φ ''
16φ 0
+ =
x x
F(x, y, y' , y'' ) = 0
F( x,φ(x),φ' (x),φ(x)'' ) = 0

27. Second-Order Differential Equation
Consider the simple, linear second-order equation
y'' -12x = 0
y'' =12x ,
y ' = ò y '' ( x ) dx = ò 12 xdx = 6 x 2
+C y = ò y ' ( x ) dx = ò (6 x 2 +C ) dx = 2 x 3
+Cx + K 
To determine C and K, we need two initial conditions, one
specify a point lying on the solution curve and the
. ,other its slope at that point, e.g
???WHY
y(0) = K y' (0) = C

28. Second-Order Differential Equation
y'' =12x
y = 2x3 +Cx + K
IF only try x=x1, and x=x2

3
y x = x + Cx +
K
3
1
( ) 2
1 1
( ) 2
y x = x + Cx +
K
2
2 2
,It cannot determine C and K

29. ©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.
e.g. X=0, y=k
Figure 2.1 Graphs of y = 2x³ + C x +K for various values of C and K.

30. To satisfy the I.C. y(0)=3
The solution curve must
pass through (0,3)
Many solution ) curves through (0,3
©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.
Figure 2.2 Graphs of y = 2x³ + C x + 3 for various values of C.

31. To satisfy the I.C. y(0)=3,
y’(0)=-1, the solution curve
must pass through (0,3)
having slope -1
©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.
Figure 2.3 Graph of y = 2x³ - x + 3.

32. Definitions and 1.1
Terminology
Solutions
General Solution: Solutions obtained from integrating
the differential equations are called general solutions. The
general solution of a nth order ordinary differential
equation contains n arbitrary constants resulting from
.integrating times
Particular Solution: Particular solutions are the
solutions obtained by assigning specific values to the
.arbitrary constants in the general solutions

33. Definitions and Terminology 1.1
DEFINITION: implicit solution
A relation G(x, y) = 0
is said to be an
implicit solution of an ODE on an interval I
provided there exists at least one function
that satisfies the relation as well as the
. differential equation on I
a relation or expression that
.defines a solution implicitly
In contrast to an explicit solution
y =f (x)
f
f
G(x, y) = 0

34. Definitions and 1.1
Terminology
DEFINITION: implicit solution
Verify by implicit differentiation that the given
equation implicitly defines a solution of the
differential equation
y2 + xy - 2x2 - 3x - 2y = C
y - 4x - 3 + (x + 2y - 2) y' = 0

35. Definitions and 1.1
Terminology
DEFINITION: implicit solution
Verify by implicit differentiation that the given
equation implicitly defines a solution of the
differential equation
y2 + xy - 2x2 - 3x - 2y = C
y - 4x - 3 + (x + 2y - 2) y' = 0
d ( y + xy - 2 x - 3 x - 2 y ) / dx =
d ( C ) /
dx
yy y xy x y
==> + + - - - =
2 ' ' 4 3 2 '
0
y x xy yy y
==> - - + + - =
4 3 ' 2 ' 2 '
0
y x x y y
4 3 ( 2 2) '
0
2 2
==> - - + + - =

36. Definitions and 1.1
Terminology
Conditions
Initial Condition: Constrains that are specified at the
initial point, generally time point, are called initial
conditions. Problems with specified initial conditions
.are called initial value problems
Boundary Condition: Constrains that are specified
at the boundary points, generally space points, are
called boundary conditions. Problems with specified
boundary conditions are called boundary value
.problems

37. Initial-Value Problem 1.2
First- and Second-Order IVPS
:Solve
:Subject to
:Solve
:Subject to
f (x, y)
dy =
dx
y(x0 ) = y0
( , , ' )
d y =
2
2
f x y y
dx
0 1
'
y(x0 ) = y0 , y (x ) = y

38. Initial-Value Problem 1.2
DEFINITION: initial value problem
An initial value problem or IVP is a
problem which consists of an n-th order
ordinary differential equation along with n
initial conditions defined at a point x0
found
in the interval of definition
I
differential equation
d n
y
= f ( x , y , y ' ,..., y
(n-1) )
n
initial conditions
dx
y x y y x y y n x y
= = - = n
.where are known constants
0 1
( 1)
0 1
'
( 0 ) 0 , ( ) ,..., ( ) -
y0 , y1,..., yn-1

39. Initial-Value Problem 1.2
THEOREM: Existence of a Unique
Solution
Let R be a rectangular a £ x £ b, c £ y £ region d
in the xy-plane
defined by (x0 , y0 ) that contains
f (x, y)
the point ¶f / ¶y
in its interior. If and
are continuous on R, Then there
exists some interval
I0 : x0 - h < x < x0 + h, h > 0
a £ x £ b
contained in y(x) and
I0
a unique function defined on
.that is a solution of the initial value problem