Differential Equation

1. Introduction to Ordinary
Differential Equations
Asst Lect Sarmad K. Ibrahim

2. Slide number 2
Ordinary Differential Equations
 Where do ODEs arise?
 Notation and Definitions
 Solution methods for 1st order
ODEs

3. Slide number 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. Slide number 4
Example – Newton’s Law of
Cooling
 This is a model of how the
temperature of an object changes as
it loses heat to the surrounding
atmosphere:
Temperature of the object: Obj
T Room Temperature: Room
T
Newton’s laws states: “The rate of change in the temperature of an
object is proportional to the difference in temperature between the object
and the room temperature”
)
( Room
Obj
Obj
T
T
dt
dT


 
Form
ODE
t
Room
init
Room
Obj e
T
T
T
T 



 )
(
Solve
ODE
Where is the initial temperature of the object.
init
T

5. Slide number 5
Notation and Definitions
 Order
 Linearity
 Homogeneity
 Initial Value/Boundary value
problems

6. Slide number 6
Order
 The order of a differential
equation is just the order of
highest derivative used.
0
2
2


dt
dy
dt
y
d
.
2nd order
3
3
dt
x
d
x
dt
dx
 3rd order

7. Slide number 7
Linearity
 The important issue is how the
unknown y appears in the equation.
A linear equation involves the
dependent variable (y) and its
derivatives by themselves. There
must be no "unusual" nonlinear
functions of y or its derivatives.
 A linear equation must have constant
coefficients, or coefficients which
depend on the independent variable
(t). If y or its derivatives appear in the
coefficient the equation is non-linear.

8. Slide number 8
Linearity - Examples
0

 y
dt
dy
is linear
0
2

 x
dt
dx
is non-linear
0
2

 t
dt
dy
is linear
0
2

 t
dt
dy
y is non-linear

9. Slide number 9
Linearity – Summary
y
2
dt
dy
dt
dy
y
dt
dy
t
2






dt
dy
y
t)
sin
3
2
(  y
y )
3
2
( 2

Linear Non-linear
2
y )
sin( y
or

10. Slide number 10
Linearity – Special Property
If a linear homogeneous ODE has solutions:
)
(t
f
y  )
(t
g
y 
and
then:
)
(
)
( t
g
b
t
f
a
y 



where a and b are constants,
is also a solution.

11. Slide number 11
Linearity – Special Property
Example:
0
2
2

 y
dt
y
d
0
sin
sin
sin
)
(sin
2
2




 t
t
t
dt
t
d
0
cos
cos
cos
)
(cos
2
2




 t
t
t
dt
t
d
0
cos
sin
cos
sin
cos
sin
)
cos
(sin
2
2









t
t
t
t
t
t
dt
t
t
d
t
y sin
 t
y cos

has solutions and
Check
t
t
y cos
sin 

Therefore is also a solution:
Check

12. Slide number 12
Homogeniety
 Put all the terms of the equation
which involve the dependent variable
on the LHS.
 Homogeneous: If there is nothing
left on the RHS the equation is
homogeneous (unforced or free)
 Nonhomogeneous: If there are
terms involving t (or constants) - but
not y - left on the RHS the equation
is nonhomogeneous (forced)

13. Slide number 13
Example
 1st order
 Linear
 Nonhomogeneous
 Initial value problem
g
dt
dv

0
)
0
( v
v 
w
dx
M
d

2
2
0
)
(
0
)
0
(
and


l
M
M
 2nd order
 Linear
 Nonhomogeneous
 Boundary value
problem

14. Slide number 14
Example
 2nd order
 Nonlinear
 Homogeneous
 Initial value problem
 2nd order
 Linear
 Homogeneous
 Initial value problem
0
sin
2
2
2

 


dt
d
0
)
0
(
,
0 0 

dt
d
θ
)
θ(

0
2
2
2

 


dt
d
0
)
0
(
,
0 0 

dt
d
θ
)
θ(


15. Slide number 15
Solution Methods - Direct
Integration
 This method works for equations
where the RHS does not depend on
the unknown:
 The general form is:
)
(t
f
dt
dy

)
(
2
2
t
f
dt
y
d


)
(t
f
dt
y
d
n
n


16. Slide number 16
Direct Integration
 y is called the unknown or
dependent variable;
 t is called the independent variable;
 “solving” means finding a formula for
y as a function of t;
 Mostly we use t for time as the
independent variable but in some
cases we use x for distance.

17. Slide number 17
Direct Integration – Example
Find the velocity of a car that is
accelerating from rest at 3 ms-2:
3

 a
dt
dv
c
t
v 

 3
t
v
c
c
v
3
0
0
3
0
0
)
0
(









If the car was initially at rest we
have the condition:

18. Slide number 18
Solution Methods - Separation
The separation method applies only to
1st order ODEs. It can be used if the
RHS can be factored into a function of t
multiplied by a function of y:
)
(
)
( y
h
t
g
dt
dy


19. Slide number 19
Separation – General Idea
First Separate:
dt
t
g
y
h
dy
)
(
)
(

Then integrate LHS with
respect to y, RHS with respect
to t.
C
dt
t
g
y
h
dy

 
 )
(
)
(

20. Slide number 20
Separation - Example
)
sin( t
y
dt
dy

Separate:
dt
t
dy
y
)
sin(
1

Now integrate:
)
cos(
)
cos(
)
cos(
)
ln(
)
sin(
1
t
c
t
Ae
y
e
y
c
t
y
dt
t
dy
y











 
